problem_id large_stringlengths 3 7 | problem_name large_stringlengths 2 63 | rating float64 0 3.5k | tags large_stringlengths 0 139 | time_limit float64 0.5 15 ⌀ | memory_limit float64 4 1.02k ⌀ | description large_stringlengths 35 9.88k | input_format large_stringlengths 28 4.25k ⌀ | output_format large_stringlengths 14 2.22k ⌀ | interaction_format large_stringclasses 171
values | note large_stringlengths 18 5.36k ⌀ | examples listlengths 0 8 | editorial_cleaned large_stringlengths 17 34k | s1_hint large_stringlengths 116 500 ⌀ | wrong_path_1 dict | wrong_path_2 dict | wrong_path_3 dict | wrong_path_4 dict | small_thinking_trace large_stringlengths 349 100k ⌀ | small_pseudo large_stringlengths 45 13.1k ⌀ | s3_narrative large_stringlengths 1.22k 54.2k ⌀ | s3_pseudocode large_stringlengths 48 21.3k ⌀ |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
1007/E | Mini Metro | 3,400 | dp | 2 | 256 | In a simplified version of a "Mini Metro" game, there is only one subway line, and all the trains go in the same direction. There are $$$n$$$ stations on the line, $$$a_i$$$ people are waiting for the train at the $$$i$$$-th station at the beginning of the game. The game starts at the beginning of the $$$0$$$-th hour. ... | The first line contains three integers $$$n$$$, $$$t$$$, and $$$k$$$ ($$$1 \leq n, t \leq 200, 1 \leq k \leq 10^9$$$) — the number of stations on the line, hours we want to survive, and capacity of each train respectively.
Each of the next $$$n$$$ lines contains three integers $$$a_i$$$, $$$b_i$$$, and $$$c_i$$$ ($$$0... | Output a single integer number — the answer to the problem. | null | Let's look at the sample. There are three stations, on the first, there are initially 2 people, 3 people on the second, and 4 people on the third. Maximal capacities of the stations are 10, 9, and 8 respectively.
One of the winning strategies is to appoint two trains to the first and the third hours. Then on the first... | [
{
"input": "3 3 10\n2 4 10\n3 3 9\n4 2 8",
"output": "2"
},
{
"input": "4 10 5\n1 1 1\n1 0 1\n0 5 8\n2 7 100",
"output": "12"
}
] | Let's enumerate the hours and the stations starting from zero. Let's add a station to the end with an infinite number of people and infinite capacity. It is obvious that it will not affect the answer. Also, every train now will be filled completely.
Let's calculate $sa[p]$, $sb[p]$ and $sc[p]$: the sum of $a[i]$, $b[i]... | The key observation is that you can think about the process station-by-station, focusing on when the last station is serviced. This leads to a DP that balances the need to clear earlier stations against preventing overflow at the current station. The technique is DP with a clever state that captures whether you are in ... | {
"exploration": "I'll build a flow network where each hour and each station becomes a node, and train capacity flows through them. The idea: for each hour h (0 to t-1) and each station i (1 to n), I have a node (h,i) representing the state of station i at hour h. Trains dispatched at hour h flow from station 1 to n,... | {
"exploration": "I'll binary search on M, the total number of trains. For a fixed M, I need to check if there exists an assignment of x_h (trains at hour h) with sum x_h ≤ M such that no station ever exceeds c_i. The key constraint is the ordering rule: a train can't take from station i if station i-1 still has peop... | null | null | null | null | Okay, all of those went nowhere. The flow network couldn't encode the precedence rule without non-linear constraints, and the binary search just pushed the hard feasibility check back on itself.
hmm — but that half-formed thought about station-by-station processing and "whether you are in the middle of an hour"... wai... | READ n, t, k
a = array of size n+2
b = array of size n+2
c = array of size n+2
for i from 1 to n:
READ a[i], b[i], c[i]
// Add virtual station n+1
a[n+1] = 0
b[n+1] = 0
c[n+1] = INFINITY
sa = array of size n+2, initialized to 0
sb = array of size n+2, initialized to 0
sc = array of size n+2, initialized to 0
for p... |
1305/H | Kuroni the Private Tutor | 3,500 | binary search, greedy | 3 | 256 | As a professional private tutor, Kuroni has to gather statistics of an exam. Kuroni has appointed you to complete this important task. You must not disappoint him.
The exam consists of $$$n$$$ questions, and $$$m$$$ students have taken the exam. Each question was worth $$$1$$$ point. Question $$$i$$$ was solved by at ... | The first line of input contains two integers ($$$1 \le n, m \le 10^{5}$$$), denoting the number of questions of the exam and the number of students respectively.
The next $$$n$$$ lines contain two integers each, with the $$$i$$$-th line containing $$$l_{i}$$$ and $$$r_{i}$$$ ($$$0 \le l_{i} \le r_{i} \le m$$$).
The ... | Output two integers: the maximum number of students who could have gotten as many points as the student with rank $$$1$$$, and the maximum possible score for rank $$$1$$$ achieving this maximum number of students. If there is no valid arrangement that fits the given data, output $$$-1$$$ $$$-1$$$. | null | For the first sample, here is one possible arrangement that fits the data:
Students $$$1$$$ and $$$2$$$ both solved problems $$$1$$$ and $$$2$$$.
Student $$$3$$$ solved problems $$$2$$$ and $$$3$$$.
Student $$$4$$$ solved problem $$$4$$$.
The total score of all students is $$$T = 7$$$. Note that the scores of the s... | [
{
"input": "5 4\n2 4\n2 3\n1 1\n0 1\n0 0\n1\n4 1\n7",
"output": "3 2"
},
{
"input": "5 6\n0 6\n0 6\n2 5\n6 6\n4 6\n1\n3 3\n30",
"output": "-1 -1"
}
] | Suppose we fix the scores of the students $b_0 \le b_1 \le … \le b_{m-1}$. We model our problem as a flow graph. Consider a flow graph where the left side contains the source and $n$ nodes denoting the problems, and the right side contains the sink and $m$ nodes denoting the students. Each problem node is connected to ... | What if you think about distributing the total score t as a flow from questions to students, with each question having lower and upper bounds on how many students solve it? The key observation is that the feasibility of a given top score and tie size can be checked by examining inequalities involving prefix sums of stu... | {
"exploration": "Okay so we have n questions each worth 1 point, m students, each question i was solved by between l_i and r_i students. Total score sum is t. We have q known rank-score pairs (p_i, s_i) with non-increasing scores. We need the maximum number of students who could tie for rank 1, and the maximum possi... | {
"exploration": "Let me set up the binary variables. For each question i (1 to n) and each student j (1 to m), x_{i,j} ∈ {0,1} is 1 if student j solved question i. Constraints: for each i, l_i ≤ Σ_j x_{i,j} ≤ r_i. For each known rank p with score s, Σ_i x_{i,p} = s. Total sum over all i,j x_{i,j} = t. I'll introduce... | {
"exploration": "I'll build a DP over students in rank order. Let dp[i][k][S] = possible? where i students processed, k of them are in the \"tied for first\" group, and S is the total score of those k top students. The remaining i-k students have strictly lower scores. I also need to track total score of all i stude... | null | null | null | So that's a few failed attempts now. The greedy assignment locked into a locally minimal distribution, the simulated annealing couldn't navigate the combinatorial constraints, and DP exploded because tracking each question's state individually is impossible at this scale.
hmm. But the flow idea — distributing total sc... | function solve():
READ n, m
arrays l[1..n], r[1..n]
for i from 1 to n:
READ l[i], r[i]
READ q
arrays known_p[1..q], known_s[1..q]
for i from 1 to q:
READ known_p[i], known_s[i]
READ t
// Precompute total sums
total_l = sum(l[1..n])
total_r = sum(r[1..n])
if t... |
1844/H | Multiple of Three Cycles | 3,400 | combinatorics, data structures, dp, dsu, math | 3 | 256 | An array $$$a_1,\dots,a_n$$$ of length $$$n$$$ is initially all blank. There are $$$n$$$ updates where one entry of $$$a$$$ is updated to some number, such that $$$a$$$ becomes a permutation of $$$1,2,\dots,n$$$ after all the updates.
After each update, find the number of ways (modulo $$$998\,244\,353$$$) to fill in t... | The first line contains a single integer $$$n$$$ ($$$3 \le n \le 3 \cdot 10^5$$$, $$$n \equiv 0 \pmod 3$$$).
The $$$i$$$-th of the next $$$n$$$ lines contains two integers $$$x_i$$$ and $$$y_i$$$, representing that the $$$i$$$-th update changes $$$a_{x_i}$$$ to $$$y_i$$$.
It is guaranteed that $$$x_1,\dots,x_n$$$ and... | Output $$$n$$$ lines: the number of ways (modulo $$$998\,244\,353$$$) after the first $$$1,2,\dots,n$$$ updates. | null | In the first sample, for example, after the $$$3$$$rd update the $$$3$$$ ways to complete the permutation $$$a = [4,\_,2,5,\_,\_]$$$ are as follows:
- $$$[4,1,2,5,6,3]$$$: The only cycle is $$$(1\,4\,5\,6\,3\,2)$$$, with length $$$6$$$.
- $$$[4,6,2,5,1,3]$$$: The cycles are $$$(1\,4\,5)$$$ and $$$(2\,6\,3)$$$, with le... | [
{
"input": "6\n3 2\n1 4\n4 5\n2 6\n5 1\n6 3",
"output": "32\n8\n3\n2\n1\n1"
},
{
"input": "3\n1 1\n2 3\n3 2",
"output": "0\n0\n0"
},
{
"input": "18\n1 2\n2 3\n3 4\n4 5\n5 6\n6 7\n7 8\n8 9\n9 10\n10 11\n11 12\n12 13\n13 14\n14 15\n15 16\n16 17\n17 18\n18 1",
"output": "671571067\n3539... | The partially formed permutation is composed of several paths and cycles, and only the length of each path/cycle modulo $3$ matters. We can use a DSU to track the number of paths/cycles of each length $\pmod 3$. If at any point a cycle whose length is not $\equiv 0 \pmod 3$ is formed, the answer is $0$.
Thus, the probl... | null | {
"exploration": "Okay so we have an array being filled in step by step, eventually becoming a permutation of 1..n. After each update, we need to count the number of ways to fill the remaining blanks so that the final permutation has all cycle lengths divisible by 3. The constraint is n ≤ 3×10^5, time limit 3.0s. We ... | {
"exploration": "I'll precompute the EGF for permutations with all cycle lengths multiples of 3. The exponential generating function for a single cycle of length k is z^k/k, and we restrict to k ≡ 0 mod 3. So the EGF for a set of such cycles is exp(∑_{m≥1} z^{3m}/(3m)). That sum is -ln(1-z^3)/3, so the EGF simplifie... | null | null | null | null | So none of those panned out. Brute force is impossible, and the generating function approach fell apart because the components aren't independent — the open paths can connect to each other through blank vertices, so you can't just multiply precomputed values.
hmm. But that partial observation about DSU and lengths mod... | MOD = 998244353
READ n
Initialize arrays parent[1..n], sz[1..n], len_mod[1..n], is_cycle[1..n]
Initialize cnt[0..2] = {n, 0, 0} // counts of paths by length mod 3
Initialize answer = 1 // will be updated after each edge
Precompute factorials fac[0..n] and inv_fac[0..n]
Precompute inv[1..n] // modular inverses
fu... |
2183/I1 | Pairs Flipping (Easy Version) | 3,500 | constructive algorithms | null | null | This is the easy version of the problem. The difference between the versions is that in this version, the restriction on the number of 1's remaining in the final string is more lenient. You can hack only if you solved all versions of this problem.
You are given a binary string $$$s_1s_2\ldots s_n$$$ containing only ch... | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.
The first line of each test case contains an integer $$$n$$$ ($$$3 \leq n \leq 2\cdot 10^6$$$).
The second line of each test case contains a binary stri... | For each test case, output $$$\lfloor\frac{n}{2}\rfloor$$$ numbers on a new line: the chosen $$$l$$$ for operation $$$1,2,\ldots,\lfloor\frac{n}{2}\rfloor (0 \leq l_x \leq n-x)$$$. You should guarantee that after all operations, there are at most $$$15$$$ 1 remaining in the string.
If there are multiple solutions, out... | null | null | [] | If $n$ is even, we only consider $s_1s_2\ldots s_{n-1}$. Now assume that $n$ is odd. Let $m=\frac{n-1}2$.
Consider dividing the $n$ numbers into the first $m$ and the last $m+1$ parts. The first $m$ are eliminated using odd lengths in $[1,m-1]$, and the last $m+1$ are eliminated using even lengths in $[1,m]$. (Assume ... | The key observation is a recursive pairing scheme that exploits symmetry. You need to think about pairing characters from opposite ends of an interval and using the distance between them as the operation length. The structure allows you to eliminate ones in a way that the interval shrinks geometrically when a direct pa... | {
"exploration": "Okay so we have n up to 2e6 total across test cases, and we need to reduce the number of 1's to at most 15 using exactly floor(n/2) operations. Each operation x picks l between 0 and n-x, and flips positions l and l+x. The hypothesis is to use DP over prefix parity states with a small period, tracki... | {
"exploration": "I'll start with a random assignment of l values for each operation x from 1 to floor(n/2). For each position i in the string, I maintain flip[i] = parity of how many times position i has been flipped so far. When I change l_x from old_l to new_l, I update flip[old_l] ^= 1, flip[old_l+x] ^= 1, flip[n... | {
"exploration": "For each operation x from 1 to floor(n/2), I'll try to greedily reduce the number of ones by flipping pairs (l, l+x) where both are 1, turning two ones into zeros. So for each x, I scan l from 1 to n-x, and whenever s[l] and s[l+x] are both 1, I set l = that index and perform the flip, then break. I... | null | null | null | Right, so nothing worked so far. The DP lost count information, random search was astronomically hopeless, and any greedy covering blew up quadratically. The core problem is the massive search space — I need a deterministic, linear-time construction that systematically reduces ones, but the allowed operations flip two ... | function solve():
READ t
for each test case:
READ n
READ string s of length n (1-indexed)
// Convert s to array of integers: s[i] = 0 or 1
a = array of size n+1
for i from 1 to n:
a[i] = integer value of s[i]
ops = array of size (n//2 + 1), initialize... |
848/E | Days of Floral Colours | 3,400 | combinatorics, divide and conquer, dp, fft, math | 7 | 256 | The Floral Clock has been standing by the side of Mirror Lake for years. Though unable to keep time, it reminds people of the passage of time and the good old days.
On the rim of the Floral Clock are 2n flowers, numbered from 1 to 2n clockwise, each of which has a colour among all n possible ones. For each colour, the... | The first and only line of input contains a lonely positive integer n (3 ≤ n ≤ 50 000) — the number of colours present on the Floral Clock. | Output one integer — the sum of beauty over all possible arrangements of flowers, modulo 998 244 353. | null | With n = 3, the following six arrangements each have a beauty of 2 × 2 = 4.
While many others, such as the left one in the figure below, have a beauty of 0. The right one is invalid, since it's asymmetric. | [
{
"input": "3",
"output": "24"
},
{
"input": "4",
"output": "4"
},
{
"input": "7",
"output": "1316"
},
{
"input": "15",
"output": "3436404"
}
] | tl;dr Just look at recurrences of g, f0, f1 and f2 and the part after f2's recurrence.
Break the circle down into semicircles.
We're basically pairing flowers under the restrictions. It's hard to deal with the whole circle, let's consider something simpler. Consider an arc of length i (segment of i flowers) and their o... | null | {
"exploration": "I'll generate all valid arrangements by brute force: assign each of the 2n positions one of n colors, respecting the constraints. For each color, its two occurrences must be at distance ≤2 or exactly n. Also, if positions u and v have the same color, then their opposites (u+n mod 2n) must also have ... | {
"exploration": "I'll model the arrangement as a perfect matching on 2n vertices (flowers) where each pair (same colour) must satisfy the distance condition: distance ≤ 2 or distance = n. Also, the opposite-colour symmetry: if flowers u and v have same colour, then their opposites (u+n mod 2n, v+n mod 2n) also have ... | {
"exploration": "I'll try to build a DP on the circle by scanning positions clockwise, keeping track of the last two positions' pairing status. Let me define dp[i][mask] where i is the current position (from 1 to 2n) and mask encodes whether the previous two positions (i-2 and i-1) are already paired with something ... | null | null | null | So all those ideas fell apart — brute-force was astronomically impossible, inclusion-exclusion blew up in complexity, and DP with local state couldn't capture the non-local opposite-pairing constraints spanning n positions.
hmm. But that last thread — "break the circle down into semicircles" — keeps tugging at me. The... | MOD = 998244353
function solve():
READ n
if n < 3:
PRINT 0
return
// Precompute factorials and inverse factorials up to n
fact = array of size n+1
inv_fact = array of size n+1
fact[0] = 1
for i from 1 to n:
fact[i] = fact[i-1] * i % MOD
inv_fact[n] = pow(fact[n]... |
2124/H | Longest Good Subsequence | 3,400 | dp,math,trees | null | null | Call an array $$$b$$$ of size $$$m$$$ good if the following conditions hold:
• $$$1 \leq b_i \leq i$$$ for each $$$1 \leq i \leq m$$$.
• There exists a permutation$$$^{\text{∗}}$$$ $$$p$$$ of size $$$m$$$ such that for each $$$1 \leq i \leq m$$$, $$$b_i$$$ is the smallest integer where $$$\min(p_{b_i}, p_{b_i+1}, \l... | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.
The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 15\,000$$$) – the length of $$$a$$$.
The second line of each test case co... | For each test case, output the length of the longest good subsequence of $$$a$$$ on a new line. | null | null | [] | Let's first determine whether a given sequence can be good. We claim that a sequence $b$ is good if for each index $1 \leq i \leq |b|$, if $b_i=x$, then one of the following must be true:
- $x=i$
- $b_x=x$ and $min(b_x,b_{x+1},\ldots,b_i)\geq x$.
To prove this, we can note that this is necessary by transitivity of $ ... | The key observation is that the condition for a good sequence can be expressed as a constraint on intervals defined by positions where a value equals its index. This suggests a dynamic programming approach that processes the array from right to left, building answers for segments where all included values are at least ... | {
"exploration": "I'll scan the array left to right, building the longest subsequence that satisfies both conditions. For each position i in the subsequence (1-indexed within the subsequence), the value must be between 1 and i. So I'll maintain a running length L and for each element a[j] in the original array, I'll ... | {
"exploration": "I'll maintain a partial permutation assignment as I build the subsequence left to right. For each chosen position i in the subsequence, I assign p_i a distinct integer from 1 to m (where m is the final subsequence length). The condition: for each chosen index i, b_i is the smallest index such that m... | null | null | null | null | Okay, so both paths failed. The greedy approach missed the structural nuance of the permutation condition, letting in sequences like [1,1,2] that are impossible. The segment tree attempt to model all valid permutations exploded in complexity—even with pruning, the state space was still factorial in the worst case.
hmm... | function solve():
READ t
for each test case:
READ n
READ array a[1..n]
// dp[i][j] = max length of good subsequence ending at index i with last value j?
// Actually, we need to implement the O(n^2) DP described in the editorial.
// Let dp[i][j] be the maximum le... |
1158/E | Strange device | 3,400 | binary search, interactive, math, trees | 1 | 256 | It is an interactive problem.
Vasya enjoys solving quizzes. He found a strange device and wants to know how it works.
This device encrypted with the tree (connected undirected graph without cycles) with $$$n$$$ vertices, numbered with integers from $$$1$$$ to $$$n$$$. To solve this quiz you should guess this tree.
F... | null | null | In the beginning, your program should read one integer $$$n$$$ — the number of vertices of the tree which encrypts the device ($$$2 \leq n \leq 1000$$$).
After that, you can make several operations in the following format. To do operation print a symbol"?" (without quotes) and $$$n$$$ integers $$$d_1, d_2, \ldots, d_n... | It is a picture of the tree which encrypt the device from the first test:
It is a table of pairwise distances between vertices in this tree:
- If you make operation where $$$d = [0, 0, 0, 0, 0]$$$, no lamp will switch on, because $$$dist(i, j) > 0$$$ for all $$$i \neq j$$$.
- If you make operation where $$$d = [1, 1,... | [
{
"input": "5\n00000\n11011\n11100\n10010",
"output": "? 0 0 0 0 0\n? 1 1 2 0 2\n? 0 0 0 1 0\n? 0 1 0 0 1\n!\n4 2\n1 5\n3 4\n4 1"
}
] | The solution will consist of two parts.
1 part
Let's divide all points into sets with equal distance to the vertex $1$. To do this, we will do the following algorithm. How to understand what set of points lies at a distance $\lfloor \frac{n}{2} \rfloor$ from the top of $1$? Let's fill $d$ with zeros and make $d_1 = \lf... | The key observation is that you can treat the distance from a fixed root as a monotone property you can binary search on for all vertices simultaneously using just two global queries. Then, to find parent-child relationships across layers, you can encode ancestor indices in binary and use the device's response as a bit... | {
"exploration": "Okay so we have an interactive tree reconstruction problem with n ≤ 1000 vertices. We can query the device by providing an array d_1..d_n of non-negative integers, and for each vertex i, lamp i lights up if there exists some vertex j ≠ i such that dist(i, j) ≤ d_j. We get back a binary string of len... | {
"exploration": "For each vertex j, I'll set d_j = k and all other d_i = 0. Then lamp i lights up exactly when dist(i, j) ≤ k. So for fixed j, I can binary search k from 0 to n-1 to find the exact dist(i, j) for every i — the smallest k where lamp i turns on is dist(i, j). With n=1000 vertices, that's n binary searc... | null | null | null | null | So none of those panned out. Random vertex probing would need a thousand queries, and binary lifting on distances lost individual edges when too many ancestors shared the same bit.
hmm — but that half-formed idea about binary searching distances from a root and encoding ancestors in binary... wait — what if I could, i... | function solve():
READ n
// Part 1: find distances from vertex 1 for all vertices
dist = array of size n+1, initialized to 0
// good distances set: initially {0, n}
good = sorted list [0, n]
// vertices_by_distance[d] will store list of vertices at exact distance d
vertices_by_distance = arr... |
1441/E | Differentiating Games | 3,400 | games, interactive | 2 | 512 | This is an interactive problem
Ginny is taking an exam on game theory. The professor is tired of hearing the same answers over and over again, so he offered Ginny to play a game instead of a standard exam.
As known from the course, a combinatorial game on a graph with multiple starting positions is a game with a dire... | null | null | In this task, interaction consists of several phases.
In the first phase, the interactor gives as an input to your program three integers $$$N$$$ ($$$1 \le N \le 1000$$$), $$$M$$$ ($$$0 \le M \le 100\,000$$$), $$$T$$$ ($$$1 \le T \le 2000$$$): the number of vertices and edges in the initial graph, and the number of ti... | In the sample test, the empty lines represent waiting for the input by the other side of the interaction. The real interactor will not print empty lines, and the solution should not print them either.
The image above illustrates the sample test. Added edges are coloured in red, and the removed edges are drawn with a d... | [
{
"input": "3 2 3\n1 2\n2 3\n\nLose\n\nCorrect\n\nWin\n\nCorrect\n\nDraw\n\nCorrect",
"output": "6\n+ 2 2\n- 1 2\n+ 2 3\n- 2 2\n+ 3 1\n+ 2 2\n? 0\n\n! 1\n\n? 1 2\n\n! 3\n\n? 5 1 3 1 3 1\n\n! 2"
}
] | Let's first solve the problem for an empty graph. To do this we need to build a construction with not too many edges, where all vertices are pairwise not equivalent.
For acyclic graph, there is not such construction. All Grundy functions of vertices should be different, so they must be integers from 0 to $n-1$. The onl... | What if you could transform the graph so that each vertex's game-theoretic behavior is uniquely encoded by its relationship to a small set of special vertices? The key observation is that adding self-loops to most vertices changes the nature of draws, and you can then use a small acyclic subgraph as a basis for binary ... | {
"exploration": "Okay, so we have an interactive problem where we can modify a directed acyclic graph by adding/removing up to 4242 edges, then we need to guess a hidden vertex in up to T=2000 phases, each with at most 20 queries of total multiset size ≤20. The graph has N≤1000 vertices, M≤100000 edges initially. Th... | {
"exploration": "I'll start by topologically sorting the initial graph, which is acyclic. For Example 1: N=3, edges 1→2, 2→3. Topological order is [1,2,3]. The chosen vertex is somewhere in this order — I'll treat its position as an unknown integer from 1 to 3.\n\nThe plan: add edges to make the first 20 vertices (o... | {
"exploration": "I'll add 4242 random edges to the graph, then for each vertex compute a 64-bit hash of its \"game signature\" — the sequence of outcomes when queried against a fixed set of random multisets. During interaction, I query those same multisets, get the outcome sequence, hash it, and match against precom... | null | null | null | So that's a few failed attempts now. Grundy XOR fails because draws break the impartial game rules, binary search on a DAG would need too many edges, and the state space is astronomically large to simulate directly. The core problem is that with cycles, the game outcome isn't a simple function of vertex properties — it... | READ N, M, T
READ edges[1..M] as pairs (a_i, b_i)
// Step 1: Identify acyclic part vertices (no outgoing edges to outside set)
// We will use 20 vertices as the acyclic part (k=20)
// Find all vertices with no outgoing edges in the original graph, or pick any 20 vertices that have no edges among themselves? Actually, ... |
715/E | Complete the Permutations | 3,400 | combinatorics, fft, graphs, math | 5 | 256 | ZS the Coder is given two permutations p and q of {1, 2, ..., n}, but some of their elements are replaced with 0. The distance between two permutations p and q is defined as the minimum number of moves required to turn p into q. A move consists of swapping exactly 2 elements of p.
ZS the Coder wants to determine the n... | The first line of the input contains a single integer n (1 ≤ n ≤ 250) — the number of elements in the permutations.
The second line contains n integers, p1, p2, ..., pn (0 ≤ pi ≤ n) — the permutation p. It is guaranteed that there is at least one way to replace zeros such that p is a permutation of {1, 2, ..., n}.
Th... | Print n integers, i-th of them should denote the answer for k = i - 1. Since the answer may be quite large, and ZS the Coder loves weird primes, print them modulo 998244353 = 223·7·17 + 1, which is a prime. | null | In the first sample case, there is the only way to replace zeros so that it takes 0 swaps to convert p into q, namely p = (1, 2, 3), q = (1, 2, 3).
There are two ways to replace zeros so that it takes 1 swap to turn p into q. One of these ways is p = (1, 2, 3), q = (3, 2, 1), then swapping 1 and 3 from p transform it ... | [
{
"input": "3\n1 0 0\n0 2 0",
"output": "1 2 1"
},
{
"input": "4\n1 0 0 3\n0 0 0 4",
"output": "0 2 6 4"
},
{
"input": "6\n1 3 2 5 4 6\n6 4 5 1 0 0",
"output": "0 0 0 0 1 1"
},
{
"input": "4\n1 2 3 4\n2 3 4 1",
"output": "0 0 0 1"
}
] | Prerequisites : Math, Graph Theory, DP, Any fast multiplication algorithm
We'll slowly unwind the problem and reduce it to something easier to count. First, we need to determine a way to tell when the distance between $p$ and $q$ is exactly $k$. This is a classic problem but I'll include it here for completeness.
Let... | What if you think of the distance in terms of cycles in a graph formed by the filled permutations? The key observation is that the number of swaps needed relates directly to the cycle count. The combinatorial challenge becomes counting ways to assign the unknown values to achieve a specific cycle count, which involves ... | {
"exploration": "Okay so we have two permutations p and q of {1..n} with some zeros, and we need to count completions such that the swap distance between the completed permutations is exactly k, for all k from 0 to n-1. The distance is the minimum number of swaps of two elements to turn p into q. For permutations, t... | {
"exploration": "I'll build a graph on vertices 1..n where each position i with both p_i and q_i non-zero gives a directed edge p_i → q_i. The distance between permutations is n minus the number of cycles in the permutation that maps p to q — specifically, if we think of q as the target, the permutation σ = q^{-1} ∘... | {
"exploration": "I'll build a bipartite graph between the zero positions in p and the zero positions in q, then try to express the inversion count as a quadratic form over assignment variables. Let me set up the variables first.\n\nLet the zero positions in p be indices i_1, i_2, ..., i_a where p[i_j] = 0. Let the z... | null | null | null | None of those got me anywhere useful. The DP blew up exponentially, inclusion-exclusion was factorial, and interpolation was hopelessly high-dimensional — all because they tried to track assignments or permutations directly, missing the underlying structure.
hmm. But that thread about cycles and distance... wait — the... | READ n
READ array p of size n
READ array q of size n
MOD = 998244353
// Step 1: Precompute factorials and inverse factorials up to n
fact = array of size n+1
invFact = array of size n+1
fact[0] = 1
for i from 1 to n:
fact[i] = fact[i-1] * i % MOD
invFact[n] = pow(fact[n], MOD-2, MOD) // modular inverse
for i from ... |
914/H | Ember and Storm's Tree Game | 3,400 | combinatorics, dp, games, trees | 3 | 512 | Ember and Storm play a game. First, Ember picks a labelled tree T of n vertices, such that the degree of every vertex is at most d. Then, Storm picks two distinct vertices u and v in this tree and writes down the labels of the vertices in the path from u to v in a sequence a1, a2... ak. Finally, Ember picks any index i... | The input consists of a single line containing three integers n, d and m (2 ≤ n ≤ 200, 1 ≤ d < n, 1 ≤ m ≤ 2·109). | Print a single number — the number of possible tuples if Ember and Storm play as described, modulo m. | null | In the first sample case, there is only one possible tree. There are two possible paths, 1 to 2 and 2 to 1. For both paths, i can only be 1, and op can take both possibilities. Therefore, the answer is 4.
In the second sample, there are no possible trees.
In the third sample, there are three possible trees. | [
{
"input": "2 1 1000000007",
"output": "4"
},
{
"input": "3 1 250",
"output": "0"
},
{
"input": "3 2 100",
"output": "36"
}
] | Ember wins if the path chosen by Storm is monotonic or bitonic. In this case, there can be two (i, op) pairs.
Let S be the set of trees having n vertices in which all paths are bitonic or monotonic. We need to find 2n(n - 1)|S|.
For every tree in S, there exists at least one vertex such that every path starting or endi... | The key observation is that the winning condition for Ember forces a strong structural restriction on the tree itself — all paths must be bitonic. This transforms the problem into counting trees with a specific global ordering property. The counting technique involves a combinatorial DP that builds trees layer by layer... | {
"exploration": "Okay so we have a game where Ember picks a tree with max degree d, Storm picks two vertices u,v and writes the path labels, then Ember picks a split index i and either flips-and-adds or negates-and-adds the suffix. Ember wins if the resulting sequence is monotonic. We need to count all tuples (T,u,v... | {
"exploration": "First, I need to simulate optimal play for a given tree T, vertices u,v, and then Ember's choice of i and operation. The number of possible tuples is the count of (T,u,v,i,op) that can occur under optimal play. Since n ≤ 200, I can try enumerating all labelled trees with max degree ≤ d, then for eac... | {
"exploration": "I'll attempt to build a generating function for bitonic paths in trees with degree bound d. Let me start by trying to count something concrete for small n.\n\nFor a fixed tree T on n vertices, a path from u to v has vertex labels along it. The path is bitonic if the sequence of labels has exactly on... | null | null | null | Okay, so I've burned through a few ideas and none stuck. Brute force is astronomically impossible, and trying to simulate optimal play directly is lost in a sea of trees. The common failure is treating the tree as arbitrary — I need to find the structural property that makes Ember's win possible, and then count only th... | READ n, d, m
// tree[i][deg]: number of trees with i vertices, root degree = deg, all paths from root increasing, max degree ≤ d
tree = 2D array of size (n+1) × (d+1), initialized to 0
tree[1][0] = 1 // single vertex root, degree 0
for i from 1 to n:
// precompute tree[i][any] = sum_{deg=1}^{d-1} tree[i][deg]
... |
1776/N | Count Permutations | 3,500 | math | 5 | 256 | You are given a string $$$s$$$ with length $$$n-1$$$ whose characters are either $$$\texttt{<}$$$ or $$$\texttt{>}$$$.
Count the permutations $$$p_1, \, p_2, \, \dots, \, p_n$$$ of $$$1, \, 2, \, \dots, \, n$$$ such that, for all $$$i = 1, \, 2, \, \dots, \, n - 1$$$, if $$$s_i$$$ is $$$\texttt{<}$$$ then $$$p_i < p_{... | The first line contains a single integer $$$n$$$ ($$$2 \le n \le 100\,000$$$).
The second line contains a string $$$s$$$ of length $$$n-1$$$; each character of $$$s$$$ is either $$$\texttt{<}$$$ or $$$\texttt{>}$$$. | Print the logarithm in base $$$2$$$ of the number of permutations satisfying the constraints described in the statement.
Your answer is considered correct if its absolute or relative error does not exceed $$$10^{-6}$$$. Formally, let your answer be $$$x$$$ and let the correct answer be $$$y$$$. Your answer is accepted... | null | In the first sample, there is only one valid permutation, that is $$$[2, 1]$$$. Since $$$\log_2(1)=0$$$, the correct output is $$$0$$$.
In the second sample, there are $$$2$$$ valid permutations, that are $$$[3, 1, 2]$$$ and $$$[2, 1, 3]$$$. Since $$$\log_2(2)=1$$$, the correct output is $$$1$$$.
In the third sample,... | [
{
"input": "2\n<",
"output": "0.0000000000"
},
{
"input": "3\n<>",
"output": "1.0000000000"
},
{
"input": "5\n><<<",
"output": "2.0000000000"
},
{
"input": "10\n<><<<<<>>",
"output": "9.8281364842"
}
] | This was meant to be the hardest problem of the mirror. It required advanced knowledge in modern combinatorics, together with some clever ideas to reach the desired precision. The solution splits in two parts: first we obtain a nontrivial (actually, pretty hard) formula for the exact answer, then we explain how to appr... | The key observation is that the constraints define a shape on a grid, and the count relates to the number of ways to fill a specific set of cells with increasing values. The formula involves summing contributions from all monotone paths across this shape, where each path's weight depends on the geometry of the cells it... | {
"exploration": "Okay so we have a string s of length n-1 with < and > characters, and we need to count permutations of 1..n that satisfy all the adjacent inequalities. The constraint is n ≤ 100,000, which is huge — we can't do anything exponential. We need to compute the log base 2 of the count, not the count itsel... | {
"exploration": "I'll set up a Markov chain that samples valid permutations. Start with a random permutation that satisfies the constraints — I can build one by inserting numbers 1..n one at a time, always placing the next number at a position that respects the inequalities. Then at each step, pick two positions i,j... | null | null | null | null | Well, both paths were dead ends — the bitmask DP was impossible to store, and the MCMC approach was far too slow and inaccurate for the precision needed. The core problem is that the number of valid permutations is astronomically huge, so we can't count them directly; we need the logarithm, which suggests we're summing... | READ n
READ s of length n-1
// Build adjacency list for directed graph: edge from larger to smaller
adj = array of empty lists of size n+1
indeg = array of zeros of size n+1
for i from 1 to n-1:
if s[i] == '<':
// i+1 > i, so edge from i+1 to i
adj[i+1].append(i)
indeg[i] += 1
else: //... |
794/G | Replace All | 3,400 | combinatorics, dp, math | 2 | 256 | Igor the analyst is at work. He learned about a feature in his text editor called "Replace All". Igor is too bored at work and thus he came up with the following problem:
Given two strings x and y which consist of the English letters 'A' and 'B' only, a pair of strings (s, t) is called good if:
- s and t consist of t... | The first line contains the string c (1 ≤ |c| ≤ 3·105).
The second line contains the string d (1 ≤ |d| ≤ 3·105).
The last line contains a single integer n (1 ≤ n ≤ 3·105). | Output a single integer: the answer to the problem, modulo 109 + 7. | null | For the first sample, there are four possible pairs of (c', d').
If (c', d') = (AA, A), then the flexibility is 0.
If (c', d') = (AB, A), then the flexibility is 0.
If (c', d') = (AA, B), then the flexibility is 2, as the pairs of binary strings (1, 11), (0, 00) are the only good pairs.
If (c', d') = (AB, B), then ... | [
{
"input": "A?\n?\n3",
"output": "2"
},
{
"input": "A\nB\n10",
"output": "2046"
}
] | We want to count the number of pairs of strings $(S, T)$ with $|S|, |T| \le N$ such that $A$ can be written as copies of $S$ and $B$ can be written as copies of $T$, where $A$ and $B$ are strings with some characters replaced by '?'.
First, let's handle constant strings $A$ and $B$.
A pair $(S, T)$ is valid iff $A$ i... | The key observation is that the condition for two replaced strings to be equal forces a strong relationship between the replacement strings S and T, turning the problem into a combinatorial count based on the difference in 'A' and 'B' counts. This count can be expressed using number-theoretic functions over the lengths... | {
"exploration": "Okay so we have two strings c and d with characters 'A', 'B', '?' and a bound n. We need to sum over all ways to replace '?' with 'A' or 'B' the number of pairs (s, t) of binary strings (lengths 1 to n) such that when we substitute s for 'A' and t for 'B' in both c' and d', the resulting strings are... | {
"exploration": "Let me try to build a DP that checks, for a fixed pair of lengths L = |s| and M = |t|, whether there exist binary strings s and t that make the replaced x and y equal. I'll set up dp[i][j] = true if the first i characters of the replaced x and the first j characters of the replaced y can be made equ... | {
"exploration": "I'll fix the lengths p = |s| and q = |t|, and set up a linear system over GF(2). For each position in the strings x and y (which are c' and d' after replacing '?' with A/B), the replacement process yields equalities between bits of s and t. Let s = s_0 s_1 ... s_{p-1} and t = t_0 t_1 ... t_{q-1}. Wh... | {
"exploration": "We need to test the KMP-based periodicity analysis on the merged string. The idea: for a fixed pair (s,t), the replacement transforms x and y into the same string. Let’s denote the transformed string from x as X(s,t) and from y as Y(s,t). The condition X(s,t)=Y(s,t) means that the concatenated strin... | null | null | Okay, all of those went nowhere. The brute force is impossible, the DP on alignment is too big, and even the equation approach still left me iterating over all (p,q) pairs, which is O(n²) and hopeless.
hmm. Wait — the equation approach got close, but I missed the core simplification. The condition that replacing A→s a... | MOD = 1_000_000_007
READ string c
READ string d
READ integer n
Lc = length(c)
Ld = length(d)
// Precompute factorials and inverse factorials up to total number of '?'
cnt_quest_c = 0
cnt_quest_d = 0
base = 0 // Δa from fixed characters
for i from 0 to Lc-1:
if c[i] == 'A':
base = base + 1
else if c... |
756/F | Long number | 3,400 | expression parsing, math, number theory | 2 | 512 | Consider the following grammar:
- <expression> ::= <term> | <expression> '+' <term>
- <term> ::= <number> | <number> '-' <number> | <number> '(' <expression> ')'
- <number> ::= <pos_digit> | <number> <digit>
- <digit> ::= '0' | <pos_digit>
- <pos_digit> ::= '1' | '2' | '3' | '4' | '5' | '6' | '7' | '8' | '9'
This gra... | The only line contains a non-empty string at most 105 characters long which is valid according to the given grammar. In particular, it means that in terms l-r l ≤ r holds. | Print single integer — the number described by the expression modulo 109 + 7. | null | null | [
{
"input": "8-11",
"output": "891011"
},
{
"input": "2(2-4+1)+2(2(17))",
"output": "100783079"
},
{
"input": "1234-5678",
"output": "745428774"
},
{
"input": "1+2+3+4-5+6+7-9",
"output": "123456789"
}
] | First of all we need to somehow parse the expression. Let's suppose we've builded a parse tree for the expression. Every number X can be seen as operation "concatenate something and X". We should understand how such an operation changes the number. YX = Y·10len(X) + X. So we can represent each number X as pair (10len(X... | The key observation is that concatenation acts like multiplying by a power of ten and adding. This suggests representing each expression as a pair of modular values. The challenge is to compute the length and value for complex constructs like ranges and repetitions efficiently, which involves summing geometric series a... | {
"exploration": "Okay so we have a grammar that describes huge decimal numbers via concatenation, ranges, and repetition. The input string is at most 10^5 characters long, and we need to output the described integer modulo 1e9+7. The approach is to parse recursively and actually build the decimal string as a big int... | {
"exploration": "I'll build a segment tree over the expression string, where each leaf is a single digit token, and internal nodes store (value mod MOD, digit_length). The merge operation is: left_val * 10^{right_len} + right_val, all mod MOD. For range blocks like \"8-11\", I'll preprocess them into a special node ... | {
"exploration": "I'll parse the expression recursively, evaluating each term to its decimal string representation as a big integer modulo 1e9+7, but since the number of digits can be exponential, I must compute the length and value modulo MOD without building the full string. For a number like \"123\", I compute its... | null | null | null | Hmm, all those ideas fell apart. Direct simulation explodes in digit count, segment trees can't handle the exponential range sizes, and recursion still forces iteration over impossible intervals like 1234-5678.
Wait — the nagging thread is that concatenation is just multiplication by a power of ten and addition. So an... | MOD = 1000000007
PHI = MOD - 1
function pow10(e):
// compute 10^e mod MOD, where e is given modulo PHI
return fast_exponentiation(10, e mod PHI, MOD)
function parse_number(s, start):
// parse digits until non-digit, return (value mod MOD, length, next_index)
val = 0
len = 0
i = start
while... |
843/D | Dynamic Shortest Path | 3,400 | graphs, shortest paths | 10 | 512 | You are given a weighted directed graph, consisting of n vertices and m edges. You should answer q queries of two types:
- 1 v — find the length of shortest path from vertex 1 to vertex v.
- 2 c l1 l2 ... lc — add 1 to weights of edges with indices l1, l2, ..., lc. | The first line of input data contains integers n, m, q (1 ≤ n, m ≤ 105, 1 ≤ q ≤ 2000) — the number of vertices and edges in the graph, and the number of requests correspondingly.
Next m lines of input data contain the descriptions of edges: i-th of them contains description of edge with index i — three integers ai, bi... | For each query of first type print the length of the shortest path from 1 to v in a separate line. Print -1, if such path does not exists. | null | The description of changes of the graph in the first sample case:The description of changes of the graph in the second sample case: | [
{
"input": "3 2 9\n1 2 0\n2 3 0\n2 1 2\n1 3\n1 2\n2 1 1\n1 3\n1 2\n2 2 1 2\n1 3\n1 2",
"output": "1\n0\n2\n1\n4\n2"
},
{
"input": "5 4 9\n2 3 1\n2 4 1\n3 4 1\n1 2 0\n1 5\n1 4\n2 1 2\n2 1 2\n1 4\n2 2 1 3\n1 4\n2 1 4\n1 4",
"output": "-1\n1\n2\n3\n4"
}
] | Firstly, let's run an usual Dijkstra from s, find distances and make them a potentials of vertices. Then for each request let's recalculate all distances and make the potentials equal to these distances. To quickly recalculate the distance between requests, we can use the fact that in a graph with potentials, all dista... | What if you could maintain a set of 'current best distances' and, after each update, only recompute distances in a limited region? The key observation is that after adjusting edge weights, the change in any shortest path length is bounded by the number of edges you just touched. This suggests a shortest path algorithm ... | {
"exploration": "Okay so we have a directed weighted graph with n up to 10^5, m up to 10^5, and q up to 2000. The total number of edge updates across all type-2 queries is at most 10^6. Time limit is 10 seconds. We need to answer shortest path queries from vertex 1 to v, with edge weight increases of +1 on specified... | {
"exploration": "I'll build a shortest-path tree rooted at vertex 1 using Dijkstra on the initial graph, then maintain it with a link-cut tree where each node stores the edge weight from its parent. For non-tree edges, I'll keep them in a separate list. When a type 2 query increments some edge weights, I update the ... | null | null | null | null | Right, so nothing worked so far. Floyd-Warshall couldn't handle the fact that distances should increase when edge weights increase, and trying to model path changes with a single detour missed the cascading effect where a whole chain of edges gets updated, shifting the optimal path through multiple non-tree edges.
hmm... | READ n, m, q
adj = array of empty lists of size n+1
edges = array of size m+1
FOR i from 1 to m:
READ a, b, c
edges[i] = (a, b, c)
ADD (b, c) to adj[a] // directed edge
dist = array of size n+1, initialized to INFINITY
dist[1] = 0
priority_queue = min-heap
PUSH (0, 1) into priority_queue
WHILE priority_q... |
1854/F | Mark and Spaceship | 3,500 | brute force, dp | 3 | 512 | Mark loves to move fast. So he made a spaceship that works in $$$4$$$-dimensional space.
He wants to use the spaceship to complete missions as fast as possible. In each mission, the spaceship starts at $$$(0, 0, 0, 0)$$$ and needs to end up at $$$(a, b, c, d)$$$. To do this, he instructs the spaceship's computer to ex... | The only line of the input contains the four integers $$$A, B, C, D$$$ ($$$0\le A,B,C,D\le 1000$$$). | Print the sum of $$$f(a, b, c, d)$$$ over the set of points described in the statement. | null | In the first sample, one has to compute $$$f(-1, 0, 0, 0)+f(0, 0, 0, 0) + f(1, 0, 0, 0) = 1 + 0 + 1 = 2$$$.
In the second sample, one has to compute the sum of $$$f(a, b, c, d)$$$ over $$$27$$$ different points $$$(a, b, c, d)$$$. Let us describe the value of $$$f(a, b, c, d)$$$ for some of them:
- It holds $$$f(-1, ... | [
{
"input": "1 0 0 0",
"output": "2"
},
{
"input": "1 1 0 1",
"output": "82"
},
{
"input": "3 2 4 1",
"output": "4616"
}
] | The $2$-dimensional case.
Let us begin by cpnsidering the $2$-dimensional version of the problem. The solution to this simpler version provides the idea of the approach for the $4$-dimensional version.
We want to reach $(a, b)$. Can we do it with exactly $k$ moves? Two simple necessary conditions are:
• $|a|+|b|\le 1 +... | The key observation is that the problem reduces to checking a simple necessary condition — Manhattan distance bound and parity — for each possible move count k, but there is a small, structured set of points where this condition is not sufficient. The task is to efficiently enumerate these exceptional points across all... | {
"exploration": "Okay so we have a 4D grid where each move i is executed i times — the first move once, second move twice, third move thrice, and so on. The spaceship starts at (0,0,0,0) and we need the minimum number of moves to reach each target (a,b,c,d) within bounds [-A,A]×[-B,B]×[-C,C]×[-D,D]. Constraints: A,B... | {
"exploration": "I'll set up a DP table dp[t][x][y][z][w] as a boolean — true if after t moves we can land at (x,y,z,w). The t-th move is executed t times, so from dp[t-1][x'][y'][z'][w'] we can transition to dp[t][x' + t*dx, y' + t*dy, z' + t*dz, w' + t*dw] for each of the 8 direction vectors (±1,0,0,0), (0,±1,0,0)... | {
"exploration": "I'll split the move sequence at some midpoint m. For the first half, moves 1 through m, each move i has 8 possible directions, and the displacement contributed is i times the unit vector in that direction. So the first half produces a set of reachable vectors (x,y,z,w) where each coordinate is a sum... | {
"exploration": "I'll set up variables x_p, x_n, y_p, y_n, z_p, z_n, w_p, w_n as non-negative integers counting positive and negative steps along each axis. For a target (a,b,c,d), we need x_p - x_n = a, y_p - y_n = b, z_p - z_n = c, w_p - w_n = d. The total number of steps is S_k = k(k+1)/2 where k is the number of... | null | null | Alright, back to square one. Every direct approach exploded — BFS state space was too big, DP had exponential blowup, meet-in-the-middle couldn't handle the grid size, and per-point checking was impossible.
hmm. That half-formed thread — "Manhattan distance bound and parity for each k" — keeps pulling. The i-th move i... | function countPoints(A, B, C, D):
L = [A, B, C, D]
maxM = A + B + C + D
dp = array of size maxM+1, initialized to 0
dp[0] = 1
for idx from 0 to 3:
bound = L[idx]
new_dp = array of size maxM+1, initialized to 0
for m from 0 to maxM:
if dp[m] == 0: continue
... |
1827/E | Bus Routes | 3,400 | binary search, constructive algorithms, dfs and similar, greedy, trees | 2.5 | 1,024 | There is a country consisting of $$$n$$$ cities and $$$n - 1$$$ bidirectional roads connecting them such that we can travel between any two cities using these roads. In other words, these cities and roads form a tree.
There are $$$m$$$ bus routes connecting the cities together. A bus route between city $$$x$$$ and cit... | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.
The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$2 \le n \le 5 \cdot 10^5, 0 \le m \le 5 \cdot 10^5$$$) — the number of cit... | For each test case, output "YES" if you can travel between any pair of cities using at most two bus routes.
Otherwise, output "NO". In the next line, output two cities $$$x$$$ and $$$y$$$ ($$$1 \le x, y \le n$$$) such that it is impossible to reach city $$$y$$$ from city $$$x$$$ using at most two bus routes.
You can ... | null | Here are the graphs of test case $$$1$$$, $$$2$$$, and $$$4$$$:
Sample 1
Sample 2
Sample 4 | [
{
"input": "4\n5 2\n1 2\n2 3\n3 4\n2 5\n1 4\n5 2\n5 1\n1 2\n2 3\n3 4\n2 5\n1 5\n2 0\n1 2\n6 3\n1 2\n2 3\n3 4\n4 5\n5 6\n1 3\n2 5\n4 6",
"output": "YES\nNO\n1 3\nNO\n1 2\nNO\n1 6"
}
] | **Hint 1**
You do not need to consider all pairs of nodes, only some of them will do.
**Hint 2**
Find an equivalent condition of the statement.
**Hint 2.1**
Let $S(u)$ be the set of all nodes reachable by u using at most one route. Consider all $S(l)$ where $l$ is a leaf in the tree.
**Solution**
First, notice t... | The key observation is that you can reduce the problem to considering only leaves. Think about the one-route reachable set for each leaf and a necessary intersection property. This points towards a covering problem on the tree where you need to find a common node. | {
"exploration": "Okay so we have a tree of n cities and m bus routes. Each bus route between x and y covers the entire simple path between x and y — any city on that path can be used as a boarding/alighting point. We need to check if every pair of cities can be connected using at most two bus routes. The constraints... | {
"exploration": "I'll try to greedily merge intervals along the tree. Represent each route as the path between x and y. For each route, I'll mark all edges on its path as \"covered\" by that route. Then, I'll try to see if any two routes can be merged to cover the whole tree, or if a single route already covers ever... | {
"exploration": "I'll build a bipartite graph: left side = original n cities, right side = m bus routes (hyperedges). For each route (x,y), I add edges from every city on the simple path between x and y to the route node. Then the distance between any two cities in this bipartite graph is twice the number of bus rou... | null | null | null | So none of those panned out. Brute-force checking all city pairs was astronomically too slow, greedy merging missed the per-pair condition, and expanding each route to all its nodes blew up the memory and time.
hmm — that nagging thread about leaves. If the condition fails, there must be some pair of cities that needs... | function solveTestCase():
READ n, m
adj = array of size n+1 of empty lists
for i from 1 to n-1:
READ u, v
adj[u].append(v)
adj[v].append(u)
routes = empty list of pairs
for i from 1 to m:
READ x, y
routes.append((x, y))
// Step 1: find all leaves
lea... |
2002/G | Lattice Optimizing | 3,400 | bitmasks,brute force,hashing,meet-in-the-middle | null | null | Consider a grid graph with $$$n$$$ rows and $$$n$$$ columns. Let the cell in row $$$x$$$ and column $$$y$$$ be $$$(x,y)$$$. There exists a directed edge from $$$(x,y)$$$ to $$$(x+1,y)$$$, with non-negative integer value $$$d_{x,y}$$$, for all $$$1\le x \lt n, 1\le y \le n$$$, and there also exists a directed edge fro... | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1\le t\le100$$$). The description of the test cases follows.
The first line of each test case contains a single integer $$$n$$$ ($$$2\le n\le20$$$) — the number of rows and columns.
Each of the next $$$n-1$$$ lines co... | For each test case, print a single integer — the maximum MEX of $$$S$$$ when you reach $$$(n,n)$$$. | null | null | [] | The problem hints strongly at meet in the middle, at each side, there will be $O(2^n)$ paths. The twist is on merging: we are given two sequence of sets $S_1,S_2,\cdots$ and $T_1,T_2,\cdots$, we have to find the maximum $\textrm{MEX}(S_i\cup T_j)$.
If we enumerate over the maximum MEX $P$, we only need to check whethe... | The key observation is that the path length is fixed and the edge values are bounded by 2n-2. This suggests a meet-in-the-middle approach where you split the path into two halves. The challenge is to efficiently combine the two halves to maximize the MEX, which requires a clever representation of the collected sets to ... | {
"exploration": "Okay so we have an n×n grid, edges going right and down with non-negative integer values up to 2n-2. We start at (1,1) with an empty set S, collect edge values as we traverse, and want to maximize the MEX of S when we reach (n,n). The constraint says n ≤ 20, and sum of n³ over test cases ≤ 8000. Tha... | {
"exploration": "I'll maintain a set S, initially empty, and a current MEX candidate m = 0. At each cell (i,j) with i+j < 2n, I have two choices: go down (add d_{i,j} to S) or go right (add r_{i,j} to S). For each choice, I compute the new MEX of S ∪ {edge value}. I pick the direction that increases MEX the most. If... | {
"exploration": "I'll model state as (i, j, missing_mask) where missing_mask is a bitmask of which numbers in 0..K are still missing from S. Start at (1,1) with all bits 0..K set to 1 (missing). When traversing an edge with value v ≤ K, clear bit v. Goal: reach (n,n) with missing_mask = 0 for largest possible K. BFS... | {
"exploration": "The idea is to test each candidate MEX t from 0 upward. For a fixed t, we need a path from (1,1) to (n,n) that collects all values 0,1,...,t-1. Build a flow network where each directed edge in the grid becomes a node, and we add a source connected to all edges leaving (1,1), a sink from all edges en... | null | null | So that's a few failed attempts now. The full DP over masks is astronomically large, the greedy approach is fundamentally wrong, BFS over missing numbers is still exponential, and the min-cut reduction doesn't capture the MEX structure.
hmm — but the path length is fixed at 2n-2 steps, and the edge values are bounded ... | function solve():
READ t
for each test case:
READ n
// read matrices D and R (1-indexed)
for i from 1 to n-1:
for j from 1 to n:
READ D[i][j]
for i from 1 to n:
for j from 1 to n-1:
READ R[i][j]
// meet-in-the-middl... |
804/F | Fake bullions | 3,400 | combinatorics, dfs and similar, dp, graphs, number theory | 3 | 512 | In Isart people don't die. There are n gangs of criminals. The i-th gang contains si evil people numerated from 0 to si - 1. Some of these people took part in a big mine robbery and picked one gold bullion each (these people are given in the input). That happened 10100 years ago and then all of the gangs escaped to a r... | The first line contains four integers n, a and b (1 ≤ b ≤ a ≤ n ≤ 5·103) — the number of gangs, the constants a and b from the statement.
Then n lines follow, each line contains a string of size n consisting of zeros and ones. The j-th character in the i-th of these lines is equal to 1, then the vertex i have a direct... | Print single integer: the number of different sets of b gangs the police can arrest modulo 109 + 7. | null | null | [
{
"input": "2 2 1\n01\n00\n5 11000\n6 100000",
"output": "2"
},
{
"input": "5 2 1\n00000\n10000\n11011\n11000\n11010\n2 00\n1 1\n6 100110\n1 0\n1 0",
"output": "5"
}
] | As the hints said this problem have two part:
First we want to solve the first part that is evaluating the mni and mxi for gangs. Obviously mni will be the number of thieves have a gold in i-th gang.
lemma1:Consider there is just a single edge from v to u; as hints said if one thief in v with index i has a gold it copi... | The key observation is that the propagation of bullions between gangs depends only on the congruence classes modulo the gcd of their sizes. This allows you to compress the strongly connected components of the tournament into a single gang, transforming the problem into a DAG where you can compute reachable bullions gre... | {
"exploration": "I'll simulate the hour-by-hour propagation directly. Each gang v has s_v people, and at hour i, person (i mod s_u) of gang u can send a bullion to person (i mod s_v) of gang v if the edge u→v exists, the sender has a bullion, and the receiver doesn't. I need to track which people have bullions at ea... | {
"exploration": "I'll assign a random 64-bit hash to each initial bullion holder. For each gang u, I maintain a bitmask or set of hashes for each index j in [0, s_u-1], initially just the hash of the real bullion if that person has one, else empty. Then I iterate rounds: for each directed edge u→v in the tournament,... | null | null | null | null | Okay, so I've burned through a few ideas and none stuck. Direct simulation is impossible over astronomical timescales, and any naive fixpoint iteration is still far too heavy, blowing up to O(n² * sum(s_i)) which is hopeless.
hmm — but that half-formed thought about congruence classes... wait — the sending rule: at ho... | function compute_powers(n, a, b, adj, s, initial_strings):
// Step 1: Build initial bullion sets and sizes
bullion = array of bitsets for i from 1 to n
for i from 1 to n:
bullion[i] = bitset of size s[i], initialized from initial_strings[i]
// 1 in string means fake bullion (initially presen... |
1097/E | Egor and an RPG game | 3,400 | constructive algorithms, greedy | 2 | 256 | One Saturday afternoon Egor was playing his favorite RPG game. While discovering new lands and territories, he came across the following sign:
Egor is a passionate player, but he is an algorithmician as well. That's why he instantly spotted four common letters in two words on the sign above — if we permute the letters... | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases.
You can only use $$$t = 1$$$ in hacks.
Next, descriptions of $$$t$$$ test cases come, each of them in the following format.
The first line of a single test case contains one integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) —... | For each test case print the answer in the following format:
In the first line print $$$k$$$ ($$$1 \leq k \leq f(n)$$$) — the number of the subsequences in the partition. In the next $$$k$$$ lines, print the descriptions of found subsequences. Each description should start with a number $$$l_i$$$ ($$$1 \leq l_i \leq n... | null | In the example, we can split:
- $$$[4, 3, 1, 2]$$$ into $$$[4, 3, 1]$$$, $$$[2]$$$
- $$$[4, 5, 6, 1, 3, 2]$$$ into $$$[4, 1]$$$, $$$[5, 6]$$$ and $$$[3, 2]$$$
- $$$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$$$ into $$$[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]$$$
Surely, there are many more answers possible. | [
{
"input": "3\n4\n4 3 1 2\n6\n4 5 6 1 3 2\n10\n1 2 3 4 5 6 7 8 9 10",
"output": "2\n3 4 3 1\n1 2\n3\n2 4 1\n2 5 6\n2 3 2\n1\n10 1 2 3 4 5 6 7 8 9 10"
}
] | Firstly, consider this testcase: $\{1, 3, 2, 6, 5, 4, 10, 9, 8, 7, 15, 14, 13, 12, 11\}$.
It's easy to see that here we need at least $5$ sequences. It's possible to generalize it and see that if $n \geq \frac{k \cdot (k+1)}{2}$ for some integer $k$, then we need at least $k$ sequences. So, maybe if $n < \frac{k \cdot ... | The key observation is that the bound is related to triangular numbers. The constructive approach alternates between two phases: one where you greedily remove a long increasing subsequence, and another where you rely on a classic decomposition theorem to cover the rest with decreasing sequences. | {
"exploration": "Okay so we have a permutation of length n, and we need to partition it into monotonic subsequences (each either increasing or decreasing). The key constraint is that the number of subsequences must be at most f(n), where f(n) is the minimum integer k such that every permutation of length n can be pa... | {
"exploration": "I'll compute the longest alternating subsequence (LAS) for the first example permutation [4,3,1,2] and extract it, then decompose into monotonic runs.\n\nFirst, what does \"alternating\" mean here? Consecutive elements alternate between increasing and decreasing. So the LAS of [4,3,1,2]: start at 4.... | null | null | null | null | So both attempts failed — greedy alternation forced premature decreasing chains that blocked later increasing ones, and trying to decompose alternating runs directly gave too many subsequences. The core problem is that blindly alternating monotonicity doesn't respect the global structure of the permutation.
Wait — tha... | function f(n):
// smallest k such that k*(k+1)/2 >= n
k = 0
while k*(k+1)//2 < n:
k = k + 1
return k
function patience_sort_assignments(arr, n):
// returns (LIS_length, assignments) where assignments[i] = pile index for arr[i]
piles = empty list of lists // each pile stores indices of ... |
1119/F | Niyaz and Small Degrees | 3,400 | data structures, dp, trees | 3 | 256 | Niyaz has a tree with $$$n$$$ vertices numerated from $$$1$$$ to $$$n$$$. A tree is a connected graph without cycles.
Each edge in this tree has strictly positive integer weight. A degree of a vertex is the number of edges adjacent to this vertex.
Niyaz does not like when vertices in the tree have too large degrees. ... | The first line contains a single integer $$$n$$$ ($$$2 \le n \le 250\,000$$$) — the number of vertices in Niyaz's tree.
Each of the next $$$(n - 1)$$$ lines contains three integers $$$a$$$, $$$b$$$, $$$c$$$ ($$$1 \le a, b \le n$$$, $$$1 \leq c \leq 10^6$$$) — the indices of the vertices connected by this edge and its ... | Print $$$n$$$ integers: for each $$$x = 0, 1, \ldots, (n-1)$$$ print the smallest total weight of such a set of edges that after one deletes the edges from the set, the degrees of all vertices become less than or equal to $$$x$$$. | null | In the first example, the vertex $$$1$$$ is connected with all other vertices. So for each $$$x$$$ you should delete the $$$(4-x)$$$ lightest edges outgoing from vertex $$$1$$$, so the answers are $$$1+2+3+4$$$, $$$1+2+3$$$, $$$1+2$$$, $$$1$$$ and $$$0$$$.
In the second example, for $$$x=0$$$ you need to delete all th... | [
{
"input": "5\n1 2 1\n1 3 2\n1 4 3\n1 5 4",
"output": "10 6 3 1 0"
},
{
"input": "5\n1 2 1\n2 3 2\n3 4 5\n4 5 14",
"output": "22 6 0 0 0"
}
] | First, let's learn how to solve for a fixed $x$ in $O(n \log n)$.
We can calculate the dp on subtrees, $dp_{v, flag}$, denoting the minimum total weight of edges to be removed, if we consider the subtree of the vertex $v$, and $flag$ is zero, if the vertex has the degree $\leq x$, and one, if $\leq x+1$.
Then to recalc... | The key observation is that the total work across all x can be bounded because the sum of degrees is O(n). This suggests focusing only on vertices whose current degree exceeds the limit. For each such vertex, you need to decide which incident edges to cut, balancing the cost of cutting an edge now versus the future ben... | {
"exploration": "Okay so we have a tree with n vertices, edges have positive integer weights. For each x from 0 to n-1, we need the minimum total weight of edges to delete so that every vertex has degree ≤ x. The brute-force approach enumerates all 2^(n-1) subsets of edges, checks degrees, and tracks the minimum wei... | {
"exploration": "I'll build a flow network for each x from 0 to n-1. For a fixed x, create a source S and sink T. For each vertex v with original degree d_v > x, add an edge from S to v with capacity d_v - x. For each original tree edge (u,v) with weight w, add two directed edges u→v and v→u, each with capacity w. T... | {
"exploration": "Root the tree at vertex 1. For a fixed x, define dp[v][k] as the minimum total weight of edges deleted in the subtree of v such that v ends with degree at most x, and exactly k edges from v to its children are kept (so deg(v) = k + (v has parent? 1 : 0) ≤ x). For each child u of v with edge weight w... | null | null | null | Alright, back to square one. Brute force is astronomically impossible, min-cut can't express the constraint, and independent DP for each x is quadratic and explodes. The common thread is that each approach tried to solve the problem from scratch for every x, ignoring that the solution for x is deeply related to the sol... | READ n
adj = array of size n+1, each element is an empty list of (neighbor, weight)
degree = array of size n+1, initialized to 0
for i from 1 to n-1:
READ a, b, c
adj[a].append((b, c))
adj[b].append((a, c))
degree[a] += 1
degree[b] += 1
// Preprocess vertices sorted by degree
vertices_by_degree = ... |
1863/I | Redundant Routes | 3,500 | constructive algorithms, dp, trees | 5 | 1,024 | You are given a tree with $$$n$$$ vertices labeled $$$1, 2, \ldots, n$$$. The length of a simple path in the tree is the number of vertices in it.
You are to select a set of simple paths of length at least $$$2$$$ each, but you cannot simultaneously select two distinct paths contained one in another. Find the largest ... | The first line contains a single integer $$$n$$$ ($$$2 \le n \le 3000$$$).
The $$$i$$$-th of the following $$$n - 1$$$ lines contains two integers $$$u_i$$$ and $$$v_i$$$ ($$$1 \le u_i, v_i \le n$$$, $$$u_i \neq v_i$$$) — the numbers of vertices connected by the $$$i$$$-th edge.
It is guaranteed that the given edges ... | Print a single integer — the answer to the problem. | null | In the first example, possible efficient timetables are $$$\{\{1, 2\}, \{1, 3\}, \{1, 4\}\}$$$ and $$$\{\{1, 2, 3\}, \{1, 2, 4\}, \{1, 3, 4\}\}$$$.
In the second example, we can choose $$$\{ \{1, 2, 3\}, \{2, 3, 4\}, \{3, 4, 6\}, \{2, 3, 5\}, \{3, 4, 5\}, \{3, 4, 7\}, \{4, 6, 7\}\}$$$. | [
{
"input": "4\n1 2\n1 3\n1 4",
"output": "3"
},
{
"input": "7\n2 1\n3 2\n4 3\n5 3\n6 4\n7 4",
"output": "7"
}
] | Let's call two paths adjacent if one can be obtained from the other by adding an edge to one end and removing an edge from the other end.
Proposition. There exists an optimal solution in which each pair of adjacent paths is either simultaneously chosen or simultaneously not chosen.
Proof. Consider an optimal solution t... | The key observation is that paths can be grouped into equivalence classes based on the 'obstacle' encountered when trying to extend them towards the root. These classes naturally form a forest structure. The selection problem on this forest becomes tractable via a greedy decomposition focusing on minimal-length classes... | {
"exploration": "Okay so we have a tree with n up to 3000, and we need to select a maximum set of simple paths (each length ≥ 2) such that no path is a proper subset of another. The constraint n ≤ 3000 suggests O(n²) or O(n² log n) might be okay — 3000² = 9×10⁶, well within 5 seconds. The DP hypothesis uses state at... | {
"exploration": "I'll generate all simple paths of length at least 2, build the containment poset, then run Dilworth via maximum matching on a bipartite graph. First, enumerate paths: for each vertex u, run DFS to depth at least 2, collecting all vertex sets along simple paths starting at u. To avoid duplicates, I'l... | null | null | null | null | Hmm, all those ideas fell apart. The edge DP failed because child subtrees aren't independent — a path extending upward from a child can nest around internal paths I already counted there, so the child's state needs to know where its endpoints are, not just a total. And the Dilworth approach blew up because the contain... | function solve():
READ n
adj = array of size n+1 of empty lists
for i from 1 to n-1:
READ u, v
adj[u].append(v)
adj[v].append(u)
// Step 1: compute depths, parents, tin, tout via DFS from root 1
depth = array of size n+1
parent = array of size n+1
tin = array of size... |
1205/F | Beauty of a Permutation | 3,400 | constructive algorithms, math | 2 | 256 | Define the beauty of a permutation of numbers from $$$1$$$ to $$$n$$$ $$$(p_1, p_2, \dots, p_n)$$$ as number of pairs $$$(L, R)$$$ such that $$$1 \le L \le R \le n$$$ and numbers $$$p_L, p_{L+1}, \dots, p_R$$$ are consecutive $$$R-L+1$$$ numbers in some order. For example, the beauty of the permutation $$$(1, 2, 5, 3, ... | The first line contains a single integer $$$q$$$ ($$$1\le q \le 10\,000$$$) — the number of queries.
Follow $$$q$$$ lines. Each line contains two integers $$$n$$$, $$$k$$$ ($$$1 \le n \le 100$$$, $$$1 \le k \le \frac{n(n+1)}{2}$$$) — the length of permutation and needed beauty respectively. | For a query output "NO", if such a permutation doesn't exist. Otherwise, output "YES", and in the next line output $$$n$$$ numbers — elements of permutation in the right order. | null | Let's look at the first example.
The first query: in $$$(1)$$$ there is only one segment consisting of consecutive numbers — the entire permutation.
The second query: in $$$(2, 4, 1, 5, 3)$$$ there are $$$6$$$ such segments: $$$[2]$$$, $$$[4]$$$, $$$[1]$$$, $$$[5]$$$, $$$[3]$$$, $$$[2, 4, 1, 5, 3]$$$.
There is no su... | [
{
"input": "4\n1 1\n5 6\n5 8\n5 10",
"output": "YES\n1 \nYES\n2 4 1 5 3 \nNO\nYES\n2 3 1 4 5"
},
{
"input": "2\n4 10\n100 1",
"output": "YES\n1 2 3 4 \nNO"
}
] | We will denote $ (a, b) $ if there exists a permutation of length $ a $ with beauty equal to $ b $. To begin with, it is obvious that the beauty of a permutation of length $ a $ is at least $ a + 1 $ for $ a> 1 $: indeed, you can take each element individually and the entire permutation completely.
** Statement 1 **: i... | What if you think of building permutations by merging smaller permutations with known beauties? The key observation is that the beauty of a permutation behaves like a recursive additive quantity under a specific merging operation. This suggests a dynamic programming approach over segment lengths and beauty values, wher... | {
"exploration": "Okay so we have permutations of 1..n, and beauty counts subarrays whose elements form a consecutive set of integers. The constraint is n ≤ 100, q ≤ 10,000, and k ≤ n(n+1)/2. We need to find if a permutation with beauty k exists and output one.\n\nThe hypothesis is to brute-force all n! permutations,... | {
"exploration": "I'll maintain dp[i][b] = a permutation of length i with beauty b, if one exists. Start with dp[1][1] = [1]. For i from 2 to n, for each existing state dp[i-1][b] with permutation P of length i-1, try inserting i at every possible position (0 to i-1, meaning before P[0], between elements, after P[i-2... | {
"exploration": "I'll try dynamic programming on segment partitions. The idea: a permutation's beauty counts segments where the values form a consecutive set. A segment [L,R] is \"good\" if max(p[L..R]) - min(p[L..R]) = R-L. I can think of building the permutation by partitioning [1..n] into blocks, where each block... | {
"exploration": "I'll start BFS from the identity permutation [1,2,...,n]. Each state is a permutation, and the operation is swapping any two adjacent elements. I'll compute the beauty of each new permutation exactly using the definition: count all subarrays where the elements form a consecutive set of integers. I'l... | null | null | Okay, all of those went nowhere. Brute force is impossible, greedy loses structure, segment trees cap beauty too low, and BFS explores an astronomical space. The common failure is trying to handle the full permutation at once — we need a way to break it into pieces whose beauties we can control and combine.
hmm. That ... | function precompute():
maxN = 100
// dp[n][k] = whether beauty k achievable for length n
dp = 2D boolean array of size (maxN+1) x (maxN*(maxN+1)/2 + 1), initialized false
// split[n][k] stores (a,b,x,y) used to achieve dp[n][k]
split = 2D array of size (maxN+1) x (maxN*(maxN+1)/2 + 1), each entry is... |
757/G | Can Bash Save the Day? | 3,400 | data structures, divide and conquer, graphs, trees | 5 | 768 | Whoa! You did a great job helping Team Rocket who managed to capture all the Pokemons sent by Bash. Meowth, part of Team Rocket, having already mastered the human language, now wants to become a master in programming as well. He agrees to free the Pokemons if Bash can answer his questions.
Initially, Meowth gives Bash... | The first line contains two integers n and q (1 ≤ n ≤ 2·105, 1 ≤ q ≤ 2·105) — the number of nodes in the tree and the number of queries, respectively.
The next line contains n space-separated integers — the sequence a1, a2, ..., an which is a permutation of 1, 2, ..., n.
Each of the next n - 1 lines contain three spa... | For each query of type 1, output a single integer in a separate line, denoting the answer to the query. | null | In the sample, the actual queries are the following:
- 1 1 5 4
- 1 1 3 3
- 2 3
- 2 2
- 1 1 3 3 | [
{
"input": "5 5\n4 5 1 3 2\n4 2 4\n1 3 9\n4 1 4\n4 5 2\n1\n1 5 4\n1\n22 20 20\n2\n38\n2\n39\n1\n36 38 38",
"output": "23\n37\n28"
}
] | We binarize the given tree T such that degree of each node in the transformed tree T' is <= 3. Then perform centroid decomposition of this transformed tree T'. The centroid tree formed would have the following properties.
1. The height of the centroid tree is O(logN)
2. Each node in the centroid tree has ≤ 3 children. | The key observation is that you can restructure the tree to have bounded degree and then apply a divide-and-conquer technique that decomposes the problem into independent contributions from each centroid. This allows you to maintain dynamic range sums efficiently. | {
"exploration": "Okay so we have a weighted tree with n nodes, a permutation a[1..n] of 1..n, and q queries. Type 1 asks for sum of distances from v to all a[i] for i in [l, r]. Type 2 swaps a[x] and a[x+1]. The answers are used to decode subsequent queries via XOR with previous answer mod 2^30. Constraints: n ≤ 2·1... | {
"exploration": "I'll set up Mo's algorithm with updates. Block size B = n^(2/3) ≈ (2×10^5)^(2/3) ≈ 3400. I'll sort queries offline: first by L/B, then by R/B, then by time (update index). Maintain a frequency array freq[node] = how many times node appears in current [L,R] range. I need a data structure that support... | null | null | null | null | Alright, back to square one. Trying to store all distances directly blew up memory, and dynamic Mo's blew up time — both because they treated the tree as a black box, ignoring its structure entirely. The tree's structure must be the key to compressing the distance information.
hmm — wait. The distance between any two ... | function main():
READ n, q
READ permutation a[1..n]
// Build tree adjacency
adj = array of size n+1 of lists of (node, weight)
for i from 1 to n-1:
READ u, v, w
adj[u].append((v, w))
adj[v].append((u, w))
// Preprocess LCA for distance queries on original tree
depth... |
1515/I | Phoenix and Diamonds | 3,400 | binary search, data structures, sortings | 5 | 1,024 | Phoenix wonders what it is like to rob diamonds from a jewelry store!
There are $$$n$$$ types of diamonds. The $$$i$$$-th type has weight $$$w_i$$$ and value $$$v_i$$$. The store initially has $$$a_i$$$ diamonds of the $$$i$$$-th type.
Each day, for $$$q$$$ days, one of the following will happen:
1. A new shipment o... | The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n \le 2 \cdot 10^5$$$; $$$1 \le q \le 10^5$$$) — the number of types of diamonds and number of days, respectively.
The next $$$n$$$ lines describe each type of diamond. The $$$i$$$-th line will contain three integers $$$a_i$$$, $$$w_i$$$, and $$$v_i$$$... | Print the answer for each query of the third type ($$$t=3$$$). | null | For the first query where $$$t=3$$$, Phoenix can fit $$$2$$$ diamonds of type $$$1$$$, with total weight $$$6$$$ and value $$$8$$$.
For the second query where $$$t=3$$$, Phoenix will first fit in $$$3$$$ diamonds of type $$$3$$$, then one diamond of type $$$1$$$ for a total weight of $$$9$$$ and a value of $$$16$$$. N... | [
{
"input": "3 5\n2 3 4\n1 5 1\n0 2 4\n3 6\n1 3 3\n3 10\n2 2 3\n3 30",
"output": "8\n16\n13"
}
] | Suppose the largest weight of an item is less than $2^k$. Call an item heavy if its weight is in the range $[2^{k-1},2^k)$ and light if its weight is in the range $(0,2^{k-1})$. Sort the items in decreasing order by value. As the thief moves left to right, his remaining capacity is nonincreasing. Consider the point whe... | The key observation is that the greedy process has a hierarchical structure based on powers of two. Think about maintaining separate data structures for items whose weights fall into different bit ranges. The recursion on capacity after removing heavy items leads to a solution that can be updated efficiently. | {
"exploration": "Okay so we have n types of diamonds, each with weight w_i, value v_i, and initial count a_i. Over q days we get updates (add or remove diamonds of a type) and queries: given capacity c, greedily take the highest-value diamond that fits, breaking ties by minimum weight, and among those any. The const... | {
"exploration": "I'll build a DP table over compressed capacities. The idea: each diamond type has weight w_i, value v_i, and count a_i. I'll treat each type as a bounded knapsack item, using binary lifting to decompose the count into O(log a_i) items, each representing a power-of-two bundle of that diamond type. Th... | {
"exploration": "I'll sort all diamond types by value descending, and for ties by weight ascending. Then build a segment tree over these sorted types. Each leaf corresponds to one type and stores a list of (weight, value) pairs for all available diamonds of that type — essentially a multiset of individual diamonds. ... | {
"exploration": "I'll sort all diamond types by value descending, and for equal value, by weight ascending — that's the greedy order. For each type i, I maintain a BIT-like structure keyed by rank in this sorted order, where each position stores a list of (weight, count) for that type. When querying with capacity c,... | null | null | So none of those panned out. Simulating the greedy process per query is impossibly slow, rebuilding DP on updates is too heavy, and the segment tree approach lost the per-type granularity needed for partial contributions — the binary search couldn't pinpoint where the capacity cut fell inside a type.
hmm — but that la... | READ n, q
types = array of size n
for i from 0 to n-1:
READ a_i, w_i, v_i
types[i] = (a: a_i, w: w_i, v: v_i, id: i)
// Sort types by decreasing v, then increasing w
sorted_indices = indices 0..n-1 sorted by:
primary: -types[i].v
secondary: types[i].w
// Map from original type id to position in sorted... |
1482/H | Exam | 3,400 | data structures, string suffix structures, trees | 2 | 512 | This year a Chunin Selection Exam is held again in Konoha, and taking part in it are $$$n$$$ ninjas named $$$s_1$$$, $$$s_2$$$, ..., $$$s_n$$$. All names are distinct. One of the exam stages consists of fights between the participants. This year the rules determining the ninjas for each fight are the following: ninjas ... | The first line consists of the only integer $$$n$$$ ($$$1 \leq n \leq 10^{6}$$$) standing for the number of examinees.
The following $$$n$$$ lines contain their names. No two names coincide, all names are non-empty and consist of lowercase English letters. The total length of all names doesn't exceed $$$10^6$$$. | Print the only integer standing for the number of fights. | null | In the first example hidan fights against dan, and hanabi fights against nabi, who also fights bi. Ninjas named hanabi and bi don't fight each other since there is the ninja called nabi who breaks the third condition for them.
In the second example the fights are held between abacaba and acaba, abacaba and abaca, acab... | [
{
"input": "5\nhidan\ndan\nhanabi\nbi\nnabi",
"output": "3"
},
{
"input": "4\nabacaba\nabaca\nacaba\naca",
"output": "4"
}
] | Fix a particular string $s$ and find all edges outgoing from it. For each position $i$ of the string let's find the value of $left_i$ being the starting position of the longest substring which ends at the position $i$ and is one of the strings $s_j$. It's easy to see that such strings are the only ones where the outgoi... | What if you think about each string as a text and all names as patterns? The key observation is that for a given string, the candidate outgoing edges are certain longest substrings at each position, but you need to exclude those that are 'covered' by another candidate. This reduces to efficiently marking and querying p... | {
"exploration": "Okay so we have n ninjas with distinct names, total length of all names ≤ 10^6. The fight condition: i and j fight if s_j is a substring of s_i, and there is no intermediate k such that s_j is a substring of s_k and s_k is a substring of s_i — meaning s_j is a \"direct\" substring of s_i with no oth... | {
"exploration": "For each string s_i, I'll build a suffix automaton. The automaton naturally represents all substrings of s_i as states. When I insert characters, each state corresponds to a set of end positions, and the suffix links form a tree of inclusions — if state A has a suffix link to state B, then the subst... | null | null | null | null | Okay so I've burned through a few ideas and none stuck. The brute-force substring enumeration is impossibly huge, and checking all pairs via automaton ancestors is still quadratic in the total length — both blow up on the constraints.
hmm. That half-formed thought about "longest substrings at each position" and a "tre... | function solve():
READ n
names = array of n strings
total_len = 0
for i from 0 to n-1:
READ names[i]
total_len += length(names[i])
// Build Aho-Corasick automaton
// trie nodes: each node has next[26], fail, is_end (index of name ending here, -1 if none)
// up[node] = neares... |
1693/E | Outermost Maximums | 3,400 | data structures, greedy | 2 | 256 | Yeri has an array of $$$n + 2$$$ non-negative integers : $$$a_0, a_1, ..., a_n, a_{n + 1}$$$.
We know that $$$a_0 = a_{n + 1} = 0$$$.
She wants to make all the elements of $$$a$$$ equal to zero in the minimum number of operations.
In one operation she can do one of the following:
- Choose the leftmost maximum eleme... | The first line contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$).
The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le n$$$). | Print a single integer — the minimum number of operations needed to make all elements of $$$a$$$ equal to zero. | null | In the first sample, you get $$$\langle 1, \underline{1}, 2, 4, 0, 2 \rangle$$$ by performing the first operation and $$$\langle 1, 4, 2, \underline{2}, 0, 2 \rangle$$$ by performing the second operation.
One way to achieve our goal is shown below. (The underlines show the last change.) $$$\langle 1, 4, 2, 4, 0, 2 \ra... | [
{
"input": "6\n1 4 2 4 0 2",
"output": "7"
},
{
"input": "5\n1 3 5 4 2",
"output": "9"
},
{
"input": "4\n0 0 0 0",
"output": "0"
}
] | Let's calculate for each element of the array, the minimum number of times it needs to change throughout the process.
Let's take a look at the first time the $i$-th element is changing. We know that $a_i$ is the maximum number and the elements smaller than $a_i$ haven't changed yet.
Denote the maximum element on its le... | What if you think about each element's journey to zero as a sequence of jumps to the nearest smaller value on its left or right? The key observation is that the optimal sequence for each element can be computed independently by a greedy rule, and the total cost is the sum of these independent journeys. This suggests a ... | {
"exploration": "Okay so we have an array of n+2 non-negative integers with a_0 = a_{n+1} = 0. We can repeatedly pick either the leftmost maximum element and set it to the max of the elements on its left, or the rightmost maximum and set it to the max of the elements on its right. Goal is minimum operations to make ... | {
"exploration": "I'll maintain a set of \"hill intervals\" — each interval is a contiguous segment [l, r] where all elements have the same height h, and the elements just outside (a[l-1] and a[r+1]) are strictly smaller than h. Initially I scan the array and whenever a[i] != a[i-1] I start a new interval, tracking i... | null | null | null | null | None of those got me anywhere useful. Simulating operations directly blew up to O(n²), and merging equal-height blocks missed the fact that each element still needs its own operation to actually become zero.
hmm. The thread about each element's journey to zero — what if I think about the array as a landscape of height... | READ n
READ a[1..n]
// Pad array with zeros at both ends for convenience
a0 = array of size n+2
a0[0] = 0
a0[n+1] = 0
for i from 1 to n:
a0[i] = a[i]
// Build next greater element on left and right
left = array of size n+2, initialized to 0
right = array of size n+2, initialized to n+1
stack = empty stack
for i ... |
1239/F | Swiper, no swiping! | 3,400 | graphs, implementation | 2 | 512 | In anticipation of new adventures Boots wanted to do a good deed. After discussion with the Map and Backpack, they decided to gift Dora a connected graph. After a long search, Boots chose $$$t$$$ graph's variants, which Dora might like. However fox Swiper wants to spoil his plan.
The Swiper knows, that Dora now is onl... | The first line contains a single integer $$$t$$$ ($$$1 \le t \le 100\,000$$$) — the number of graph variants.
The first line of each variant contains integers $$$n$$$, $$$m$$$ ($$$1 \le n \le 500\,000$$$, $$$0 \le m \le 500\,000$$$), the number of vertexes and edges in the graph.
Then $$$m$$$ lines follow, each conta... | For each variant:
- In case the answer exists, print "Yes" and then the answer itself.The first line should contain an integer $$$c$$$ ($$$1 < c < n$$$), the number of vertices the Crook can steal, without Dora noticing the loss. On the next line print $$$c$$$ distinct integers, the indices of the graph's vertices in ... | null | The picture below shows the third variant from the example test. The set of the vertices the Crook can steal is denoted with bold. | [
{
"input": "3\n3 3\n1 2\n2 3\n3 1\n\n6 6\n1 2\n1 3\n2 3\n2 5\n2 6\n2 4\n\n8 12\n1 2\n1 3\n2 3\n1 4\n4 5\n5 1\n3 6\n3 7\n3 8\n6 1\n7 1\n8 1",
"output": "No\nYes\n3\n4 5 6\nYes\n3\n6 7 8"
}
] | We will look only on set of vertex, that will stay after deleting.
Let node-$x$ — node with degree $mod\ 3$ equal $x$.
Let's consider some cases:
• Node-$0$ exists. It is the answer, except the case, when our graph consists of one node.
• Cycle on nodes-$2$ exists. So exists irreducible cycle on nodes-$2$. It is the an... | The key observation is that the condition on degree modulo 3 after deletion imposes a strong structural constraint on the graph. Think about classifying vertices by their degree modulo 3 and searching for small, well-structured subgraphs that can be removed as a set. The solution involves a case analysis based on these... | {
"exploration": "Okay so we have a connected graph with n vertices and m edges. The constraint is that sum of n and sum of m over all test cases are each at most 500,000. We need to find a non-empty proper subset S of vertices to steal such that after removing S and all incident edges, every remaining vertex has its... | {
"exploration": "I'll try dynamic programming on a tree decomposition. Since the graph is connected, I can root it arbitrarily, say at node 1, and treat it as a tree. For each vertex v, I'll compute a DP state representing possible (deg_S(v) mod 3) values that can be achieved for the subtree rooted at v, where S is ... | {
"exploration": "I'll set up a randomized local search with simulated annealing. The state is a partition of vertices into stolen set S and kept set K. For each kept vertex v, I compute deg_S(v) — the number of stolen neighbors. The deficit for v is the distance from deg_S(v) to the nearest multiple of 3: deficit(v)... | null | null | null | None of those got me anywhere useful. The greedy couldn't handle intermediate states, the DP blew up in complexity, and random search was hopelessly unreliable.
hmm. But that thread about degree modulo 3 — it's not just a classification. The condition says after deleting a set S, for every remaining vertex v, its degr... | function solve():
READ t
for test in 1..t:
READ n, m
adj = array of size n+1 of empty lists
deg = array of size n+1 initialized to 0
for i in 1..m:
READ a, b
adj[a].append(b)
adj[b].append(a)
deg[a] += 1
deg[b] += 1
... |
1936/E | Yet Yet Another Permutation Problem | 3,400 | divide and conquer, fft, math | 5 | 1,024 | You are given a permutation $$$p$$$ of length $$$n$$$.
Please count the number of permutations $$$q$$$ of length $$$n$$$ which satisfy the following:
- for each $$$1 \le i < n$$$, $$$\max(q_1,\ldots,q_i) \neq \max(p_1,\ldots,p_i)$$$.
Since the answer may be large, output the answer modulo $$$998\,244\,353$$$. | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.
The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^5$$$).
The second line of each test case contains $$$n$$$ ... | For each test case, print a single integer — the answer modulo $$$998\,244\,353$$$. | null | In the first test case, $$$p = [2, 1]$$$. The only suitable $$$q$$$ is $$$[1, 2]$$$. Indeed, we need to satisfy the inequality $$$q_1 \neq p_1$$$. It only holds for $$$q = [1, 2]$$$.
In the second test case, $$$p = [1, 2, 3]$$$. So $$$q$$$ has to satisfy two inequalities: $$$q_1 \neq p_1$$$ and $$$\max(q_1, q_2) \neq ... | [
{
"input": "6\n2\n2 1\n3\n1 2 3\n3\n3 1 2\n4\n2 4 1 3\n5\n3 5 1 4 2\n15\n6 13 2 8 7 11 1 3 9 15 4 5 12 10 14",
"output": "1\n3\n2\n4\n18\n424488915"
}
] | We found it difficult to directly calculate all valid permutations. Consider calculating all invalid permutations and subtract it from $n!$.
Let's first make some notes. We note $P_i=max(p_1,...,p_i),Q_i=max(q_1,...,q_i)$,and $last_i=max(j)(P_j \neq P_i)$.And note the values of all extreme points is $val[1,...,m]$ and ... | The key observation is that the condition only depends on the positions where the prefix maximum of p increases. For each such 'extreme', you can think about placing the corresponding value in q to force a violation. The number of ways to choose a subset of extremes to violate, while respecting the ordering constraints... | {
"exploration": "Okay so we have a permutation p of length n, and we need to count permutations q of the same length such that for every prefix i from 1 to n-1, the maximum of q's first i elements is never equal to the maximum of p's first i elements. The constraint is sum of n over all test cases ≤ 2×10^5, time lim... | {
"exploration": "I'll define dp[i][m] as the number of valid permutations of length i where the maximum among the first i elements is exactly m. The condition we need to avoid is that for each prefix i, the maximum in q's prefix is not equal to the maximum in p's prefix. Let me precompute the prefix maxima of p: M[i... | {
"exploration": "I'll maintain a set S of unused numbers {1..n}, and for each position i from 1 to n, compute M_p = max(p_1..p_i), exclude M_p from S, then pick the smallest remaining element in S as q_i. If S becomes empty after exclusion, backtrack to the most recent position where an alternative choice existed an... | null | null | null | Okay, all of those went nowhere. Brute force is impossible, DP is quadratic, and greedy misses valid permutations because the choice at one position isn't locally constrained—it affects the entire future set of maxima.
hmm. The condition only cares about prefix maxima. For each i, we require max(q[1..i]) ≠ max(p[1..i]... | MOD = 998244353
function solve():
READ t
for each test case:
READ n
READ p[1..n]
// find record positions and values
records = []
current_max = 0
for i from 1 to n:
if p[i] > current_max:
current_max = p[i]
rec... |
1553/I | Stairs | 3,400 | combinatorics, divide and conquer, dp, fft, math | 10 | 1,024 | For a permutation $$$p$$$ of numbers $$$1$$$ through $$$n$$$, we define a stair array $$$a$$$ as follows: $$$a_i$$$ is length of the longest segment of permutation which contains position $$$i$$$ and is made of consecutive values in sorted order: $$$[x, x+1, \ldots, y-1, y]$$$ or $$$[y, y-1, \ldots, x+1, x]$$$ for some... | The first line of input contains integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the length of a stair array $$$a$$$.
The second line of input contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$). | Print the number of permutations which have stair array equal to $$$a$$$. Since the number can be big, compute it modulo $$$998\,244\,353$$$. | null | null | [
{
"input": "6\n3 3 3 1 1 1",
"output": "6"
},
{
"input": "7\n4 4 4 4 3 3 3",
"output": "6"
},
{
"input": "1\n1",
"output": "1"
},
{
"input": "8\n2 2 2 2 2 2 1 1",
"output": "370"
},
{
"input": "4\n3 2 3 1",
"output": "0"
}
] | First of all, we need to transform the stair array into some other structure. Let's show that for each number, there is only one longest stair covering it, and these longest stairs are disjoint.
Suppose that some number $x$ belongs to two different ascending stairs. We can easily expand these stairs to the left and to ... | The key observation is that the longest stairs are unique and partition the permutation. The count reduces to arranging these segments with constraints that prevent them from merging, which can be tackled with a combinatorial DP that tracks adjacency violations. The structure allows for efficient merging using convolut... | {
"exploration": "I'll try to directly enumerate all permutations of size n and check if they produce the given stair array a. For each permutation p, compute a_i as the longest segment containing i that is a consecutive monotonic block (either increasing or decreasing). Compare with the given a. Count matches.\n\nLe... | {
"exploration": "Let me partition the array into maximal blocks of equal a_i. For Example 1: a = [3,3,3,1,1,1]. Blocks: B1 = positions 1-3 with value 3, B2 = positions 4-6 with value 1. Each block corresponds to a stair — a contiguous segment of the permutation that is a consecutive run either ascending or descendin... | {
"exploration": "I'll build a segment tree over positions 1..n where each leaf stores the set of values that could legally go at that position. Initially, for each position i, the feasible set is all values v such that there exists a monotonic consecutive-value segment of length a_i covering i with v at position i. ... | null | null | null | Alright, back to square one. Backtracking is astronomically impossible, greedy can't handle global constraints, and a naive segment tree blows up in memory — all because I was trying to build the permutation directly instead of understanding the structure that the stair array imposes.
hmm. That half-formed thought — "... | MOD = 998244353
function solve():
READ n
READ array a[1..n]
// Build Cartesian tree of maximums (like segment tree for range maximum)
// stack will store indices of increasing a values
stack = empty stack
parent = array of size n+1, initialized to 0
left_child = array of size n+1, initiali... |
1830/E | Bully Sort | 3,500 | data structures, math | 10 | 1,024 | On a permutation $$$p$$$ of length $$$n$$$, we define a bully swap as follows:
- Let $$$i$$$ be the index of the largest element $$$p_i$$$ such that $$$p_i \neq i$$$.
- Let $$$j$$$ be the index of the smallest element $$$p_j$$$ such that $$$i < j$$$.
- Swap $$$p_i$$$ and $$$p_j$$$.
We define $$$f(p)$$$ as the number ... | The first line of the input contains two integers $$$n$$$ and $$$q$$$ ($$$2 \le n \le 5 \cdot 10^5$$$, $$$1 \le q \le 5 \cdot 10^4$$$) — the length of the permutation and the number of updates.
The second line of input contains $$$n$$$ integer $$$p_1,p_2,\ldots,p_n$$$ ($$$1 \leq p_i \leq n$$$) — the permutation $$$p$$... | After each update, output $$$f(p)$$$. | null | After the first update, we have $$$f(p)=5$$$. The $$$5$$$ bully swaps are illustrated below.
- $$$[\mathbf{1}, 2, \mathbf{8}, 5, 3, 4, 7, 6]$$$,
- $$$[1, 2, \mathbf{3}, 5, \mathbf{8}, 4, 7, 6]$$$,
- $$$[1, 2, 3, 5, \mathbf{4}, \mathbf{8}, 7, 6]$$$,
- $$$[1, 2, 3, 5, 4, \mathbf{6}, 7, \mathbf{8}]$$$,
- $$$[1, 2, 3, \ma... | [
{
"input": "8 5\n6 2 1 5 3 4 7 8\n1 8\n2 3\n4 7\n7 8\n3 6",
"output": "5\n6\n9\n8\n7"
}
] | First of all, we notice that if some element moves left, it will never move right.
**Proof**
Proving this is not hard, imagine $S$ to be the set of suffix minimas. Then if an element is in $S$ we know that $p_x \le x$.
Since after every bully swap an element cannot disappear from $S$ and after each bully swap, the $... | The key observation is that the process cleanly separates into two independent contributions: one depends on how far elements are from their target positions in a specific direction, and the other is a classic combinatorial measure of disorder. To compute the answer after swaps, you need to dynamically maintain both qu... | {
"exploration": "I'll maintain two max-heaps: one for elements where p_i ≠ i (candidates for the largest such i), and a min-heap for elements to the right of the current i. On each bully swap, I pop the largest i from the first heap, then find the smallest j > i from the second heap, swap them, update the heaps, and... | {
"exploration": "First, I need to understand the bully swap definition: find the largest element p_i such that p_i ≠ i, call its index i; then find the smallest element p_j with j > i; swap them. This is repeated until sorted. The hypothesis says to model as minimum adjacent swaps with a custom cost function — but t... | {
"exploration": "I'll maintain a BIT (Fenwick tree) where each node stores an ordered set of values. For each position i, I query how many elements to the left have value greater than p_i. Then I sum these counts across all positions and multiply by some scaling factor based on swap distances.\n\nLet me start by bui... | null | null | null | None of those got me anywhere useful. Simulating swaps directly is impossibly slow, and static measures like inversion counts don't match the process dynamics — the sum of left-greater counts was 8 while f(p)=4, and after updates the relationship changed non-monotonically.
hmm — but that nagging thread about two indep... | function main():
READ n, q
READ array p[1..n]
// Fenwick tree with ordered_set in each node
bit = struct with:
n = n
tree = array of ordered_set of size n+1 // each ordered_set stores values
function bit.update(pos, val, sign):
for i = pos to n step i & -i:
... |
1601/F | Two Sorts | 3,400 | binary search, dfs and similar, math, meet-in-the-middle | 4 | 512 | Integers from $$$1$$$ to $$$n$$$ (inclusive) were sorted lexicographically (considering integers as strings). As a result, array $$$a_1, a_2, \dots, a_n$$$ was obtained.
Calculate value of $$$(\sum_{i = 1}^n ((i - a_i) \mod 998244353)) \mod 10^9 + 7$$$.
$$$x \mod y$$$ here means the remainder after division $$$x$$$ b... | The first line contains the single integer $$$n$$$ ($$$1 \leq n \leq 10^{12}$$$). | Print one integer — the required sum. | null | A string $$$a$$$ is lexicographically smaller than a string $$$b$$$ if and only if one of the following holds:
- $$$a$$$ is a prefix of $$$b$$$, but $$$a \ne b$$$;
- in the first position where $$$a$$$ and $$$b$$$ differ, the string $$$a$$$ has a letter that appears earlier in the alphabet than the corresponding lette... | [
{
"input": "3",
"output": "0"
},
{
"input": "12",
"output": "994733045"
},
{
"input": "21",
"output": "978932159"
},
{
"input": "1000000000000",
"output": "289817887"
}
] | Suppose $b$ is an inverse permutation of $a$, that is $a_{b_{i}} = b_{a_i} = i$, that is $b_i$ is an index of $i$ in the lexicographical sorting. Rewrite desired sum replacing $i \to b_i$: $\sum_{i=1 \ldots n} ((i - a_i) \bmod p) = \sum_{i=1 \ldots n} ((b_i - a_{b_i}) \bmod p) = \sum_{i=1 \ldots n} ((b_i - i) \bmod p)$... | The key observation is that the inverse permutation of lexicographic order has a structured, piecewise-linear relationship with the original number. This structure allows you to group numbers by their common prefix and treat the varying suffix digits as independent variables in a linear expression. The technique involv... | {
"exploration": "Okay so we have integers 1 through n sorted lexicographically as strings, producing a permutation array a where a_i is the i-th number in that sorted order. The constraint is n ≤ 10^12. We need to compute sum_{i=1}^n ((i - a_i) mod 998244353) mod (10^9+7). The hypothesis is to directly generate all ... | {
"exploration": "I'll do a DFS over the implicit trie of decimal strings. For each node representing prefix p, I'll compute the range of integers sharing that prefix, then for each integer x in that range, compute its lexicographic rank by counting all smaller strings via a separate counting pass. The rank for x bec... | {
"exploration": "I'll try to compute the lexicographic order position of each integer i from 1 to n by building a DP over digit positions that counts how many numbers ≤ n have a given prefix. The idea: for each integer i, I can determine its rank a_i by traversing the lexicographic trie of numbers as strings, counti... | {
"exploration": "For each i from 1 to n, I'll binary search over the range [1, n] to find its lexicographic rank a_i. The comparator: given candidate x, check if x is lexicographically less than i by converting both to strings and comparing. To speed up, I'll precompute powers of 10 and extract digits with integer d... | null | null | Alright, back to square one. Every attempt to compute ranks individually or sort explicitly exploded because n is 10^12 — any per-number operation is impossible. The structure of lexicographic order isn't random, though. It's a depth-first traversal of a 10-ary tree (digits 1-9 for the first, 0-9 after). The rank of a ... | function count_numbers_with_prefix(prefix, n):
// returns number of integers x (1 ≤ x ≤ n) whose decimal representation starts with string of prefix
let pref_val = integer value of prefix string
let len_pref = length(prefix)
let len_n = length(str(n))
total = 0
for length from len_pref to len_n:... |
739/D | Recover a functional graph | 3,400 | graph matchings | 2 | 256 | Functional graph is a directed graph in which all vertices have outdegree equal to 1. Loops are allowed.
Some vertices of a functional graph lay on a cycle. From the others we can come to a cycle by making a finite number of steps along the edges (we consider only finite functional graphs in this problem).
Let's comp... | The first line contains single integer n (1 ≤ n ≤ 300) — the number of vertices in the graph.
Each of the next n lines contain two integers — precyclei (0 ≤ precyclei ≤ n - 1) and cyclei (1 ≤ cyclei ≤ n). There could be question marks instead of some of these values. | In case there is no solution, print -1.
Otherwise, print n integers. i-th of them is the number of vertex to which the edge form the i-th vertex go.
The vertices should be in the same order as they go in input data.
If there are multiple solutions, print any of them. | null | null | [
{
"input": "3\n0 3\n0 3\n? ?",
"output": "2 3 1"
},
{
"input": "5\n3 2\n? ?\n? ?\n? ?\n? ?",
"output": "5 3 2 2 4"
},
{
"input": "8\n? 3\n? ?\n0 2\n0 2\n0 3\n0 3\n0 3\n3 3",
"output": "5 1 4 3 6 7 5 2"
},
{
"input": "1\n? ?",
"output": "1"
},
{
"input": "6\n0 3\n0... | Let's think what has to hold after we put numbers in place of question marks:
1. number of vertices with precycle = 0 and cycle = y should be divisible by y.
2. if there exists a vertex with precycle = x > 0 and cycle = y, then there should also exist a vertex with precycle = x - 1 and cycle = y.
Now looking at vertice... | The key observation is that the constraints for the graph form local predecessor relationships and global divisibility requirements on cycle lengths. This suggests modeling it as a flow or matching problem, where you must satisfy a set of 'demand' nodes with a set of 'supply' vertices. | {
"exploration": "Okay so we have a functional graph — every vertex has outdegree 1. We're given precycle and cycle values for each vertex, some unknown. precycle is distance to a cycle vertex, cycle is the length of the cycle reached. We need to construct edges or output -1. n ≤ 300, time 2s, memory 256MB. That's sm... | {
"exploration": "I'll start by placing all vertices with known (0, y) into cycles of length y. For each remaining vertex with known (x, y) where x > 0, I'll build a chain of length x ending at a cycle vertex, all sharing the same y. Let me trace Example 1.\n\nExample 1: n=3, vertices:\n 1: (0, 3)\n 2: (0, 3)\n 3:... | {
"exploration": "I'll set up binary variables x_{i, (p,c)} for each vertex i and each possible (precycle, cycle) pair it could take. For known vertices, the variable is fixed to 1 for that pair. For unknown vertices with question marks, I create variables for all (p,c) pairs that appear in the input or could be infe... | null | null | null | Okay, so every approach hit a wall. Greedy assignment broke on divisibility, chain-building locked me into wrong predecessor choices, and the LP was too abstract to guarantee a concrete graph. The core problem is that I need to simultaneously assign each vertex a specific (precycle, cycle) role and wire up edges so tha... | function solve():
READ n
precycle = array of size n
cycle = array of size n
for i from 0 to n-1:
READ token1, token2
if token1 == '?':
precycle[i] = -1
else:
precycle[i] = integer(token1)
if token2 == '?':
cycle[i] = -1
else:
... |
1852/E | Rivalries | 3,400 | constructive algorithms, data structures, greedy | 2 | 512 | Ntarsis has an array $$$a$$$ of length $$$n$$$.
The power of a subarray $$$a_l \dots a_r$$$ ($$$1 \leq l \leq r \leq n$$$) is defined as:
- The largest value $$$x$$$ such that $$$a_l \dots a_r$$$ contains $$$x$$$ and neither $$$a_1 \dots a_{l-1}$$$ nor $$$a_{r+1} \dots a_n$$$ contains $$$x$$$.
- If no such $$$x$$$ ex... | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.
The first line of each test case has a single integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$).
The next line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a... | For each test case, output $$$n$$$ integers $$$b_1, b_2, \ldots, b_n$$$ — a valid rival to $$$a$$$ such that $$$b_1 + b_2 + \cdots + b_n$$$ is maximal.
If there exist multiple rivals with the maximum sum, output any of them. | null | For the first test case, one rival with the maximal sum is $$$[2, 4, 2, 3, 3]$$$.
$$$[2, 4, 2, 3, 3]$$$ can be shown to be a rival to $$$[1, 4, 1, 3, 3]$$$.
All possible subarrays of $$$a$$$ and $$$b$$$ and their corresponding powers are listed below:
- The power of $$$a[1:1] = [1] = 0$$$, the power of $$$b[1:1] = [... | [
{
"input": "7\n5\n1 4 1 3 3\n5\n1 4 1 8 8\n5\n2 1 1 1 2\n8\n3 2 3 5 2 2 5 3\n8\n1 1 1 1 4 3 3 3\n10\n1 9 5 9 8 1 5 8 9 1\n16\n1 1 1 1 5 5 5 5 9 9 9 9 7 7 7 7",
"output": "2 4 2 3 3\n3 4 3 8 8\n2 1 2 1 2\n4 2 4 5 5 2 5 4\n1 2 2 1 4 3 2 3\n7 9 5 9 8 9 5 8 9 7\n1 8 8 1 5 8 8 5 9 9 9 9 7 8 8 7"
}
] | **Hint 1**
For each distinct value, only the leftmost and rightmost positions actually have an effect on the power.
**Hint 2**
Think of each distinct value as an interval corresponding to its leftmost and rightmost positions, and let that distinct value be the value of its interval. If interval $x$ strictly contains... | What if you think of each distinct number as an interval from its first to last occurrence? The key observation is that the power condition only depends on these endpoints. The rival array can be built by assigning values based on which intervals cover each position, but you must be careful about intervals that are dom... | {
"exploration": "Okay so we have an array a of length n, and for every subarray we define its \"power\" as the largest value that appears in that subarray but nowhere outside it. We need to construct a rival array b with positive integers that has exactly the same power for every subarray, while maximizing the sum o... | {
"exploration": "Start by setting b_i = a_i for all i. Then I need to detect subarrays where the power of b differs from the power of a, and fix them by raising some b_j values upward.\n\nLet me first understand what \"power\" actually means. For a subarray a[l..r], the power is the largest value x such that x appea... | null | null | null | null | Right, so nothing worked so far. The flow approach blew up because it tried to model every subarray individually, and the iterative refinement couldn't even enumerate all the constraints.
Hmm. That half-thought about intervals... wait — the power of a subarray is the largest value x whose entire occurrence set within ... | function solve():
READ t
for each test case:
READ n
READ a[1..n]
// map value to list of positions
map = empty map from integer to list of integers
for i from 1 to n:
map[a[i]].append(i)
// collect distinct values in descending order
distinct... |
913/H | Don't Exceed | 3,400 | math, probabilities | 4 | 256 | You generate real numbers s1, s2, ..., sn as follows:
- s0 = 0;
- si = si - 1 + ti, where ti is a real number chosen independently uniformly at random between 0 and 1, inclusive.
You are given real numbers x1, x2, ..., xn. You are interested in the probability that si ≤ xi is true for all i simultaneously.
It can be... | The first line contains integer n (1 ≤ n ≤ 30).
The next n lines contain real numbers x1, x2, ..., xn, given with at most six digits after the decimal point (0 < xi ≤ n). | Print a single integer, the answer to the problem. | null | In the first example, the sought probability is 1 since the sum of i real numbers which don't exceed 1 doesn't exceed i.
In the second example, the probability is x1 itself.
In the third example, the sought probability is 3 / 8. | [
{
"input": "4\n1.00\n2\n3.000000\n4.0",
"output": "1"
},
{
"input": "1\n0.50216",
"output": "342677322"
},
{
"input": "2\n0.5\n1.0",
"output": "623902721"
},
{
"input": "6\n0.77\n1.234567\n2.1\n1.890\n2.9999\n3.77",
"output": "859831967"
}
] | If ti were random integers and not reals, it would be natural to solve the problem using dynamic programming: f(i, j) — the probability that the required inequalities are satisfied if si = j. But in case of real numbers, the probability that a random real number equals to something is zero.
What to do then? A natural s... | The key observation is that the distribution of each partial sum si, conditioned on staying under the barriers, is piecewise polynomial in x. You need to propagate this distribution layer by layer, integrating over the previous density within a sliding window of length 1. Think about representing and updating these pol... | {
"exploration": "Okay so we have n independent uniform [0,1] variables t_i, and s_i = sum_{j=1}^i t_j. We need the probability that s_i ≤ x_i for all i simultaneously. The x_i are given with up to 6 decimal places, n ≤ 30. This is the volume of a polytope in the n-dimensional unit cube [0,1]^n defined by the linear ... | {
"exploration": "I'll discretize the possible values of each s_i into buckets of width ε. Since each t_i is uniform [0,1], s_i ranges from 0 to n, so with n=30 I'll use something like ε = 10^{-6} or finer. Let me pick ε = 10^{-6} for now — that gives about n/ε = 30 × 10^6 = 3×10^7 buckets. Each bucket represents a s... | {
"exploration": "I'll work with the characteristic function approach. For each step i, the unconditional sum s_i has distribution given by the i-fold convolution of uniform [0,1] densities. The characteristic function of a single uniform [0,1] is φ(ω) = (e^{iω} - 1)/(iω). For the unconditional s_i, the characteristi... | null | null | null | Well, every path hit a wall — explicit integration blew up in dimension, discretization drowned in granularity, and Fourier methods collapsed under the need to truncate. The core problem is that the density of each partial sum, conditioned on staying under the barriers, is a function with bounded support that gets conv... | READ n
READ array x[1..n]
// breakpoints for each layer: sorted list of points where polynomial pieces change
// For layer i, breakpoints are of the form x_i - k for integer k, also 0 and x_i.
// We'll store pieces as intervals [L, R) with polynomial coefficients.
// Initialize f for i=0: single piece at [0,0] with p... |
1280/F | Intergalactic Sliding Puzzle | 3,400 | combinatorics, constructive algorithms, math | 1 | 256 | You are an intergalactic surgeon and you have an alien patient. For the purposes of this problem, we can and we will model this patient's body using a $$$2 \times (2k + 1)$$$ rectangular grid. The alien has $$$4k + 1$$$ distinct organs, numbered $$$1$$$ to $$$4k + 1$$$.
In healthy such aliens, the organs are arranged ... | The first line of input contains a single integer $$$t$$$ ($$$1 \le t \le 4$$$) denoting the number of test cases. The next lines contain descriptions of the test cases.
Each test case consists of three lines. The first line contains a single integer $$$k$$$ ($$$1 \le k \le 15$$$) which determines the size of the grid... | For each test case, first, print a single line containing either:
- SURGERY COMPLETE if it is possible to place back all internal organs in the correct locations;
- SURGERY FAILED if it is impossible.
If it is impossible, then this is the only line of output for the test case. However, if it is possible, output a few... | null | There are three shortcuts defined in the first sample output:
- R = SrS
- S = rr
- I = lldll
The sequence of moves is IR and it expands to:
- IR
- lldllR
- lldllSrS
- lldllrrrS
- lldllrrrrr | [
{
"input": "2\n3\n1 2 3 5 6 E 7\n8 9 10 4 11 12 13\n11\n34 45 6 22 16 43 38 44 5 4 41 14 7 29 28 19 9 18 42 8 17 33 1\nE 15 40 36 31 24 10 2 21 11 32 23 30 27 35 25 13 12 39 37 26 20 3",
"output": "SURGERY COMPLETE\nIR\nR SrS\nS rr\nI lldll\nDONE\nSURGERY FAILED"
}
] | The "shortcuts" thing in the output section is basically a way for you to define subroutines, i.e., you can create simpler (useful) operations, and then you can combine them into more complex operations.
Now, to solve the problem, we may represent the grid by the "circular permutation" obtained by going around the grid... | What if you think of the grid as a circular list of cells? The key observation is that the allowed moves correspond to specific permutations on this circle. The parity of the permutation relative to the target state determines feasibility, and any even permutation can be constructed from a fundamental 3-cycle operation... | {
"exploration": "I'll try placing organs one by one into their target positions using the empty space as a sliding buffer. For each organ from 1 to 4k+1, I'll navigate the empty space next to it, then slide it step by step toward its target cell, keeping already-placed organs locked in place.\n\nLet me trace this on... | {
"exploration": "Okay so we have a 2×(2k+1) grid with 4k+1 distinct organs and one empty cell. The empty cell can swap with left/right neighbors anywhere, but vertical swaps only in the leftmost, rightmost, or centermost column. We need to find a sequence of moves from the given configuration to the target configura... | {
"exploration": "First, I need to understand the macro building approach. The idea is to define shortcuts that perform specific permutations on the grid, like rotating a block of cells or moving the empty space along a fixed path, and then compose them to solve the whole puzzle. For k=3, the grid is 2×7. The target ... | null | null | null | Alright, every approach hit a wall. Greedy placement trapped the empty cell, BFS was astronomically huge, and macro building couldn't decide feasibility. The core problem is knowing whether the given scrambled configuration is even reachable under these restrictive moves.
Hmm. The moves are restrictive — vertical move... | function solve():
READ t
for each test case from 1 to t:
READ k
n = 2 * k + 1
total_cells = 4 * k + 2
// Read first row
row1 = list of size n
for i from 0 to n-1:
READ token
if token == 'E':
row1[i] = 0
empty... |
1299/E | So Mean | 3,400 | interactive, math | 4 | 256 | This problem is interactive.
We have hidden a permutation $$$p_1, p_2, \dots, p_n$$$ of numbers from $$$1$$$ to $$$n$$$ from you, where $$$n$$$ is even. You can try to guess it using the following queries:
$$$?$$$ $$$k$$$ $$$a_1$$$ $$$a_2$$$ $$$\dots$$$ $$$a_k$$$.
In response, you will learn if the average of elemen... | The first line contains a single integer $$$n$$$ ($$$2 \le n \le 800$$$, $$$n$$$ is even). | null | You begin the interaction by reading $$$n$$$.
To ask a question about elements on positions $$$a_1, a_2, \dots, a_k$$$, in a separate line output
$$$?$$$ $$$k$$$ $$$a_1$$$ $$$a_2$$$ ... $$$a_k$$$
Numbers in the query have to satisfy $$$1 \le a_i \le n$$$, and all $$$a_i$$$ have to be different. Don't forget to 'flus... | null | [
{
"input": "2\n1 2",
"output": "? 1 2\n? 1 1\n! 1 2"
}
] | Let's solve this problem in several steps.
1. First, let's ask a query about each group of $n-1$ numbers. Note that $1 + 2 + \dots + (i-1) + (i+1) + \dots + n = \frac{n(n+1)}{2} - i \equiv \frac{1\cdot 2}{2} - i \bmod (n-1)$. Therefore, answer will be YES only for numbers $1$ and $n$. Find the positions where $1$ and $... | What if you could identify the smallest and largest numbers by looking at the sum of almost all elements? The key observation is that the average being integer relates to the sum modulo the query size, and this lets you pinpoint extreme values. After that, you can treat the permutation as unknown residues modulo severa... | {
"exploration": "Okay so we have an interactive problem with a hidden permutation of 1..n, n even, n ≤ 800. We can query any subset of distinct indices and learn whether the average of their values is an integer — equivalently, whether their sum is divisible by k (the subset size). We get up to 18n queries. The perm... | {
"exploration": "I'll build a graph on n vertices, one per position. For each pair (i,j), I query {i,j} — that's a size-2 query. The response tells me whether (p_i + p_j) is even. If it's even, p_i and p_j have the same parity; if odd, opposite parity. I'll run this for all n(n-1)/2 pairs to get a complete parity gr... | {
"exploration": "I'll try to build a meet-in-the-middle engine that maintains, for each half of the positions, a set of possible (sum, size) pairs modulo various k. For each query of size k spanning both halves, I check whether there exists a combination from the two halves whose total sum is 0 mod k. Over multiple ... | null | null | null | Alright, back to square one. Binary search can't compare values, the parity graph explodes in queries, and meet-in-the-middle is astronomically impossible — all because I was trying to infer absolute values directly from a single, weak modular condition.
Hmm. But that modular condition is about the sum modulo k. If I ... | function solve():
READ n
// Step 1: Find positions of 1 and n using queries of size n-1
total_sum = n * (n + 1) // 2
candidates = empty list
for i from 1 to n:
query_set = [1, 2, ..., n] excluding i
OUTPUT ? (n-1) query_set
READ response
if response == 1:
... |
853/E | Lada Malina | 3,400 | data structures, geometry | 5 | 256 | After long-term research and lots of experiments leading Megapolian automobile manufacturer «AutoVoz» released a brand new car model named «Lada Malina». One of the most impressive features of «Lada Malina» is its highly efficient environment-friendly engines.
Consider car as a point in Oxy plane. Car is equipped with... | The first line of input contains three integers k, n, q (2 ≤ k ≤ 10, 1 ≤ n ≤ 105, 1 ≤ q ≤ 105), the number of engines of «Lada Malina», number of factories producing «Lada Malina» and number of options of an exposition time and location respectively.
The following k lines contain the descriptions of «Lada Malina» engi... | For each possible option of the exposition output the number of cars that will be able to get to the exposition location by the moment of its beginning. | null | Images describing sample tests are given below. Exposition options are denoted with crosses, factories are denoted with points. Each factory is labeled with a number of cars that it has.
First sample test explanation:
- Car from the first factory is not able to get to the exposition location in time.
- Car from the s... | [
{
"input": "2 4 1\n1 1\n-1 1\n2 3 1\n2 -2 1\n-2 1 1\n-2 -2 1\n0 0 2",
"output": "3"
},
{
"input": "3 4 3\n2 0\n-1 1\n-1 -2\n-3 0 6\n1 -2 1\n-3 -7 3\n3 2 2\n-1 -4 1\n0 4 2\n6 0 1",
"output": "4\n9\n0"
}
] | In the original contest there were subtasks for this problem and it's more convinient to understand the editorial going through these subtasks, so we will leave them here.
The key part of a solution is to understand what are the locations that may be accessed from the origin in T seconds. First observation is that we s... | What if you transform the coordinates so that the reachable region for a single day becomes axis-aligned? The key observation is that the set of points reachable from the origin in one day is a convex polygon defined by the engine vectors. This polygon can be mapped to a rectangle via a linear transformation that depen... | {
"exploration": "Okay so we have k engines (2 ≤ k ≤ 10), each with a velocity vector (vxi, vyi). We can set each engine's level wi ∈ [-1, 1]. The car's displacement over t days is t * Σ wi * vi. For a factory at f and exposition at p after t days, we need f + t * Σ wi * vi = p, i.e., Σ wi * vi = (p - f)/t. So for ea... | {
"exploration": "I'll discretize each w_i into m equally spaced values from -1 to 1. With k=2 engines and m=21 (step 0.1), that's 21^2 = 441 weight combinations per factory-query pair. For each factory f and query (p, t), I compute the target displacement d = (px - fx, py - fy), then iterate all (w1, w2) combination... | {
"exploration": "For each factory i at (fx_i, fy_i) and each query j at (px_j, py_j, t_j), we need to check if there exist real numbers w_1,...,w_k in [-1,1] such that:\n fx_i + t_j * (w_1*vx_1 + ... + w_k*vx_k) = px_j,\n fy_i + t_j * (w_1*vy_1 + ... + w_k*vy_k) = py_j.\nRearranging, we need (px_j - fx_i)/t_j = Σ ... | null | null | null | Okay, so every approach hit a wall — per-factory checks were O(nq), discretizing the weight space exploded exponentially, and even the dual-space reduction to polygon counting was still too heavy for the constraints.
hmm... but that last thread about mapping the reachable polygon to a rectangle. The engine vectors are... | READ k, n, q
vx = array of size k
vy = array of size k
for i from 0 to k-1:
READ vx[i], vy[i]
// Precompute S_i = sum_j | (vy_i, -vx_i) dot (vx_j, vy_j) |
S = array of size k
for i from 0 to k-1:
sum = 0
for j from 0 to k-1:
dot = vy[i] * vx[j] + (-vx[i]) * vy[j] // = vy_i * vx_j - vx_i * vy_j
... |
1801/G | A task for substrings | 3,400 | data structures, string suffix structures, strings | 4 | 1,024 | To do this, he took the string $$$t$$$ and a set of $$$n$$$ strings $$$s_1$$$, $$$s_2$$$, $$$s_3$$$, ..., $$$s_n$$$. Philip has $$$m$$$ queries, in the $$$i$$$th of them, Philip wants to take a substring of the string $$$t$$$ from $$$l_i$$$th to $$$r_i$$$th character, and count the number of its substrings that match s... | The first line contains two positive integers $$$n$$$ and $$$m$$$ ($$$1 \le n, m \le 500\,000$$$) — the number of rows in the set and the number of queries.
The second line contains a single string $$$t$$$ consisting of lowercase letters of the English alphabet ($$$1 \le |t| \le 5 \cdot 10^6$$$).
The following $$$n$$... | In a single line, print $$$m$$$ integers, $$$i$$$th of them should be equal to the answers to the $$$i$$$th query. | null | In the first example, the first query requires the entire string to count the number of substrings that are included in the set. The substrings $$$[1, 3]$$$ and $$$[4, 6]$$$ coincide with the string "aba". The substrings match with the string "a" $$$[1, 1]$$$, $$$[3, 3]$$$, $$$[5, 5]$$$, $$$[7, 7]$$$. The substring $$$... | [
{
"input": "3 5\nabacaba\naba\na\nac\n1 7\n1 3\n2 7\n2 5\n4 5",
"output": "7 3 5 3 1"
},
{
"input": "4 4\nabcdca\nab\nca\nbcd\nopenolympiad\n1 5\n2 2\n2 6\n1 6",
"output": "2 0 2 3"
}
] | Let's use the Aho-Korasik structure to store strings from $S$. Let's build compressed suffix links on it. This way it is a little more optimal to find all the lines from $S$ ending in this position $t$.
Denote by $pref[i]$ the number of substrings of $S$ in the prefix $t$ of length $i$.
Denote by $suf[i]$ the number ... | The key observation is that you can precompute, for each position in t, how many pattern strings end there. This lets you answer queries about substrings using prefix sums, but you must carefully subtract occurrences that cross the query boundaries. What if you could find, for each query, a single pattern string that c... | {
"exploration": "I'll build a suffix array for t, then for each query (l, r) I'll binary search to find the range of suffixes that start within [l, r] and whose prefixes match any pattern from the set. The idea is: precompute all occurrences of each pattern in t using the suffix array and LCP, then for a query, coun... | {
"exploration": "I'll build an Aho-Corasick automaton from the set of patterns s_i, then use heavy-light decomposition on the suffix link tree. The idea is: for each query (l, r), we want to count occurrences of patterns that appear as substrings of t[l..r]. With the automaton, we can process t and track which state... | null | null | null | null | Okay, so both the suffix array and the Aho-Corasick HLD ran into brick walls — the first because it still required iterating over all patterns per query, and the second because range updates on the automaton state tree would incorrectly count depths that aren't actual pattern endings.
Hmm. That nagging thought about p... | function solve():
READ n, m
READ string t
length_T = length(t)
// Build Aho-Corasick automaton for patterns
// g: list of nodes, each node has:
// nx[26]: transition
// p: suffix link
// pp: nearest terminal suffix link (or -1)
// len: length of string to this node
// ... |
2029/I | Variance Challenge | 3,400 | flows,graphs,greedy | null | null | Kevin has recently learned the definition of variance. For an array $$$a$$$ of length $$$n$$$, the variance of $$$a$$$ is defined as follows:
• Let $$$x=\dfrac{1}{n}\displaystyle\sum_{i=1}^n a_i$$$, i.e., $$$x$$$ is the mean of the array $$$a$$$;
• Then, the variance of $$$a$$$ is $$$$$$ V(a)=\frac{1}{n}\sum_{i=1}^n... | Each test contains multiple test cases. The first line of the input contains a single integer $$$t$$$ ($$$1\le t\le 100$$$) — the number of test cases. The description of test cases follows.
The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1\le n,m\le 5000$$$, $$$\color{red}{n... | For each test case, output $$$m$$$ integers in a single line, the $$$p$$$-th integer denoting the minimum possible variance of $$$a$$$ when exactly $$$p$$$ operations are performed, multiplied by $$$n^2$$$. | null | null | [] | Key Observation. Suppose we have an array $b$ and a function $f(x)=\sum (b_i-x)^2$, then the minimum value of $f(x)$ is the variance of $b$.
Proof. This is a quadratic function of $x$, and the its symmetry axis is $x=\frac{1}{n}\sum b_i$. So the minimum value is $f\left(\frac{1}{n}\sum b_i\right)$. That is exactly the... | The key observation is that the variance is a quadratic function of the mean. What if you fix a target mean and consider the cost of adjusting the array to reach it? The operations become interval additions, and the cost reduction per operation has a structure you can exploit greedily. | {
"exploration": "Okay so we have an array a of length n, we can add k to any contiguous subarray per operation, and we need the minimum variance after exactly p operations for each p from 1 to m. The constraint says n·m ≤ 2×10^4 over all test cases, and sum of n·m over all tests ≤ 2×10^4. That's tiny — the product n... | {
"exploration": "First, I need to understand what the variance becomes after adding k to some intervals. The variance V(a) = (1/n) Σ (a_i - x)^2, where x = (Σ a_i)/n. Multiplying by n^2 gives n Σ (a_i - x)^2 = n Σ a_i^2 - (Σ a_i)^2. So I need to minimize n Σ a_i^2 - (Σ a_i)^2 after exactly p operations. Each operati... | null | null | null | null | None of those got me anywhere useful. The direct DP over intervals couldn't handle operations that cross the split boundary, and the convex hull trick fell apart because the total sum T couples everything — you can't compute the cost for a prefix without knowing the final global sum.
Hmm. But that coupling — the varia... | function solve_test_case():
READ n, m, k
READ array a[1..n]
// Precompute constants
S = sum(a[1..n])
Q = sum(a[i] * a[i] for i from 1 to n)
base = n * Q - S * S
// Initialize d_i and M_i
d[1..n] = array of size n
for i from 1 to n:
d[i] = n * a[i] - S
M[1..n] = array of... |
833/E | Caramel Clouds | 3,400 | data structures, dp, sortings | 3 | 256 | It is well-known that the best decoration for a flower bed in Sweetland are vanilla muffins. Seedlings of this plant need sun to grow up. Slastyona has m seedlings, and the j-th seedling needs at least kj minutes of sunlight to grow up.
Most of the time it's sunny in Sweetland, but sometimes some caramel clouds come, ... | The first line contains two integers n and C (0 ≤ n ≤ 3·105, 0 ≤ C ≤ 109) – the number of caramel clouds and the number of candies Slastyona has.
The next n lines contain three integers each: li, ri, ci (0 ≤ li < ri ≤ 109, 0 ≤ ci ≤ 109), describing one caramel cloud.
The next line contains single integer m (1 ≤ m ≤ 3... | For each seedling print one integer – the minimum minute Slastyona can grow it up. | null | Consider the first example. For each k it is optimal to dispel clouds 1 and 3. Then the remaining cloud will give shadow on time segment [1..6]. So, intervals [0..1] and [6..inf) are sunny.
In the second example for k = 1 it is not necessary to dispel anything, and for k = 5 the best strategy is to dispel clouds 2 and... | [
{
"input": "3 5\n1 7 1\n1 6 2\n1 7 1\n3\n7\n2\n5",
"output": "12\n7\n10"
},
{
"input": "3 15\n1 4 17\n2 8 6\n4 8 9\n2\n5\n1",
"output": "8\n1"
},
{
"input": "2 10\n3 7 9\n10 90 10\n2\n10\n100",
"output": "10\n104"
}
] | Prequisites: scanline.
The key idea is that only minutes which are covered by no more than two clouds can contribute to the answer.
Let us demonstrate how to solve the problem for a fixed k using scanline. Let's create 2·n events: for the start and the end of each cloud. We will then go through all the events in ascend... | The key observation is that only minutes covered by at most two clouds can contribute to the sunny total after removal. The problem reduces to tracking, via a sweep line, the amount of time covered by exactly one or two clouds, and then selecting up to two clouds to remove to maximize the sum of their covered times, su... | {
"exploration": "I'll brute-force check all cloud pairs — for each seedling, try removing 0, 1, or 2 clouds (any pair) that cost ≤ C, compute the earliest time the seedling accumulates kj sunny minutes, and take the minimum.\n\nFirst, I need to understand the timeline. The sun is blocked whenever at least one undisp... | {
"exploration": "I'll discretize time at all l_i and r_i points, sort them, and process intervals between consecutive event times. At each event, I maintain which clouds are currently active. The DP state is dp[removed][mask] = maximum sunny minutes accumulated so far, where removed ∈ {0,1,2} and mask encodes which ... | null | null | null | null | Alright, back to square one. Brute force over pairs was impossible, and DP over time exploded because the state of active clouds was too large — I can't track every combination of clouds overlapping at a given moment.
hmm — but the regulations say I can remove at most two clouds. That means the only minutes that matte... | READ n, C
clouds = array of size n
events = empty list
FOR i FROM 0 TO n-1:
READ l, r, c
clouds[i].l = l
clouds[i].r = r
clouds[i].c = c
clouds[i].id = i
ADD (l, START, i) to events
ADD (r, END, i) to events
READ m
queries = array of size m
FOR j FROM 0 TO m-1:
READ k
queries[j].k =... |
917/E | Upside Down | 3,400 | data structures, string suffix structures, strings, trees | 3 | 512 | As we all know, Eleven has special abilities. Thus, Hopper convinced her to close the gate to the Upside Down World with her mind. Upside down monsters like to move between the worlds, so they are going to attack Hopper and Eleven in order to make them stop. The monsters live in the vines. The vines form a tree with n ... | The first line of input contains three integers, n, m and q (2 ≤ n ≤ 105, 1 ≤ m, q ≤ 105).
The next n - 1 lines contain the tunnels (edges). Each line contains two integers v and u (1 ≤ v, u ≤ n, v ≠ u) and a lowercase English letter c, meaning there's a tunnel connecting junctions v and u written c in it. It is guara... | Print q lines. i-th line should contain a single integer, the power the i-th monster gains after moving. | null | null | [
{
"input": "6 4 5\n1 6 b\n2 3 a\n1 2 b\n5 3 b\n4 5 b\na\nb\nbb\naa\n1 2 1\n6 2 3\n1 6 2\n4 5 4\n1 6 2",
"output": "0\n1\n1\n0\n1"
},
{
"input": "10 6 7\n1 3 s\n10 1 d\n2 6 s\n5 2 d\n7 4 l\n8 9 d\n8 10 l\n7 2 d\n8 7 l\ndl\ndssld\nd\nd\nl\nsl\n4 5 4\n3 7 5\n10 6 2\n3 1 4\n7 5 6\n10 9 4\n9 8 4",
"o... | Assume ti is reverse of si. Use centroid-decomposition. When solving the problem for subtree S, assume its centroid is c. For a fixed query, assume v and u are both in S and path from v to u goes through the centroid c (this happens exactly one time for each v and u). Assume x is string of path from c to v and y is str... | The key observation is that counting pattern occurrences on arbitrary tree paths requires a divide-and-conquer technique that breaks paths at a central point. You need to combine string automata to handle patterns crossing the join, and cleverly manage contributions from reversed halves. | {
"exploration": "Okay so we have a tree with n vertices, edges labeled with lowercase letters, m pattern strings with total length ≤ 10^5, and q queries each asking for the number of occurrences of pattern s_k on the path from i to j. The constraint n ≤ 10^5, total pattern length ≤ 10^5, q ≤ 10^5. We need to count p... | {
"exploration": "Let me build the HLD with segment trees of hashes. First, I'll do the heavy-light decomposition: pick a root (say 1), compute subtree sizes, and for each node, the heavy child is the one with the largest subtree. Each edge belongs to the node deeper in the tree, so I assign the edge's character to t... | null | null | null | null | Well, brute-force string extraction and KMP is hopelessly slow, and even HLD with hashes still forces us to check every possible starting position along the path, which is still O(n) per query. The core issue is that patterns can be long, and occurrences can start anywhere, so we can't just compress the path into a sin... | function solve():
READ n, m, q
adj = array of size n+1 of empty lists
for i from 1 to n-1:
READ u, v, c
adj[u].append((v, c))
adj[v].append((u, c))
patterns = array of size m+1 of strings
total_len = 0
for i from 1 to m:
READ patterns[i]
total_len += len(... |
1220/G | Geolocation | 3,400 | geometry | 2 | 256 | You are working for the Gryzzl company, headquartered in Pawnee, Indiana.
The new national park has been opened near Pawnee recently and you are to implement a geolocation system, so people won't get lost. The concept you developed is innovative and minimalistic. There will be $$$n$$$ antennas located somewhere in the... | The first line of input contains a single integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) which is the number of antennas.
The following $$$n$$$ lines contain coordinates of antennas, $$$i$$$-th line contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \leq x_i,y_i \leq 10^8$$$). It is guaranteed that no two antennas coincid... | For each query output $$$k$$$, the number of possible a user's locations matching the given input and then output the list of these locations in lexicographic order.
It is guaranteed that the sum of all $$$k$$$ over all points does not exceed $$$10^6$$$. | null | As you see in the second example, although initially a user's location is picked to have non-negative coordinates, you have to output all possible integer locations. | [
{
"input": "3\n0 0\n0 1\n1 0\n1\n1 1 2",
"output": "1 1 1"
},
{
"input": "4\n0 0\n0 1\n1 0\n1 1\n2\n0 1 1 2\n2 5 5 8",
"output": "4 0 0 0 1 1 0 1 1 \n4 -1 -1 -1 2 2 -1 2 2"
}
] | Let's look on the sum of squared distances from unknown point $(x;y)$ to all known points $(x_i;y_i)$:
$$\sum\limits_{i=1}^n d_i^2 = \sum\limits_{i=1}^n\left((x-x_i)^2+(y-y_i)^2\right)= n(x^2+y^2)-2x\sum\limits_{i=1}^n x_i - 2y\sum\limits_{i=1}^n y_i + \sum\limits_{i=1}^n (x_i^2+y_i^2)$$
If we switch to new coordinates... | The key observation is that the sum of all squared distances is invariant under permutation, which leads to a simple condition on the unknown point's coordinates. What if you transform coordinates to make the linear terms vanish? The problem reduces to finding integer points on a circle and verifying them via matching ... | {
"exploration": "Okay so we have n antennas at known coordinates, and for each query we get an unordered multiset of squared distances from some unknown user location (x,y) to those antennas. The user coordinates are integers between 0 and 10^8. We need to find all possible (x,y) that could produce that exact multis... | {
"exploration": "I'll build the Voronoi diagram of the n antennas, then for each query, I'll try to assign the largest distance to each possible \"farthest antenna\" candidate. For a fixed assignment of the largest distance to some antenna A_far, the user location must lie on the circle centered at A_far with radius... | null | null | null | null | Alright, back to square one. The coordinate search was impossible because we'd have to iterate over a 10^8 by 10^8 grid, and the Voronoi idea collapsed because matching distances to antennas is a combinatorial nightmare — we can't check all n! permutations.
hmm. Wait — the sum of squared distances is invariant under p... | READ n
READ antennas[1..n] as (x_i, y_i)
sumX = 0, sumY = 0
FOR i FROM 1 TO n:
sumX += antennas[i].x
sumY += antennas[i].y
// Precompute K' = sum_i ((sumX - n*x_i)^2 + (sumY - n*y_i)^2)
K_prime = 0
FOR i FROM 1 TO n:
dx = sumX - n * antennas[i].x
dy = sumY - n * antennas[i].y
K_prime += dx*dx + dy... |
1268/E | Happy Cactus | 3,400 | dp | 3 | 256 | You are given a cactus graph, in this graph each edge lies on at most one simple cycle.
It is given as $$$m$$$ edges $$$a_i, b_i$$$, weight of $$$i$$$-th edge is $$$i$$$.
Let's call a path in cactus increasing if the weights of edges on this path are increasing.
Let's call a pair of vertices $$$(u,v)$$$ happy if the... | The first line of input contains two integers $$$n,m$$$ ($$$1 \leq n, m \leq 500\,000$$$): the number of vertices and edges in the given cactus.
The next $$$m$$$ lines contain a description of cactus edges, $$$i$$$-th of them contain two integers $$$a_i, b_i$$$ ($$$1 \leq a_i, b_i \leq n, a_i \neq b_i$$$).
It is guar... | Print $$$n$$$ integers, required values for vertices $$$1,2,\ldots,n$$$. | null | null | [
{
"input": "3 3\n1 2\n2 3\n3 1",
"output": "2 2 2"
},
{
"input": "5 4\n1 2\n2 3\n3 4\n4 5",
"output": "4 4 3 2 1"
}
] | At first, let's solve for the tree. Let $dp_v$ be the number of answer for vertex $v$.
Let's look at edges in the order of decreasing weight.
How $dp$ is changing when you are looking at edge $i$?
I claim that $dp'_v = dp_v$ for $v \neq a_i $and $v \neq b_i$.
And $dp'_{a_i} = dp'_{b_i} = dp_{a_i} + dp_{b_i}$.
Why? I li... | What if you process the edges in a specific order and maintain connectivity components? The key observation is that when an edge connects two vertices already in the same component, you need to account for the unique path between them in the cactus. The subtraction depends on the structure of that path relative to the ... | {
"exploration": "Okay so we have a cactus graph — each edge lies on at most one simple cycle. The edges are numbered 1 through m, and their weight is their index. We need to count, for each vertex u, how many other vertices v have an increasing path from u to v (edge weights strictly increasing along the path). The ... | {
"exploration": "I'll build an n×n matrix dp[i][j] where dp[i][j] stores the maximum weight of the last edge on any increasing path from i to j. Initialize dp[i][j] = -1 for all pairs, then for each edge k with weight k connecting a_k and b_k, set dp[a_k][b_k] = k and dp[b_k][a_k] = k. Then triple loop: for k from 1... | {
"exploration": "I'll build an Euler tour of the cactus by doing a DFS, assigning each vertex an in-time and out-time so every subtree becomes a contiguous interval. Then I'll try to use a segment tree over these intervals to propagate reachability via increasing-weight paths.\n\nFirst, the DFS on Example 1: n=3, m=... | {
"exploration": "I'll build a heavy-light decomposition of the cactus, and for each node maintain a persistent set of edge weights from which an increasing path can start. For each node, I'll query its ancestors along heavy paths to count how many reachable vertices exist by strictly increasing edge weight sequences... | null | null | So all those ideas fell apart — the state space was just too big, because tracking reachability with a weight threshold for every vertex pair blows up to O(m²), and any structure that tries to propagate reachable intervals along edges ends up merging O(n) elements per node, which is O(n²) again. The core problem is tha... | function solve():
READ n, m
edges = array of size m
for i from 0 to m-1:
READ a, b
edges[i] = (a-1, b-1) // 0-index vertices
// Build adjacency list of cactus
adj = array of n empty lists
for i from 0 to m-1:
(u, v) = edges[i]
adj[u].append((v, i))
adj[... |
627/F | Island Puzzle | 3,400 | dfs and similar, dsu, graphs, trees | 2 | 256 | A remote island chain contains n islands, with some bidirectional bridges between them. The current bridge network forms a tree. In other words, a total of n - 1 bridges connect pairs of islands in a way that it's possible to reach any island from any other island using the bridge network. The center of each island con... | The first line contains a single integer n (2 ≤ n ≤ 200 000) — the total number of islands.
The second line contains n space-separated integers ai (0 ≤ ai ≤ n - 1) — the statue currently located on the i-th island. If ai = 0, then the island has no statue. It is guaranteed that the ai are distinct.
The third line con... | Print a single line of integers:
If the rearrangement can be done in the existing network, output 0 t, where t is the number of moves necessary to perform the rearrangement.
Otherwise, print u, v, and t (1 ≤ u < v ≤ n) — the two endpoints of the new bridge, and the minimum number of statue movements needed to perform... | null | In the first sample, the islanders can build a bridge connecting islands 1 and 3 and then make the following sequence of moves: first move statue 1 from island 1 to island 2, then move statue 2 from island 3 to island 1, and finally move statue 1 from island 2 to island 3 for a total of 3 moves.
In the second sample, ... | [
{
"input": "3\n1 0 2\n2 0 1\n1 2\n2 3",
"output": "1 3 3"
},
{
"input": "2\n1 0\n0 1\n1 2",
"output": "0 1"
},
{
"input": "4\n0 1 2 3\n0 2 3 1\n1 2\n1 3\n1 4",
"output": "-1"
}
] | First, if we never move the empty pedestal through any cycle, then moving the empty pedestal to and from any given position cannot change the location of the statues, as performing a move in the opposite direction as the previous undoes the previous move.
Thus, in our graph with one cycle, we can only do the following... | What if you think about the statues that are already in their correct positions? The key observation is that the statues which need to be moved must all lie along a single contiguous chain in the tree. This suggests a structural property you can verify with a single DFS to classify vertices. | {
"exploration": "Okay so we have n islands, a tree of n-1 bridges, one empty pedestal, and distinct statues labeled 1..n-1. The operation is: pick a statue adjacent to the empty pedestal, slide it across the bridge into the empty spot. This is exactly the 15-puzzle on a tree — the empty space moves, and statues shif... | {
"exploration": "I'll build a flow network. Source connects to islands where a_i ≠ 0 and a_i ≠ b_i (statue needs to move). Sink connects to islands where b_i ≠ 0 and a_i ≠ b_i (island desires a different statue). Each statue that needs to move must travel from its current island to its target island along tree paths... | null | null | null | null | None of those got me anywhere useful. BFS from each misplaced statue exploded combinatorially, and the flow model completely ignored the sequential nature of statue moves — you can't just teleport statues to their destinations; the empty pedestal has to physically travel along edges, swapping with one statue at a time.... | function solve():
READ n
READ arrays a[1..n], b[1..n]
adjacency = list of size n+1, each an empty list
for i from 1 to n-1:
READ u, v
adjacency[u].append(v)
adjacency[v].append(u)
// find start position of empty pedestal (value 0 in a)
start = 0
for i from 1 to n:
... |
1936/F | Grand Finale: Circles | 3,300 | binary search, geometry | 2 | 512 | You are given $$$n$$$ circles on the plane. The $$$i$$$-th of these circles is given by a tuple of integers $$$(x_i, y_i, r_i)$$$, where $$$(x_i, y_i)$$$ are the coordinates of its center, and $$$r_i$$$ is the radius of the circle.
Please find a circle $$$C$$$ which meets the following conditions:
- $$$C$$$ is contai... | The first line contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$) — the number of circles.
The $$$i$$$-th of the following $$$n$$$ lines contains three integers $$$x_i$$$, $$$y_i$$$, $$$r_i$$$ ($$$-10^6 \le x_i,y_i \le 10^6$$$, $$$1 \le r_i \le 2 \cdot 10^6$$$).
It is guaranteed that there is a circle with a ... | Output three real values, $$$x$$$, $$$y$$$, and $$$r$$$ — the coordinates of the center and the radius of the circle. | null | A two-dimensional plot depicting the first test case is given below. The output circle $$$C$$$ is dashed with blue lines.
A two-dimensional plot depicting the second test case is given below. The output circle $$$C$$$ is dashed with blue lines. | [
{
"input": "4\n1 1 3\n-1 1 3\n1 -1 2\n-1 -1 2",
"output": "0.0000000000000000 -0.7637626158259733 0.9724747683480533"
},
{
"input": "4\n41580 -23621 95642\n-41580 -23621 95642\n0 47821 95642\n0 0 109750",
"output": "0.0000000000000000 0.0000000000000000 47821.0000000000000000"
}
] | For a given center coordinate $(x,y)$, we can model the objective function $r=f(x,y)$ to maximize as follows.
$$f(x,y)=\min_{1 \le i \le n} {\left({r_i-\sqrt{(x_i-x)^2+(y_i-y)^2}}\right)}$$
Formally, this can be modeled as follows:
$$\begin{split} \max & \ r \\ \text{s.t.} & \ \sqrt{(x_i-x)^2+(y_i-y)^2} \le r_i-r \ \ \... | The key observation is that the optimal circle is defined by at most 3 of the given circles. This suggests an incremental approach where you build a candidate solution from a small subset and efficiently update it when a constraint is violated. Think of a randomized method that processes circles in a random order, simi... | {
"exploration": "Okay so we have n circles, each with center (x_i, y_i) and radius r_i. We need to find the largest circle C that fits inside all of them — that is, for every i, the distance from C's center to (x_i, y_i) plus C's radius must be ≤ r_i. The constraint is n ≤ 10^5, coordinates up to 10^6, radii up to 2... | {
"exploration": "I'll approximate each circle by a regular polygon with K vertices. For a circle centered at (x_i, y_i) with radius r_i, the condition that a point (x, y) with radius r lies inside means the distance from (x, y) to (x_i, y_i) plus r ≤ r_i. For each vertex direction θ_j = 2πj/K, the tangent half-plane... | {
"exploration": "I'll maintain a bounding box for the optimal center. Initially, the center must lie within the intersection of all circles. A simple bound: the optimal center (x,y) must satisfy |x - x_i| ≤ r_i and |y - y_i| ≤ r_i for each circle i, so I can initialize the search region as the intersection of all th... | null | null | null | Okay, so every approach hit a wall — coordinate searches blew the time budget, polygon approximations needed impossible precision, and spatial subdivision exploded because the feasible region is microscopic relative to the bounding box. The common thread is they all tried to search the continuous space directly, which ... | function solve():
READ n
circles = array of size n
for i from 0 to n-1:
READ x[i], y[i], r[i]
// Binary search on the answer radius
low = 0.0
high = min(r) // maximum possible radius is the smallest original radius
for iteration from 1 to 80: // enough for 1e-18 precision
... |
1530/H | Turing's Award | 3,400 | data structures, dp | 10 | 512 | Alan Turing is standing on a tape divided into cells that is infinite in both directions.
Cells are numbered with consecutive integers from left to right. Alan is initially standing in cell $$$0$$$. Every cell $$$x$$$ has cell $$$x - 1$$$ on the left and cell $$$x + 1$$$ on the right.
Each cell can either contain an ... | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 1000$$$). Description of the test cases follows.
Each test case is given in two lines. The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 15\,000$$$) — the length of the perm... | For each test case output a single integer — the largest possible value of Turing's award.
Hacks are not allowed for this problem. | null | Longest increasing subsequence of sequence $$$b$$$ is the longest increasing sequence that can be obtained from $$$b$$$ by deletion of several (possibly, zero or all) elements.
In the first test case, Alan can make a decision only at time $$$2$$$. If Alan stays in cell $$$0$$$, sequence $$$b$$$ will be equal to $$$[2]... | [
{
"input": "4\n2\n1 2\n4\n4 1 2 3\n7\n3 6 5 7 4 1 2\n7\n5 2 3 7 6 1 4",
"output": "2\n3\n4\n4"
}
] | Let's look at Alan's moves in reverse. The process can be reformulated as follows. Initially, $a_n$ is written down into the cell where Alan is located. Then, for each $i$ from $n-1$ down to $1$, first Alan decides whether to stay or move to a neighboring cell, and then $a_i$ is written down into Alan's cell only if th... | What if you think about the process in reverse? The key observation is that the final sequence is built by appending numbers to either end, and the constraints for moving between ends depend on the number of 'useful' elements kept. This suggests a DP where you track the best possible boundaries of an increasing subsequ... | {
"exploration": "Okay so we have a process where Alan walks left/right on an infinite tape, writing numbers from a random permutation into cells, overwriting old values. After all n numbers are written, we read the non-empty cells left to right to form sequence b, and the award is the LIS length of b. We need to max... | {
"exploration": "Let me pick a candidate interval length and try to simulate with two pointers. For each interval [L,R], I'll maintain Alan's current position and try to place numbers greedily. The idea is to always move toward the nearest empty cell within [L,R] that can accept the current a_i while keeping the pla... | {
"exploration": "I'll define dp[i][j] as the maximum LIS length achievable using a subsequence from the first i numbers, where the last chosen number is a_j (with j ≤ i), and I'll also track the cell position where a_j was placed. For each state I need to know if I can extend it with a_k (k > i) where a_k > a_j, and... | {
"exploration": "We can think of each time i as placing value a_i into some cell, with the constraint that the cell chosen must be either the same cell as the previous placement or one step left/right. This is like a walk on the integer line where at each step we write a number. The final sequence b is the numbers w... | null | null | So that's a few failed attempts now. Direct simulation explodes, greedy fails on future unknowns, and any DP that tries to track the full spatial configuration is either exponential or impossibly large.
hmm. Wait — the final sequence b is just the integers left in cells, read left to right. But Alan starts at 0 and ca... | function solve():
READ t
for each test case:
READ n
READ a[1..n]
// dp arrays: fL[k][i] = INF, fR[k][i] = -INF
// We only need two layers for k and k+1, but k is at most O(sqrt(n))
// Estimate max possible k: 2*sqrt(n) + 5
maxK = 2 * int(sqrt(n)) + 5
// In... |
1787/I | Treasure Hunt | 3,400 | data structures, divide and conquer, two pointers | 2 | 512 | Define the beauty value of a sequence $$$b_1,b_2,\ldots,b_c$$$ as the maximum value of $$$\sum\limits_{i=1}^{q}b_i + \sum\limits_{i=s}^{t}b_i$$$, where $$$q$$$, $$$s$$$, $$$t$$$ are all integers and $$$s > q$$$ or $$$t\leq q$$$. Note that $$$b_i = 0$$$ when $$$i<1$$$ or $$$i>c$$$, $$$\sum\limits_{i=s}^{t}b_i = 0$$$ whe... | Each test contains multiple test cases. The first line contains an integer $$$T$$$ ($$$1 \le T \le 10^4$$$) — the number of test cases.
The first line contains an integer $$$n$$$ ($$$1\le n\le 10^6$$$) — the length of $$$a$$$.
The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$-10^6 \leq a_i \leq ... | For each test case, print a line containing a single integer — the answer modulo $$$998\,244\,353$$$. | null | In the second test case, for the subsequence $$$[-26,43,-41,34,13]$$$, when $$$q=5$$$, $$$s=2$$$, $$$t=5$$$, $$$\sum\limits_{i=1}^{q}b_i + \sum\limits_{i=s}^{t}b_i = 23 + 49 = 72$$$.
In the third test case, there is only one non-empty consecutive subsequence $$$[74]$$$. When $$$q=1$$$, $$$s=1$$$, $$$t=1$$$, $$$\sum\li... | [
{
"input": "4\n7\n80 59 100 -52 -86 -62 75\n8\n-48 -14 -26 43 -41 34 13 55\n1\n74\n20\n56 -60 62 13 88 -48 64 36 -10 19 94 25 -69 88 87 79 -70 74 -26 59",
"output": "5924\n2548\n148\n98887"
}
] | First, we observe that: $\max\limits_{l>p\,\texttt{or}\,r\leq p}{(\sum\limits_{i=1}^{p}{a_i}+\sum\limits_{i=l}^{r}{a_i})}=\max\limits_{p=1}^{n}{\sum\limits_{i=1}^{p}{a_i}}+\max\limits_{l,r}{\sum\limits_{i=l}^{r}{a_i}}$.
Proof: We can proof it by contradiction. Define $g(l,r,p)$ as $\sum\limits_{i=1}^{p}{a_i}+\sum\limit... | What if you could separate the problem into two independent maximization problems? The key observation is that the beauty value for any segment is the sum of two classic maximum-sum queries. This suggests a divide-and-conquer approach where you compute contributions across a midpoint, but the challenge is efficiently c... | {
"exploration": "Okay so we have an array a of length n, and we need to sum the \"beauty value\" over all non-empty subsegments. The beauty value of a sequence b of length c is defined as the maximum over all integer q, s, t of (sum of first q elements) + (sum of elements from s to t), with the condition that s > q ... | {
"exploration": "I'll set up dp[l][r] as the beauty value for subsegment a[l..r]. For each fixed l, I'll extend r from l to n, maintaining the best prefix sum and best subarray sum within the segment. When adding a[r], I need to recompute the maximum of (prefix sum up to q) + (subarray sum in [s,t]) with the non-ove... | {
"exploration": "I'll build a segment tree where each node stores: total sum, max prefix sum, max suffix sum, max subarray sum, and the beauty value for its segment. The merge needs to compute the beauty value for the combined segment from the two children's data. The beauty definition: pick q, and pick [s,t] such t... | null | null | null | Okay, so every approach hit a wall — brute force was astronomically slow, DP was cubic, and a naive segment tree merge was quadratic. The core issue is that for each of the O(n²) subsegments, I was trying to compute the beauty from scratch, which is O(n) per segment, leading to O(n³) total. I need a way to compute cont... | function solve():
MOD = 998244353
READ T
for each test case:
READ n
READ array a[1..n]
// Compute prefix sums
pref = array of size n+1, pref[0] = 0
for i from 1 to n:
pref[i] = pref[i-1] + a[i]
// Compute max subarray sum for all suffixes ... |
865/F | Egg Roulette | 3,300 | bitmasks, brute force, divide and conquer, math, meet-in-the-middle | 3 | 256 | The game of Egg Roulette is played between two players. Initially 2R raw eggs and 2C cooked eggs are placed randomly into a carton. The shells are left on so there is no way to distinguish a raw egg from a cooked egg. One at a time, a player will select an egg, and then smash the egg on his/her forehead. If the egg was... | The first line of input will contain integers R and C (1 ≤ R, C ≤ 20, R + C ≤ 30).
The second line of input will contain the string S of length 2(R + C) consisting only of characters 'A', 'B', '?'. | Print the number of valid orderings that minimize unfairness and match S. | null | In the first test case, the minimum unfairness is 0, and the orderings that achieve it are "ABBA" and "BAAB", neither of which match S. Note that an ordering such as "ABBB" would also have an unfairness of 0, but is invalid because it does not contain the same number of 'A's as 'B's.
In the second example, the only ma... | [
{
"input": "1 1\n??BB",
"output": "0"
},
{
"input": "2 4\n?BA??B??A???",
"output": "1"
},
{
"input": "4 14\n????A??BB?????????????AB????????????",
"output": "314"
}
] | We can permute any prefix of an ordering containing at most R - 1 'A's or R - 1 'B's without changing the unfairness. This is because within that prefix one of the players cannot lose the game. This means that every ordering there is a unique ordering with the same unfairness that begins with exactly R - 1 'A's followe... | The key observation is that many orderings are equivalent in terms of win probability because early turns where a player cannot yet lose are interchangeable. This suggests focusing on a compressed representation of the game state. Think about a meet-in-the-middle approach on the remaining turns after fixing a prefix th... | {
"exploration": "I'll simulate the game for each possible ordering that matches S, compute win probabilities exactly by enumerating all egg arrangements, then find the minimum unfairness and count matching orderings.\n\nFor a fixed ordering of length 2(R+C) with R+C A's and R+C B's, the game outcome depends on the p... | {
"exploration": "I'll model this as a DP over the state (i, a_raw, b_raw, raw_rem, cooked_rem) where i is the turn index, a_raw is how many raw eggs player A has broken, b_raw for player B, raw_rem is remaining raw eggs, cooked_rem is remaining cooked eggs. The ordering string T of length 2(R+C) tells whose turn it ... | null | null | null | null | None of those got me anywhere useful — brute-force enumeration is astronomically impossible, and even trying to reason about win probabilities directly collapses under the combinatorial explosion of possible egg sequences.
hmm. The win probability only depends on who gets the R-th raw egg first, which is determined by... | function binomial(n, k):
if k < 0 or k > n: return 0
res = 1
for i from 1 to k:
res = res * (n - i + 1) / i
return res
function solve():
READ R, C
READ S (string of length N = 2*(R+C))
N = 2*(R+C)
total_weight = 0
weight = array of size N
non_zero_indices = empty list
... |
1039/E | Summer Oenothera Exhibition | 3,400 | data structures | 7 | 512 | While some people enjoy spending their time solving programming contests, Dina prefers taking beautiful pictures. As soon as Byteland Botanical Garden announced Summer Oenothera Exhibition she decided to test her new camera there.
The exhibition consists of $$$l = 10^{100}$$$ Oenothera species arranged in a row and co... | The first line contains three positive integer $$$n$$$, $$$w$$$, $$$q$$$ ($$$1 \leq n, q \leq 100\,000$$$, $$$1 \leq w \leq 10^9$$$) — the number of taken photos, the number of flowers on a single photo and the number of queries.
Next line contains $$$n$$$ non-negative integers $$$x_i$$$ ($$$0 \le x_i \le 10^9$$$) — t... | Print $$$q$$$ integers — for each width of the truncated photo $$$k_i$$$, the minimum number of cuts that is possible. | null | null | [
{
"input": "3 6 5\n2 4 0\n1 2 3 4 5",
"output": "0\n0\n1\n1\n2"
},
{
"input": "6 4 3\n1 2 3 4 3 2\n1 2 3",
"output": "0\n1\n2"
}
] | Let's reconsider this task in terms of segments: we need to split sequence of photos-segments into minimum number of contiguous groups such that there exists a subsegment in each group of length $k$, which contains inside of each segment of the group.
It's easy to see that if we move right edge of each segment to the l... | What if you think about each photo as an interval of possible left endpoints for the truncated k-length segment? The key observation is that grouping photos into a scene corresponds to finding a single point common to all their intervals. The challenge is to compute the greedy grouping quickly for many k values by expl... | {
"exploration": "Okay so we have n photos, each covering a window of width w, with left endpoint x_i. We can truncate each photo to a segment of exactly k flowers, meaning we pick some offset within the photo's w-width window to keep a k-width subsegment. A scene is a contiguous block of photos where the set of flow... | {
"exploration": "For a fixed k and a hypothesized maximum cuts C, I'll greedily scan the photos left to right, maintaining the intersection of intervals [x_i, x_i + w - k] (the range of possible left endpoints for a k-length subsegment within photo i). When the intersection becomes empty, I force a cut before the cu... | {
"exploration": "I'll sort all queries by k, then process them in increasing order. For each photo i, its interval I_i(k) = [x_i + (w - k), x_i + w - 1]? Wait, let me re-derive. The original photo is [x_i, x_i + w - 1]. We truncate to exactly k flowers. The truncated segment must be a contiguous subsegment of length... | null | null | null | None of those got me anywhere useful. The DP was quadratic per query, binary search still needed O(n) per check, and Mo's algorithm couldn't handle global updates efficiently — all because they were treating each query independently, recomputing everything from scratch for each k.
hmm. But that clue about intervals of... | READ n, w, q
READ x[1..n]
READ k[1..q]
// Step 1: Compute merge thresholds using priority queue on adjacent segments
thresholds = empty list
// Each segment: left index, right index, min_val, max_val
// Use arrays to represent linked list of segments
prev = array of size n+2, initialized to -1
next = array of size n+2... |
1552/I | Organizing a Music Festival | 3,400 | dfs and similar, math | 1 | 256 | You are the organizer of the famous "Zurich Music Festival". There will be $$$n$$$ singers who will perform at the festival, identified by the integers $$$1$$$, $$$2$$$, $$$\dots$$$, $$$n$$$. You must choose in which order they are going to perform on stage.
You have $$$m$$$ friends and each of them has a set of favou... | The first line contains two integers $$$n$$$ and $$$m$$$ ($$$1\le n,\,m\le 100$$$) — the number of singers and the number of friends correspondingly.
The $$$i$$$-th of the next $$$m$$$ lines contains the integer $$$q_i$$$ ($$$1\le q_i\le n$$$) — the number of favorite singers of the $$$i$$$-th friend – followed by the... | Print the number of valid orderings of the singers modulo $$$998\,244\,353$$$. | null | Explanation of the first sample: There are $$$3$$$ singers and only $$$1$$$ friend. The friend likes the two singers $$$1$$$ and $$$3$$$. Thus, the $$$4$$$ valid orderings are:
- 1 3 2
- 2 1 3
- 2 3 1
- 3 1 2
Explanation of the second sample: There are $$$5$$$ singers and $$$5$$$ friends. One can show that no orderin... | [
{
"input": "3 1\n2 1 3",
"output": "4"
},
{
"input": "5 5\n2 1 2\n2 2 3\n2 3 4\n2 4 5\n2 1 5",
"output": "0"
},
{
"input": "100 1\n1 50",
"output": "35305197"
},
{
"input": "5 1\n5 1 2 3 4 5",
"output": "120"
},
{
"input": "2 5\n1 2\n1 2\n1 2\n1 1\n1 1",
"outp... | Let $S_i := \{s_{i,1}, \, s_{i,2}, \, \dots, \, s_{i, q_i}\}$. Consider the graph on the subsets $S_1, \, S_2, \, \dots, \, S_k$ such that $S_i$ is adjacent to $S_j$ if and only if $S_i \cap S_j \ne \varnothing, \, S_i, \, S_j$. Without loss of generality we can assume that the sets $S_1, \, S_2, \, \dots, \, S_m$ are ... | The key observation is that each friend's constraint forces a block of singers to be contiguous. This creates a complex system of overlapping blocks. Think about grouping singers based on which exact subsets of friends like them. The valid global orderings are determined by how these groups can be arranged linearly whi... | {
"exploration": "Okay so we have n singers and m friends, each friend has a set of favorite singers they want to see consecutively. n and m are both up to 100. We need to count permutations of 1..n where each friend's set appears as a contiguous block. The brute-force approach would generate all n! permutations and ... | {
"exploration": "I'll build a directed graph where an edge X→Y means singer X must appear before singer Y. For each friend's set of favorite singers, since they must appear consecutively, all singers in that set must form a contiguous block. That means for any singer c not in the set, c must either appear entirely b... | null | null | null | null | None of those got me anywhere useful. Enumerating permutations is impossible, and topological sort on pairwise precedences can't enforce contiguity — it only cares about relative order, not adjacency.
hmm. The contiguity constraints are the real challenge. Each friend demands their liked singers form a single block in... | MOD = 998244353
function main():
READ n, m
friends = list of m sets
for i from 1 to m:
READ q_i
S = empty set
for j from 1 to q_i:
READ s
add s to S
friends[i] = S
// Build PQ-tree with leaves 1..n
root = new P-node with children = [leaf(1), ... |
1276/F | Asterisk Substrings | 3,400 | string suffix structures | 5 | 512 | Consider a string $$$s$$$ of $$$n$$$ lowercase English letters. Let $$$t_i$$$ be the string obtained by replacing the $$$i$$$-th character of $$$s$$$ with an asterisk character *. For example, when $$$s = \mathtt{abc}$$$, we have $$$t_1 = \tt{*bc}$$$, $$$t_2 = \tt{a*c}$$$, and $$$t_3 = \tt{ab*}$$$.
Given a string $$$s... | The only line contains a string $$$s$$$ of $$$n$$$ lowercase English letters ($$$1 \leq n \leq 10^5$$$). | Print a single integer — the number of distinct strings of $$$s, t_1, \ldots, t_n$$$. | null | For the sample case, the distinct substrings are (empty string), a, b, c, *, ab, a*, bc, b*, *b, *c, abc, ab*, a*c, *bc. | [
{
"input": "abc",
"output": "15"
}
] | There are two types of substrings we have to count: with and without the *. The substrings without * are just all substrings of the intial string $s$, which can be counted in $O(n)$ or $O(n \log n)$ using any standard suffix structure.
We now want to count the substrings containing *. Consider all such substrings of th... | What if you think about the asterisk as splitting the substring into a prefix and a suffix? The key observation is that the number of distinct suffixes you can attach after the * depends on the set of positions where the prefix appears. This leads to a data structure problem on the suffix tree where you need to merge s... | {
"exploration": "Okay so we have a string s of length n up to 10^5. We form n+1 strings: s itself, and for each position i, t_i which is s with the i-th character replaced by *. We need to count all distinct substrings across all these n+1 strings, including the empty string. The * is just a regular character, no sp... | {
"exploration": "I'll set up a DP where dp[i][j] means the substring of s ending at position i can match a pattern ending with an asterisk at position j in some t_k. For each asterisk position k from 1 to n, I'll try to align substrings of s with substrings of t_k. The idea is to iterate over all possible (i, j) pai... | {
"exploration": "I'll build a suffix automaton (SAM) for the original string s, then for each position i where we replace with an asterisk, I'll try to merge a modified version of the SAM into the original. The idea is that the asterisk acts as a wildcard matching any single character, so at the state corresponding ... | null | null | null | Okay, so brute-force is impossible, dynamic programming on positions explodes, and trying to shove asterisks into a suffix automaton creates phantom strings with multiple stars—none of those worked because they either generated too many substrings or counted strings that don't actually exist in any t_i.
hmm... but tha... | function solve():
READ s
n = length(s)
// Build SAM of s
// states: 0..sz-1, each with len, link, next[26]
// also compute terminal states (suffix links from last)
sz = 1
last = 0
len = array of size 2*n, initialized to 0
link = array of size 2*n, initialized to -1
next = ar... |
1310/F | Bad Cryptography | 3,400 | math, number theory | 3 | 512 | In modern cryptography much is tied to the algorithmic complexity of solving several problems. One of such problems is a discrete logarithm problem. It is formulated as follows:
Let's fix a finite field and two it's elements $$$a$$$ and $$$b$$$. One need to fun such $$$x$$$ that $$$a^x = b$$$ or detect there is no suc... | In the first line of input there is single integer $$$t$$$ ($$$1 \le t \le 100$$$) — number of pairs, for which you need to find the discrete logarithm.
In each of next $$$t$$$ line there is a pair of integers $$$a$$$ $$$b$$$ ($$$1 \le a, b < 2^{64}$$$). | For each pair you should print one integer $$$x$$$ ($$$0 \le x < 2^{64}$$$), such that $$$a^{\odot x} = b$$$, or -1 if no such x exists. It can be shown, that if any such $$$x$$$ exists, there is one inside given bounds. If there are several good values, you can output any of them. | null | null | [
{
"input": "7\n2 2\n1 1\n2 3\n8 10\n8 2\n321321321321 2\n123214213213 4356903202345442785",
"output": "1\n1\n2\n4\n-1\n6148914691236517205\n68943624821423112"
}
] | One of the well-known algorithms for the discrete logarithm problem is baby step giant step algorithm based on the meet in the middle idea. It can solve the problem in $O(\sqrt{|F|})$ time, which is definitely too much for the field of size $2^64$.
Multiplicative group of field has size $F = 2^64 - 1 = 3 \cdot 5 \cdot ... | The key observation is that the multiplicative group of this field has a surprisingly small and smooth order. What if you could break the problem into several independent subproblems, each of a manageable size, and then combine the answers? This suggests a meet-in-the-middle approach on subgroups, not the whole group. | {
"exploration": "Okay so we have the nimber field on numbers 0 to 2^64-1, with addition being XOR and multiplication defined by those recursive properties. The group size is 2^64-1 nonzero elements. We need to solve discrete log: given a, b, find x such that a^x = b under nimber multiplication, or return -1. t ≤ 100... | {
"exploration": "For each test case, start with p = 1 (since a^0 = 1 in any group), x = 0. Then loop: if p == b, return x. Otherwise, multiply p = p ⊙ a, increment x, and check if p has been seen before. Use a hash set to store visited powers. If we see a repeat before hitting b, return -1.\n\nLet me trace Example 1... | null | null | null | null | Well, every approach hit a wall: BSGS needed impossible memory, and brute force would take eons. The group is too big to handle directly. But that nagging thought — the group order is smooth — keeps pulling me back. If the order factors nicely, maybe we can solve the discrete log modulo each prime power factor separate... | function nimber_multiply(a, b):
// Precomputed multiplication table for small numbers (e.g., up to 2^8)
if a < 256 and b < 256:
return precomputed[a][b]
// Decompose into high and low halves
shift = 32 // half of 64 bits
mask = (1 << shift) - 1
a1 = a >> shift
a2 = a & mask
b1 =... |
1704/H2 | Game of AI (hard version) | 3,500 | combinatorics, fft, math | 12 | 1,024 | This is the hard version of this problem. The difference between easy and hard versions is the constraint on $$$k$$$ and the time limit. Notice that you need to calculate the answer for all positive integers $$$n \in [1,k]$$$ in this version. You can make hacks only if both versions of the problem are solved.
Cirno is... | The only line contains two positive integers $$$k$$$ and $$$M$$$ ($$$1\le k\le 10^5$$$, $$$2\le M\le 10^9$$$ ). It is guaranteed that $$$2^{18}$$$ is a divisor of $$$M-1$$$ and $$$M$$$ is a prime number. | Output $$$k$$$ lines, where the $$$i$$$-th line contains a non-negative integer, which is the answer for $$$n=i$$$ modulo $$$M$$$. | null | For $$$n=1$$$, no valid sequence $$$a$$$ exists. We regard the answer as $$$0$$$.
For $$$n=2$$$, there is only one possible array $$$a$$$: $$$[2, 1]$$$.
- For array $$$a$$$ is $$$[2, 1]$$$ and permutation $$$p$$$ is $$$[1, 2]$$$, the sequence $$$b$$$ will be $$$[1, 1]$$$ after all rounds have finished. The details fo... | [
{
"input": "8 998244353",
"output": "0\n2\n24\n360\n6800\n153150\n4057452\n123391016"
}
] | Given $b$, The conclusion of possible array $a$ remains. If we fix array $a$, we will find that for each $i$, $b_i$ will only be $j$ that $a_j=i$ or $j=i$. For any pair $(x,y)$, if $a_x=y$, we will find that $b_x=x$ and $b_y=y$ at the same time is impossible. All arrays $b$ which satisfies the above conditions are vali... | The key observation is that the sequence a defines a functional graph on n nodes, and the final configuration b is constrained by an independent set condition on this graph. The counting problem reduces to summing over all functional graphs a weighted by a product over in-degrees of nodes not in the independent set. Th... | {
"exploration": "I'll enumerate all possible sequences a and permutations p for small n, then count the resulting b sequences. For each n, loop over all n! permutations p and all (n-1)^n possible sequences a (since a_i ≠ i, so n-1 choices per position). For each pair (p, a), simulate the game and record b, then coun... | {
"exploration": "First, I need to understand what the DP state should capture. The process involves bots occupying towers, and bots being eliminated when they lose all towers. The final configuration b is determined by the choices of permutation p and attack targets a. The hypothesis suggests DP over attack order an... | {
"exploration": "I'll try to derive a recurrence for f(n) = number of valid (a,b) pairs for given n. The description says a is any sequence of length n with a_i ≠ i, and p is any permutation of [1..n]. The process then yields b. We need to count distinct (a,b) pairs across all p. For each n, I need to compute f(n) m... | null | null | null | So none of those panned out. Brute force exploded combinatorially, DP over states was astronomically large, and my vague recurrence didn't actually compute anything.
hmm — but that half-formed thought about functional graphs and independent sets — wait, let's actually build that. For a given n, sequence a defines a fu... | MOD = M
k = READ integer
M = READ integer
// Precompute factorials and inverse factorials up to 2*k
maxN = 2 * k
fact = array of size maxN+1
invFact = array of size maxN+1
fact[0] = 1
for i from 1 to maxN:
fact[i] = fact[i-1] * i % MOD
invFact[maxN] = pow(fact[maxN], MOD-2, MOD) // modular inverse
for i from max... |
1580/E | Railway Construction | 3,400 | brute force, constructive algorithms, data structures, graphs, shortest paths | 2.5 | 512 | Because the railway system in Gensokyo is often congested, as an enthusiastic engineer, Kawasiro Nitori plans to construct more railway to ease the congestion.
There are $$$n$$$ stations numbered from $$$1$$$ to $$$n$$$ and $$$m$$$ two-way railways in Gensokyo. Every two-way railway connects two different stations and... | The first line contains three integers $$$n$$$, $$$m$$$, and $$$q$$$ ($$$1 \le n \le 2 \cdot 10^5$$$, $$$1 \le m \le 3 \cdot 10^5$$$, $$$0 \le q \le 2\cdot10^5$$$).
The second line contains $$$n$$$ integers $$$w_1,w_2,\ldots,w_n$$$ ($$$1 \le w_i \le 10^9$$$).
Each of the next $$$m$$$ lines contains three integers $$$... | Print $$$q+1$$$ lines, and the $$$i$$$-th of these lines contains one integer, denoting the minimal cost of railway construction after the $$$i-1$$$-th incident (especially, the $$$0$$$-th incident means no incident occurred). | null | In the second example, Nitori can build railways as follows: $$$1 \rightarrow 2$$$, $$$1 \rightarrow 3$$$, $$$1 \rightarrow 4$$$, $$$2 \rightarrow 8$$$, and the cost is $$$14 + 14 + 14 + 4 = 46$$$. | [
{
"input": "5 5 1\n1 1 1 1 1\n1 2 1\n2 3 1\n2 4 1\n3 5 1\n4 5 1\n1 2",
"output": "3\n9"
},
{
"input": "8 11 0\n14 4 16 15 1 3 1 14\n4 2 1\n1 2 3\n7 5 4\n2 3 1\n8 6 2\n8 5 5\n5 4 5\n7 6 7\n3 5 5\n1 6 6\n8 1 4",
"output": "46"
},
{
"input": "10 16 8\n29 1 75 73 51 69 24 17 1 97\n1 2 18\n2 ... | For convenience, we first define $dis[u]$ as the length of the shortest path between node 1 and node $u$ and "distance" of node $u$ as $dis[u]$, and call node $u$ is "deeper" than node $v$ if and only if $dis[u] \gt dis[v]$. Similarly, we call node $u$ is "lower" than node $v$ if and only if $dis[u] \lt dis[v]$. We wil... | What if you consider the graph of all edges that lie on some shortest path from the source? The key observation is that the requirement for two vertex-disjoint shortest paths forces a specific structural condition on the in-degrees in this DAG. The minimal construction cost then becomes a greedy selection problem based... | {
"exploration": "Okay so we have n stations, m undirected edges with lengths, and we can add directed edges from any station u at cost w_u (plus incident increases). The goal: after construction, for every station t ≠ 1, there are at least two vertex-disjoint shortest paths from 1 to t, and the shortest distance fro... | {
"exploration": "I'll model this as a 2-SAT instance. First, compute shortest distances from station 1 to all others using Dijkstra on the original graph — that gives me dist[v] for each v. A potential directed edge u→v is valid only if dist[u] < dist[v], because adding it must not change any shortest path distances... | null | null | null | null | Alright, back to square one. The flow approach missed that a single added edge can serve multiple terminals, and the 2-SAT idea broke because the constraints aren't about each node independently — they're about the structure of the shortest-paths DAG as a whole.
hmm. That half-formed thought about edges on shortest pa... | READ n, m, q
READ w[1..n]
// Build graph and compute shortest distances from node 1
adj = adjacency list of size n+1
edges = list of m edges (u, v, d)
for i from 1 to m:
READ u, v, d
add (v, d) to adj[u]
add (u, d) to adj[v]
edges[i] = (u, v, d)
dist[1..n] = array initialized to INFINITY
dist[1] = 0
p... |
1943/E2 | MEX Game 2 (Hard Version) | 3,300 | binary search, greedy, two pointers | 2 | 256 | This is the hard version of the problem. The only difference between the two versions is the constraint on $$$t$$$, $$$m$$$ and the sum of $$$m$$$. You can make hacks only if both versions of the problem are solved.
Alice and Bob play yet another game on an array $$$a$$$ of size $$$n$$$. Alice starts with an empty arr... | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^5$$$) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers $$$m$$$ and $$$k$$$ ($$$1 \le m \le 2 \cdot 10^5, 1 \le k \le 10^9$$$).
The ... | For each test case, find Alice's score if both players play optimally. | null | In the first test case, the array $$$a$$$ is $$$[0, 0, 0, 0, 1, 1, 1, 1, 1]$$$. A possible game with a score of $$$2$$$ is as follows:
1. Alice chooses the element $$$0$$$. After this move, $$$a = [0, 0, 0, 1, 1, 1, 1, 1]$$$ and $$$c=[0]$$$.
2. Bob chooses to remove the $$$3$$$ elements $$$0$$$, $$$0$$$ and $$$1$$$. A... | [
{
"input": "5\n1 4\n4 5\n2 1000000000\n1000000000 1000000000 1000000000\n3 2\n2 3 100 1\n1 1\n2 2\n3 1\n1 1 1 1",
"output": "2\n1\n3\n2\n1"
}
] | **Solution**
Instead of doing the check for $ans \geq i$ in $O(m^3)$, we will do it in $O(m)$.
For an array $f$ of length $n$. Let $s=f_1+f_2+\ldots+f_n$ be the sum of the element. $f$ will be called flat if $f_i = \lfloor \frac{s+i-1}{n} \rfloor$. That is, it has the form $x, \ldots, x, x+1, \ldots x+1$. All flat ar... | What if you focus on the portion of the array Bob actually gets to affect? The key observation is that Bob's operations only meaningfully impact a suffix of the frequencies, and the resulting structure has a predictable, repeating pattern that you can analyze iteratively. | {
"exploration": "I'll consider a DP where state i represents the smallest MEX Alice can guarantee if we only consider elements from 0 to i, given the frequencies and Bob's deletions. The idea: process numbers in increasing order, and at each step, decide whether Alice can secure that number in her array c. Let dp[i]... | {
"exploration": "I'll binary search the answer X — can Alice guarantee MEX at least X? For a candidate X, Alice needs to collect all numbers from 0 to X-1 at least once each into c, while Bob tries to delete them before she can pick them. Bob can delete up to k elements per turn, but he only gets a turn after each o... | null | null | null | null | None of those got me anywhere useful — the per-value condition ignored that Bob's deletions can be spread across multiple values, and the min-cost flow approach collapsed under the combinatorial explosion of trying to allocate deletions per turn.
hmm. Wait — Bob's deletions only matter for values Alice actually needs ... | function solve():
READ t
for each test case:
READ m, k
READ array f of length m+1
// compute max_pos for each value v
max_pos = array of size m+1
for v from 0 to m:
max_pos[v] = (f[v] + k - 1) // k // ceil(f[v] / k)
// build min-heap of pairs (v, max... |
1949/A | Grove | 3,300 | brute force, dfs and similar, dp, geometry, probabilities | 4 | 256 | You want to plant trees in a square lawn of size $$$n \times n$$$ whose corners have Cartesian coordinates $$$(0, 0)$$$, $$$(n, 0)$$$, $$$(0, n)$$$, and $$$(n, n)$$$. Trees can only be planted at locations with integer coordinates. Every tree will grow roots within a disk of radius $$$r$$$ centered at the location wher... | The first and only line contains an integer $$$n$$$ ($$$1 \leq n \leq 20$$$) and a real number $$$r$$$ ($$$0 < r \leq n/2$$$) — the length of the sides of the lawn, and the radius of the disks where each tree will grow roots. The real number $$$r$$$ is given in decimal notation with at least $$$1$$$ and at most $$$3$$$... | In the first line, print the maximum number $$$m$$$ of trees that can be planted.
In the next $$$m$$$ lines, print a configuration that maximizes the number of trees. Specifically, in the $$$(i+1)$$$-th line, print two integers $$$x$$$ and $$$y$$$ — the coordinates of the location where the $$$i$$$-th tree should be p... | null | For the first sample, the sample output is shown in the following figure. Note that this is not the only configuration that maximizes the number of trees.
For the second sample, the sample output is shown in the following figure. Note that this is not the only configuration that maximizes the number of trees. | [
{
"input": "6 1.241",
"output": "2\n4 2\n2 4"
},
{
"input": "9 2.0",
"output": "4\n2 2\n7 2\n2 6\n6 6"
}
] | Let $\delta = \lceil r \rceil$. Note that all trees need to be planted at least $\delta$ away from the boundary of the lawn, so their coordinates $(x, y)$ have to satisfy $\delta \leq x, y \leq n-\delta$.
For integers $0 \leq a_\delta, \dots, a_{n-\delta} \leq n-\delta$, denote by $f(a_\delta, \dots, a_{n-\delta})$ any... | The key observation is that the problem has a strong row-by-row structure you can exploit. What if you fix an ordering on the rows and consider placing trees from right to left? This leads to a recursive state that depends only on the current frontier of allowed x-coordinates for each row. | {
"exploration": "Okay so we have an n×n square lawn, n up to 20, and a radius r up to n/2. Trees go at integer coordinates, their disks of radius r must be fully inside the lawn and disks can only touch at boundaries — meaning the Euclidean distance between any two tree centers must be at least 2r. We need to maximi... | {
"exploration": "I'll scan rows left-to-right, bottom-to-top first. Points must be at integer coordinates (x,y) with both x and y between 0 and n inclusive. The disk of radius r must be fully inside the lawn, so the center must be at least r from each boundary: x ∈ [r, n-r], y ∈ [r, n-r]. Two disks can touch at boun... | {
"exploration": "I'll construct the graph explicitly. For a given n and r, the candidate points are all integer coordinates (x, y) such that the disk of radius r centered at (x, y) is fully contained in the [0, n] × [0, n] square. That means x must satisfy r ≤ x ≤ n - r, and similarly for y. Since coordinates must b... | null | null | null | None of those got me anywhere useful — brute force is astronomically huge, greedy locks in bad placements, and treating it as a general MIS on 361 vertices is hopelessly exponential. The common failure is trying to reason about the whole grid at once; the constraints are too many.
Wait — that half-formed thought about... | function solve():
READ n, r
delta = ceil(r)
left = delta
right = n - delta
if left > right:
PRINT 0
return
// plantable rows and columns are the same set
rows = list from left to right inclusive
cols = list from left to right inclusive
R = length(rows)
C = length... |
1023/G | Pisces | 3,400 | data structures, flows, trees | 5 | 512 | A group of researchers are studying fish population in a natural system of lakes and rivers. The system contains $$$n$$$ lakes connected by $$$n - 1$$$ rivers. Each river has integer length (in kilometers) and can be traversed in both directions. It is possible to travel between any pair of lakes by traversing the rive... | The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 10^5$$$) — the number of lakes in the system.
The next $$$n - 1$$$ lines describe the rivers. The $$$i$$$-th of these lines contains three integers $$$u_i$$$, $$$v_i$$$, $$$l_i$$$ ($$$1 \leq u_i, v_i \leq n$$$, $$$u_i \neq v_i$$$, $$$1 \leq l_i \leq 10... | Print one integer — the smallest total number of fish not contradicting the observations. | null | In the first example, there could be one fish swimming through lakes $$$2$$$, $$$1$$$, and $$$4$$$, and the second fish swimming through lakes $$$3$$$, $$$1$$$, and $$$2$$$.
In the second example a single fish can not possibly be part of all observations simultaneously, but two fish swimming $$$2 \to 1 \to 4$$$ and $$... | [
{
"input": "4\n1 2 1\n1 3 1\n1 4 1\n5\n1 1 2\n1 1 3\n2 2 1\n3 1 4\n3 1 2",
"output": "2"
},
{
"input": "5\n1 2 1\n1 3 1\n1 4 1\n1 5 1\n4\n1 1 2\n2 1 3\n3 1 4\n4 1 5",
"output": "2"
},
{
"input": "5\n2 5 1\n5 1 1\n2 4 1\n5 3 3\n6\n5 2 4\n2 1 1\n2 1 3\n2 2 4\n4 7 5\n4 1 2",
"output": "... | First, how do we find the answer in any time complexity? Let us construct a partially ordered set where each element is a single fish in one of the observations. For two elements $x = (v_1, d_1)$ and $y = (v_2, d_2)$ we put $x < y$ if $x$ and $y$ could possibly be two occurences of the same fish one after the other, th... | What if you think of each fish as a chain of compatible observations? The key observation is that the minimal number of fish equals the size of a largest set of observations that cannot be connected by any single fish's travel. This leads to a tree DP that tracks, for each vertex, the best 'forbidden' time intervals. T... | {
"exploration": "We have a tree with n ≤ 10^5 lakes and k ≤ 10^5 observations. Each observation says at day d_j, lake p_j must have at least f_j distinct fish. Fish are indistinguishable, can travel at speed 1 km/day along tree edges, and can wait anywhere. We need the minimum total number of fish consistent with al... | {
"exploration": "I'll construct a flow network with k observation nodes plus a source S and sink T. Each observation j has a demand of f_j fish that must pass through it. I'll add a lower bound of f_j on the flow into node j. To model fish movement: if a fish can travel from observation i at (d_i, p_i) to observatio... | {
"exploration": "I'll binary search on the total number of fish F, and for a given F, check if all observations can be satisfied using Hall's marriage theorem. The idea is to model each fish as a \"resource\" that can be at certain lakes at certain times, and each observation as a demand for f_j distinct fish at lak... | null | null | null | Well, every path hit a computational wall — greedy assignment was quadratic, flow was impossible to build, and Hall's condition checking exploded the network. The core issue is that pairwise reachability between observations creates O(k²) edges, which is fundamentally too many.
hmm. That half-formed thought about fish... | function solve():
READ n
adj = array of size n+1 of empty lists
for i from 1 to n-1:
READ u, v, l
adj[u].append((v, l))
adj[v].append((u, l))
READ k
observations = array of size n+1 of empty lists
for i from 1 to k:
READ d, f, p
observations[p].append((d,... |
1764/H | Doremy's Paint 2 | 3,400 | data structures | 4 | 256 | Doremy has $$$n$$$ buckets of paint which is represented by an array $$$a$$$ of length $$$n$$$. Bucket $$$i$$$ contains paint with color $$$a_i$$$. Initially, $$$a_i=i$$$.
Doremy has $$$m$$$ segments $$$[l_i,r_i]$$$ ($$$1 \le l_i \le r_i \le n$$$). Each segment describes an operation. Operation $$$i$$$ is performed as... | The first line of input contains three integers $$$n$$$, $$$m$$$, and $$$k$$$ ($$$1\le n,m\le 2\cdot 10^5$$$, $$$1 \le k \le m$$$) — the length of the array $$$a$$$, the total number of operations, and the integer that Doremy selects.
The $$$i$$$-th line of the following $$$m$$$ lines contains two integers $$$l_i$$$, ... | Output $$$m$$$ integers. The $$$(x+1)$$$-th integer should be the number of distinct colors in the array if we start from the initial array and perform operations $$$x \bmod m +1, (x+1) \bmod m + 1, \ldots, (x+k-1) \bmod m +1$$$. | null | In the first test case, the picture below shows the resulting array for the values of $$$x=0,1,2$$$ respectively. | [
{
"input": "7 5 5\n3 5\n2 3\n4 6\n5 7\n1 2",
"output": "3 3 3 3 2"
},
{
"input": "10 9 4\n1 1\n2 3\n3 4\n7 9\n6 8\n5 7\n2 4\n9 10\n1 3",
"output": "6 6 7 6 5 5 7 7 7"
}
] | The main idea of the solution is that we can try to the answers for position $x,x+1,\ldots,x+k$ all at once. We do this by representing the interval $[x+i,x+k+i)$ as $[x+i,x+k) \cup [x+k,x+k+i)$ and perform sweep line on both halves of interval.
Firstly, after applying some operations on the array $a$, we will have $a_... | What if you think about the array as a sequence of non-decreasing values and track when adjacent values differ? The key observation is that applying an operation merges a range of values into a single equivalence class. You can compute the effect of adding or removing an operation from a block by updating these classes... | {
"exploration": "Okay so we have n buckets initially a[i]=i, m operations each of the form (l, r) meaning for all j in (l, r] set a[j] = a[l]. We need to answer m queries: for each starting offset x from 0 to m-1, apply k consecutive operations (wrapping around modulo m) to the initial array and count distinct color... | {
"exploration": "I'll treat each operation as a function f_i: [1..n] → [1..n] that maps each position to its new color. Initially a_j = j, so the identity function is f_0(j) = j. Operation i with [l_i, r_i] sets a_j := a_{l_i} for all j in (l_i, r_i]. This means f_i(j) = f_{i-1}(l_i) if l_i < j ≤ r_i, and f_i(j) = f... | null | null | null | null | Well, every path hit the same wall: trying to simulate all m windows directly, even with clever data structures, still ends up O(m²) because each window is a fresh start from the initial array. Binary lifting on operations seemed promising, but representing the state of the entire array after composition is too heavy —... | function solve():
READ n, m, k
l = array of size m
r = array of size m
for i from 0 to m-1:
READ l[i], r[i]
// discretization using all l and r from relevant windows
// We will process windows in chunks of size k, each window [x, x+k)
// For each chunk, we consider operations in [x,... |
1870/G | MEXanization | 3,300 | data structures | 2 | 512 | Let's define $$$f(S)$$$. Let $$$S$$$ be a multiset (i.e., it can contain repeated elements) of non-negative integers. In one operation, you can choose any non-empty subset of $$$S$$$ (which can also contain repeated elements), remove this subset (all elements in it) from $$$S$$$, and add the MEX of the removed subset t... | The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then follows the description of the test cases.
The first line of each test case contains an integer $$$n$$$ ($$$1 \leq n \leq 2 \cdot 10^5$$$) — the size of array $$$a$$$.
The second line of each test case contain... | For each test case, output $$$n$$$ numbers: $$$f(S)$$$ for each of the $$$n$$$ prefixes of array $$$a$$$. | null | Consider the first test case. For a prefix of length $$$1$$$, the initial multiset is $$$\{179\}$$$. If we do nothing, we get $$$179$$$.
For a prefix of length $$$2$$$, the initial multiset is $$$\{57, 179\}$$$. Using the following sequence of operations, we can obtain $$$2$$$.
1. Apply the operation to $$$\{57\}$$$,... | [
{
"input": "4\n8\n179 57 2 0 2 3 2 3\n1\n0\n3\n1 0 3\n8\n1 0 1 2 4 3 0 2",
"output": "179 2 3 3 3 4 4 5 \n1 \n1 2 2 \n1 2 2 3 3 5 5 5"
}
] | Let's start by introducing a more convenient way to store numbers - an array $cnt$, where $cnt[x]$ represents the number of occurrences of $x$ in the prefix for which we are finding the answer. All $x$ such that $x > n$ will be added to $cnt[0]$ because applying the operation to $\{x\}$ in such cases is optimal.
Let's ... | What if you think about the multiset as a collection of resources? The key observation is that you can simulate a process of repeatedly 'creating' the smallest missing number by consuming available smaller numbers, but you need to track deficits that propagate leftwards. This leads to a greedy, incremental construction... | {
"exploration": "Okay so we have an operation on a multiset S: pick any non-empty subset, remove it, and add its MEX back. We can repeat this until one number remains, and f(S) is the maximum possible final number. We need f for each prefix of array a. The constraint: sum of n over all test cases ≤ 2×10^5, and each ... | {
"exploration": "I'll model each prefix's multiset as a flow network where each integer value is a node, and we have a source connected to each value node with capacity equal to its frequency in the prefix, and each value node connects to a sink through a node that represents \"taking a subset\" — but the operation ... | null | null | null | null | Okay, so I've burned through a few ideas and none stuck. Direct simulation is impossible, and my attempts to model it as a simple greedy on frequencies kept failing because the operations are non-local—you can combine elements from anywhere, and the MEX of a subset can create a new element that changes the entire resou... | function solve():
READ t
for each test case:
READ n
READ array a of size n
// cnt[x] will store frequency of x in current prefix
// We only need up to n because any x > n can be treated as 0
cnt = array of size n+1, initialized to 0
// extra counts for numbers > n... |
2053/G | Naive String Splits | 3,400 | binary search,brute force,greedy,hashing,math,number theory,strings | null | null | And I will: love the world that you've adored; wish the smile that you've longed for. Your hand in mine as we explore, please take me to tomorrow's shore.
— Faye Wong, As Wished
Cocoly has a string $$$t$$$ of length $$$m$$$, consisting of lowercase English letters, and he would like to split it into parts. He calls a... | Each test contains multiple test cases. The first line contains an integer $$$T$$$ ($$$1 \leq T \leq 10^5$$$) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers $$$n$$$ and $$$m$$$ ($$$2 \leq n \leq m \leq 5 \cdot 10^6$$$) — the lengths of $$$s... | For each test case, output a single binary string $$$r$$$ of length $$$n - 1$$$: for each $$$1 \leq i \lt n$$$, if the $$$i$$$-th pair is beautiful, $$$r_i=\texttt{1}$$$; otherwise, $$$r_i=\texttt{0}$$$. Do not output spaces. | null | null | [] | Let's call the prefix $s_1$ the short string and the suffix $s_2$ the long string. If $\lvert s_1\rvert\gt \lvert s_2\rvert$, swapping the order doesn't affect the answer.
Consider a greedy approach. We first match a few short strings until we can't match anymore. If we can't match, we try discarding a few short strin... | The key observation is that the matching process can be viewed as greedily consuming the string t with the shorter piece, and when you get stuck, you only need to consider a constant number of specific backtracking points. The structure of how the longer string relates to the shorter one—specifically, how many copies o... | {
"exploration": "Okay so we have string s of length n, string t of length m. For each split point i from 1 to n-1, we form x = s[1..i] and y = s[i+1..n], and we need to check if t can be partitioned into pieces that are each either x or y. The constraint: sum of m over all test cases ≤ 10^7, n ≤ m ≤ 5×10^6 per test ... | {
"exploration": "For each split point i from 1 to n-1, I need to check whether the pair (x = s[0..i-1], y = s[i..n-1]) can partition t. The lengths are len_x = i, len_y = n-i. The condition is: there exist non-negative integers a, b such that a·len_x + b·len_y = m, AND there exists an interleaving of a copies of x a... | {
"exploration": "For each split point i from 1 to n-1, I'll take x = s[0..i-1] and y = s[i..n-1], then build an Aho-Corasick automaton with these two patterns. The automaton has a trie for x and y, failure links, and output flags marking which pattern(s) end at each node. Then I simulate t through the automaton whil... | null | null | null | Alright, back to square one I guess. Every path hit a wall: DP was quadratic, BFS on remainders exploded combinatorially, and backtracking without memoization was exponential. The core issue is that checking all interleavings of two strings to see if they can tile t is fundamentally exponential if you don't exploit str... | function solve():
READ T
Precompute pw[0..MAXM] where pw[i] = base^i mod P
for each test case:
READ n, m
READ s (length n)
READ t (length m)
Prepend ' ' to s and t for 1-indexing: s = ' ' + s, t = ' ' + t
Precompute hash arrays S[0..n], T[0..m]:
S[0] = 0, ... |
1930/H | Interactive Mex Tree | 3,300 | constructive algorithms, dfs and similar, interactive, trees | 5 | 256 | This is an interactive problem.
Alice has a tree $$$T$$$ consisting of $$$n$$$ nodes, numbered from $$$1$$$ to $$$n$$$. Alice will show $$$T$$$ to Bob. After observing $$$T$$$, Bob needs to tell Alice two permutations $$$p_1$$$ and $$$p_2$$$ of $$$[1, 2, \ldots, n]$$$.
Then, Alice will play $$$q$$$ rounds of the foll... | null | null | Each test contains multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Read it. The description of the test cases follows.
The first line of each test case contains two positive integers $$$n$$$ and $$$q$$$ ($$$2 \leq n \leq 10^5$$$, $$$1 \leq q ... | In the first test, the interaction proceeds as follows.
SolutionJuryExplanation1There are 1 test cases.3 1The tree $$$T$$$ consists of $$$3$$$ nodes, and Alice will play for only one round.1 2First edge of $$$T$$$2 3Second edge of $$$T$$$1 2 3The permutation $$$p_1$$$2 1 3The permutation $$$p_2$$$Alice shuffles $$$a$$... | [
{
"input": "1\n3 1\n1 2\n2 3\n\n\n2 3\n\n1\n\n0\n\n1",
"output": "1 2 3\n2 1 3\n\n? 1 2 3\n\n? 2 1 3\n\n! 0"
}
] | **Hint 1**
$\operatorname{MEX}$ of the path from $u$ to $v$ will be the minimum value over all the nodes of $T$ which do not lie on the path from $u$ to $v$.
**Hint 2**
$p_1$ and $p_2$ are associated with the Euler tour
**Hint 3**
$p_1$ is the permutation of $[1,2, \ldots n]$ sorted in increasing order on the basi... | The key observation is that the MEX is the minimum value outside the path. Think about how to partition the nodes outside a given path into a constant number of groups that correspond to contiguous ranges in some ordering of the tree. This suggests using Euler tour properties to design the two permutations. | {
"exploration": "I'll try to find the MEX on the path between u and v by using the queries to check candidate values one by one. The MEX is the smallest non-negative integer not present on the path. Since values are a permutation of 0..n-1, the MEX is at most n. I can test candidate x from 0 upward: if x is on the p... | {
"exploration": "I'll build p1 and p2 from centroid decomposition. First, compute the centroid tree: recursively find centroid of the current component, make it root of that subtree in the centroid tree, then recurse on each connected component after removing the centroid. For each centroid c, I'll store the list of... | null | null | null | null | Okay, so both random sampling and centroid decomposition failed because they couldn't isolate the path's complement—the values not on the path—with a constant number of queries. The random approach couldn't guarantee contiguous segments, and the decomposition couldn't distinguish off-path values from the minimum alone.... | function main():
READ t
for test_case from 1 to t:
READ n, q
adj = array of size n+1 of empty lists
for i from 1 to n-1:
READ u, v
adj[u].append(v)
adj[v].append(u)
// First DFS from root 1
tin1 = array of size n+1
tout1 = arra... |
2127/G1 | Inter Active (Easy Version) | 3,400 | binary search,constructive algorithms,interactive,probabilities | null | null | This is the easy version of the problem. The difference between the versions is that in this version, you can make at most $$$15\cdot n$$$ queries. You can hack only if you solved all versions of this problem.
Ali loved Bahamin's gift (from problem E) so much that he illegally traveled from Qazvin to Liverpool to have... | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 500$$$). The description of the test cases follows.
The only line of each test case contains a single integer $$$n$$$ ($$$4 \leq n \leq 100$$$) — the length of $$$p$$$.
It is guaranteed that the sum of $$$... | null | null | null | [] | **Hints**
**Step 0**
Consider the following graph: $i\to p_i$. We will see the permutation as the graph.
When asking a permutation $q$, the result is sum of whether $q^{-1}(i) \lt q^{-1}(p_i)$.
**Hint 1**
Why is there a condition about $k$? If we do not have $k$, can the permutation $[2, 1, 4, 3]$ be guessed out?
... | The key observation is that you can design a pair of complementary queries to isolate a single edge in the permutation graph. What if you fix the target value at a special position and then compare the total count of 'forward' edges under two opposite orderings of the remaining candidates? This hints at a binary search... | {
"exploration": "Okay so we have a hidden permutation p of length n ≥ 4 with no fixed points (p_i ≠ i for all i). We first choose a constant k ≤ n, then we can query with any permutation q. The response is the number of pairs (i, j) with i < j, p_{q_i} = q_j, and i ≠ k. So the query counts edges in the functional gr... | {
"exploration": "I'll pick k = n, and use the identity permutation as baseline. Let me trace through what happens with a small concrete example to understand the query response.\n\nLet n = 4, and suppose the hidden permutation p = [2, 4, 1, 3] (meaning p₁=2, p₂=4, p₃=1, p₄=3). Note p_i ≠ i holds.\n\nI set k = n = 4.... | {
"exploration": "I'll set k=1 for simplicity, so the query ignores the first position in the query permutation q. For each query, I'll construct q as a specific permutation designed to extract information about p. The idea is to treat each query as a linear equation: the response is the count of pairs (i,j) with i<j... | null | null | null | So none of those panned out. Every approach that tried to directly measure edges or solve equations needed O(n²) queries, blowing past the 15n budget by a factor of six or more.
Hmm. That nagging thread about isolating a single edge by comparing two complementary queries... wait — what if we could, for a fixed value v... | function solve():
READ t
for each test case:
READ n
k = floor((n + 1) / 4) + 1
PRINT k // output k as required by interaction
p = array of size n+1 // 1-indexed, to store final permutation
for v from 1 to n:
candidates = list of all values from 1 to n exce... |
1817/E | Half-sum | 3,400 | brute force, divide and conquer, greedy | 4 | 256 | You're given a multiset of non-negative integers $$$\{a_1, a_2, \dots, a_n\}$$$.
In one step you take two elements $$$x$$$ and $$$y$$$ of the multiset, remove them and insert their mean value $$$\frac{x + y}{2}$$$ back into the multiset.
You repeat the step described above until you are left with only two numbers $$$... | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 100$$$). Description of the test cases follows.
The first line of each test case contains a single integer $$$n$$$ ($$$2 \le n \le 10^6$$$) — the size of the multiset.
The second line contains $$$n$$$ inte... | For each test case, output a single integer, the answer to the problem modulo $$$10^9+7$$$.
Formally, let $$$M = 10^9+7$$$. It can be shown that the answer can be expressed as an irreducible fraction $$$\frac{p}{q}$$$, where $$$p$$$ and $$$q$$$ are integers and $$$q \not \equiv 0 \pmod{M}$$$. Output the integer equal ... | null | In the first case, you can't do any operations, so the answer is $$$|7-3|=4$$$.
In the second case, one of the optimal sequence of operations:
1. Substitute $$$1$$$ and $$$2$$$ with $$$1.5$$$;
2. Substitute $$$10$$$ and $$$11$$$ with $$$10.5$$$;
3. The difference between $$$1.5$$$ and $$$10.5$$$ is $$$9$$$.
In the t... | [
{
"input": "5\n2\n7 3\n4\n1 2 10 11\n3\n1 2 3\n6\n64 32 64 16 64 0\n4\n1 1 1 1",
"output": "4\n9\n500000005\n59\n0"
}
] | Similar to the Huffman encoding, the process described in the statement can be represented as constructing a tree, in which on every step we take two trees with weights $a$ and $b$, and connect them with a shared root into a single tree with weight $\frac{a+b}{2}$.
In the end of the process, we have two trees with weig... | The key observation is that the process is analogous to building two binary trees where each leaf's contribution is scaled by a power of 1/2. The structure of an optimal tree is highly unbalanced. Think about sorting the numbers and trying all possible ways to split them into two groups, then computing the weighted sum... | {
"exploration": "Okay so we have a multiset of n numbers, and we repeatedly take two numbers, remove them, and insert their average. We do this until only two numbers remain, and we want to maximize the absolute difference between those final two numbers. The constraint is sum of n over all test cases ≤ 10^6, time 4... | {
"exploration": "I'll represent the merging process as a binary tree where leaves are the original numbers and internal nodes are means. The final two numbers A and B correspond to the two children of the root — each is the mean of some subset of leaves. The operation is associative in the sense that any binary tree... | null | null | null | null | Okay, all of those went nowhere. Greedy pairing collapsed the spread, and simulated annealing couldn't even scratch the surface of the state space.
hmm. But that thread about binary trees and powers of 1/2... wait — let me formalize it. Each time I take the mean, I'm essentially building a node whose value is the aver... | function solve():
MOD = 10^9 + 7
inv2 = (MOD + 1) // 2 // modular inverse of 2
READ t
for test_case from 1 to t:
READ n
READ a[1..n]
sort a ascending
// prefix sums
pre = array of size n+1, pre[0] = 0
for i from 1 to n:
pre[i] = (pre[i-1] + a... |
1148/G | Gold Experience | 3,300 | constructive algorithms, graphs, math, number theory, probabilities | 2 | 1,024 | Consider an undirected graph $$$G$$$ with $$$n$$$ vertices. There is a value $$$a_i$$$ in each vertex.
Two vertices $$$i$$$ and $$$j$$$ are connected with an edge if and only if $$$gcd(a_i, a_j) > 1$$$, where $$$gcd(x, y)$$$ denotes the greatest common divisor (GCD) of integers $$$x$$$ and $$$y$$$.
Consider a set of ... | The first line contains integers $$$n$$$ and $$$k$$$ ($$$6 \leq 2 \cdot k \leq n \leq 10^5$$$) — the number of vertices and parameter $$$k$$$.
The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$2 \le a_i \le 10^7$$$) — the values in the vertices. | Print exactly $$$k$$$ distinct integers — the indices of the vertices in the chosen set in any order. | null | In the first test case, set $$$\{2, 4, 5\}$$$ is an example of set where no vertices are fair. The vertex $$$2$$$ does not share an edge with vertex $$$4$$$ since $$$gcd(15, 8) = 1$$$. The vertex $$$4$$$ does not share an edge with vertex $$$2$$$. The vertex $$$5$$$ does not share an edge with vertex $$$2$$$.
In the s... | [
{
"input": "6 3\n6 15 10 8 14 12",
"output": "1 3 6"
},
{
"input": "8 4\n11 15 10 6 21 15 10 6",
"output": "1 2 3 4"
},
{
"input": "10 5\n3003 17017 3230 49742 546 41990 17765 570 21945 36465",
"output": "4 6 9 10 1"
}
] | Firstly, let's discuss how to find the answer in $O(N^2)$ time. Let's assume $k$ is even. Pick any set on $k$ vertices. If it is a clique, just print it as an answer, if it is not, there must be two vertices in this set that don't share an edge. Just remove these two vertices and replace them with 2 arbitrary ones from... | The key observation is to first find a small 'seed' of vertices that are not pairwise connected in a specific chain. Then, the rest of the construction reduces to a monotonic property about the size of independent sets in the complement graph, which you can determine efficiently using inclusion-exclusion over prime fac... | {
"exploration": "I'll start with a single vertex and greedily expand to a clique — pick a vertex, then repeatedly add any vertex that's connected to all vertices currently in the set. If the clique reaches size k, we're done with all-fair. If it stalls before k, maybe the complement idea works.\n\nLet me build the g... | {
"exploration": "Okay so we have n vertices, each with a value a_i up to 10^7. Two vertices are connected if gcd(a_i, a_j) > 1. We need to pick exactly k vertices (2k ≤ n ≤ 10^5) such that either all k are pairwise connected (a clique in the GCD graph) or all k are pairwise disconnected (an independent set). The pro... | null | null | null | null | None of those got me anywhere useful. The greedy clique expansion kept locking in vertices that blocked later essential ones, and the random sampling was computationally impossible for k up to 50,000 — evaluating all pairs in a candidate set would be O(k²), which is far beyond the limit.
hmm. The problem guarantees a ... | function solve():
READ n, k
READ a[1..n]
// Precompute smallest prime factor up to 10^7
MAX_A = 10^7
spf = array of size MAX_A+1
for i from 2 to MAX_A:
if spf[i] == 0:
spf[i] = i
for j from i*i to MAX_A step i:
if spf[j] == 0:
... |
1190/F | Tokitsukaze and Powers | 3,400 | number theory, probabilities | 3 | 256 | Tokitsukaze is playing a room escape game designed by SkywalkerT. In this game, she needs to find out hidden clues in the room to reveal a way to escape.
After a while, she realizes that the only way to run away is to open the digital door lock since she accidentally went into a secret compartment and found some clues... | The only line contains three integers $$$n$$$, $$$m$$$ and $$$p$$$ ($$$1 \leq n \leq 5 \times 10^5$$$, $$$1 \leq p < m \leq 10^{18}$$$).
It is guaranteed that $$$m$$$ is a positive integer power of a single prime number. | If the number of possible different passwords is less than $$$n$$$, print a single integer $$$-1$$$.
Otherwise, print $$$n$$$ distinct integers ranged from $$$0$$$ to $$$(m - 1)$$$ as passwords. You can print these integers in any order. Besides, if there are multiple solutions, print any. | null | In the first example, there is no possible password.
In each of the last three examples, the given integer $$$n$$$ equals to the number of possible different passwords for the given integers $$$m$$$ and $$$p$$$, so if the order of numbers in the output is ignored, the solution is unique as shown above. | [
{
"input": "1 2 1",
"output": "-1"
},
{
"input": "3 5 1",
"output": "2 4 3"
},
{
"input": "2 5 4",
"output": "2 3"
},
{
"input": "4 9 8",
"output": "2 4 7 5"
}
] | Firstly, let me briefly review this problem for you. Given integers $n$, $m$ and $p$ ($n > 0$, $m > 1$, $p \neq 0$), we denote $S = \{x | x \in \mathbb{Z}, 0 \leq x < m, \gcd(m, x) = 1\}$ and $T = \{p^e \bmod m | e \in \mathbb{Z}, e \geq 0\}$, where $\gcd(m, x)$ means the greatest common divisor of $m$ and $x$, and you... | The key observation is that the forbidden set T forms a multiplicative subgroup of the group of units modulo m. The size of this subgroup is the order of p modulo m, which divides φ(m). The problem reduces to finding elements in the group of units that are not in this subgroup. This suggests thinking in terms of group ... | {
"exploration": "Okay so we have a number m that is a positive integer power of a single prime — so m = q^t for some prime q and t ≥ 1. The passwords are integers in [0, m-1] that must be coprime to m (so not divisible by q) and also must NOT be of the form p^e mod m for any non-negative integers e, k (i.e., x canno... | {
"exploration": "I'll build a hash table of all p^e mod m for e from 0 up to some bound, then for each candidate x from 0 to m-1, first check gcd(x,m)==1, then check if x is in the hash table (meaning it's a power of p). The valid passwords are those coprime to m and NOT in the hash table. Then I need to output the ... | {
"exploration": "I'll randomly sample integers x in [0, m-1], check if gcd(x, m) > 1 — if so, skip. For each candidate coprime to m, I need to test whether x is in the subgroup generated by p, i.e., whether there exists some e such that p^e ≡ x (mod m). The idea is: in a cyclic group of order φ(m), an element x is i... | {
"exploration": "I'll decompose the problem modulo m = q^k into subproblems modulo q and modulo q^{k-1}. Since m is a prime power, the Chinese Remainder Theorem doesn't directly apply — CRT requires coprime moduli, and q and q^{k-1} share factor q. But I can use the structure of Z/mZ: every residue x mod q^k can be ... | null | null | Alright, back to square one. Every path failed because I kept trying to directly enumerate or test elements against the exponential condition, which is either impossible at this scale or mathematically flawed for non-cyclic groups. The prime factor reconstruction missed that 1 is always forbidden, the binary search nee... | function mod_mul(a, b, m):
// returns (a * b) % m for 64-bit m, using 128-bit if available or split multiplication
if m <= 2^63:
return (a * b) % m
else:
// use binary splitting or built-in __int128
return ((__int128)a * b) % m
function mod_pow(base, exp, m):
result = 1 % m
... |
1887/F | Minimum Segments | 3,400 | constructive algorithms | 1 | 256 | You had a sequence $$$a_1, a_2, \ldots, a_n$$$ consisting of integers from $$$1$$$ to $$$n$$$, not necessarily distinct. For some unknown reason, you decided to calculate the following characteristic of the sequence:
- Let $$$r_i$$$ ($$$1 \le i \le n$$$) be the smallest $$$j \ge i$$$ such that on the subsegment $$$a_i... | Each test consist of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases. The description of test cases follows.
The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the length of the lost sequence $$... | For each test case, output the following:
- If there is no sequence $$$a$$$ with the characteristic $$$r$$$, print "No".
- Otherwise, print "Yes" on the first line, and on the second line, print any sequence of integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$) that matches the characteristic $$$r$$$. | null | In the first test case, the sequence $$$a = [1, 2, 1]$$$ is suitable. The integers $$$1$$$ and $$$2$$$ appear on the subsegments $$$[1, 2]$$$ and $$$[2, 3]$$$.
In the second test case, it can be proved that there is no suitable sequence $$$a$$$. | [
{
"input": "5\n3\n2 3 4\n5\n2 3 5 4 6\n1\n1\n3\n1 3 4\n8\n3 6 6 6 8 9 9 9",
"output": "Yes\n1 2 1\nNo\nYes\n1 \nNo\nYes\n1 3 5 3 5 1 1 3"
}
] | Let's consider a sequence $a_1, a_2, \ldots, a_n$. Let $nxt_i$ be the smallest $j > i$ such that $a_j = a_i$, or $n+1$ if such $j$ does not exist. It is claimed that the characteristic of the sequence can be uniquely determined from the values of $nxt_1, nxt_2, \ldots, nxt_n$. Let's write down the conditions on $nxt_i$... | What if you think about the next occurrence of each element? The key observation is that the characteristic imposes strict interval constraints on where the next equal element for each position can appear. A greedy assignment respecting these intervals, while ensuring every number beyond a certain point is used as a ta... | {
"exploration": "Okay so we have an array r[1..n] where r_i is the smallest j ≥ i such that the subsegment a[i..j] contains every distinct number that appears anywhere in the full sequence a. If no such j exists, r_i = n+1. We need to reconstruct any valid a[1..n] with values in [1,n] that produces exactly this r, o... | {
"exploration": "For each i from 1 to n, I need to check if the segment [i, r_i-1] contains all distinct numbers from the original sequence a. The approach is to build a segment tree that tracks, for each position j, the last occurrence index of the value a_j. Then the condition \"all distinct numbers appear in [i, ... | {
"exploration": "For each i from 1 to n, we need to assign a value a_i such that the smallest j≥i covering all distinct numbers is exactly r_i. The flow idea: treat each distinct value as a token that must appear at least once in the sequence. The r_i condition forces that by position r_i, all tokens must have been ... | null | null | null | So none of those panned out. The direct greedy assignment ignored the structure of r entirely, and backtracking is impossible for n up to 2×10^5.
hmm. That half-formed thought — about next occurrences and interval constraints — is tugging at me. r_i tells me the earliest index where all distinct numbers from the whole... | function solve():
READ t
for each test case:
READ n
READ r[1..n]
// Check if r is non-decreasing
possible = true
for i from 1 to n-1:
if r[i] > r[i+1]:
possible = false
if not possible:
PRINT "No"
contin... |
1896/H1 | Cyclic Hamming (Easy Version) | 3,400 | brute force, divide and conquer, dp, fft, math, number theory | 2 | 256 | This is the easy version of the problem. The only difference between the two versions is the constraint on $$$k$$$. You can make hacks only if all versions of the problem are solved.
In this statement, all strings are $$$0$$$-indexed.
For two strings $$$a$$$, $$$b$$$ of the same length $$$p$$$, we define the followin... | The first line of the input contains a single integer $$$k$$$ ($$$1 \le k \le 7$$$).
The second line of the input contains string $$$s$$$ of size $$$2^{k+1}$$$, consisting of the characters '0', '1' and '?'.
The third line of the input contains string $$$t$$$ of size $$$2^{k+1}$$$, consisting of the characters '0', '... | Print a single integer — the answer to the problem modulo $$$998\,244\,353$$$. | null | In the first example, we can check that the condition $$$h(s, c) \ge 2^k$$$ for all cyclic shift $$$c$$$ of $$$t$$$ is satisfied. In particular:
- for $$$c = \mathtt{0101}$$$, $$$h(s, c) = h(\mathtt{0110}, \mathtt{0101}) = 2 \ge 2^1$$$;
- for $$$c = \mathtt{1010}$$$, $$$h(s, c) = h(\mathtt{0110}, \mathtt{1010}) = 2 \g... | [
{
"input": "1\n0011\n0101",
"output": "1"
},
{
"input": "1\n0011\n0110",
"output": "0"
},
{
"input": "1\n0??1\n01??",
"output": "2"
},
{
"input": "2\n000?????\n01010101",
"output": "3"
},
{
"input": "2\n0???????\n1???????",
"output": "68"
},
{
"input":... | **Hint 1**
For any $c$ which is a cyclic shift of $t$, what will happen if $h(s,c) \gt 2^k$?
**Hint 2**
Try finding some useful relationship between $1$-s of $s$ and $1$-s of $c$.
**Hint**
There are exactly $n/2$ positions $i$ such that $s[i]=c[i]=1$.
**Hint 3**
Think about polynomial multiplication.
**Hint 1**... | What if you think about the condition h(s,c) ≥ 2^k for all cyclic shifts c in terms of polynomial multiplication? The key observation is that the requirement forces a very rigid algebraic structure: the product of the string polynomials, after a reversal, must be divisible by specific cyclotomic polynomials (x^{2^p}+1)... | {
"exploration": "I'll brute-force enumerate all completions of s and t that satisfy the exact 0/1 balance (each has exactly 2^k zeros and 2^k ones). For each pair (s_complete, t_complete), I'll check the condition: for all cyclic shifts c of t_complete, h(s_complete, c) ≥ 2^k. If any shift violates it (distance < 2^... | {
"exploration": "Let me try inclusion-exclusion over the set of cyclic shifts where h(s,c) < 2^k. The condition we need is that for ALL 2^{k+1} cyclic shifts c of t, we have h(s,c) ≥ 2^k. So the forbidden event for a given shift c is h(s,c) < 2^k, meaning at most 2^k - 1 mismatches between s and that shift of t.\n\n... | {
"exploration": "Let me set up the FFT-based linear system. The strings s and t have length N = 2^{k+1}. For k ≤ 7, N ≤ 256. The cyclic shift condition means for each shift d from 0 to N-1, the Hamming distance h(s, c_d) ≥ 2^k = N/2, where c_d is t shifted by d. The Hamming distance between s and a shifted t can be ... | {
"exploration": "I'll build a flow network with 2^{k+1} copies of the t-side, one per cyclic shift. For each shift d from 0 to 2^{k+1}-1, I create a bipartite graph: source connects to each position i of s (capacity 1), position i connects to position j of t if under shift d, s[i] could equal t[j] (meaning either bo... | null | null | Okay, so every approach hit a wall — brute force is astronomically impossible, inclusion-exclusion explodes exponentially, and the quadratic constraints from the hamming distance condition block any simple linear algebra or flow model. The core issue is that the condition must hold for *all* cyclic shifts, which seems ... | READ k
n = 1 << k
N = 2 * n
READ s of length N
READ t of length N
MOD = 998244353
// reverse bits of index i with total bits = k+1
function reverse_bits(x, bits):
res = 0
for b from 0 to bits-1:
if (x >> b) & 1:
res |= 1 << (bits-1-b)
return res
// reorder s and t according to bit-rev... |
1876/E | Ball-Stackable | 3,300 | constructive algorithms, data structures, dp, trees | 2 | 512 | With a problem title like that, there is no way this is going to be a graph problem.
Chaneka has a graph with $$$n$$$ vertices and $$$n-1$$$ edges. Some of the edges are directed and some of the edges are undirected. Edge $$$i$$$ connects vertex $$$u_i$$$ to vertex $$$v_i$$$. If $$$t_i=0$$$, edge $$$i$$$ is undirected... | The first line contains a single integer $$$n$$$ ($$$2\leq n\leq10^5$$$) — the number of vertices in the graph.
The $$$i$$$-th of the next $$$n-1$$$ lines contains three integers $$$u_i$$$, $$$v_i$$$, and $$$t_i$$$ ($$$1 \leq u_i,v_i \leq n$$$; $$$0\leq t_i\leq1$$$) — an undirected edge connecting vectices $$$u_i$$$ a... | A single line containing $$$-1$$$ if it is impossible.
Otherwise, the output consists of $$$n$$$ lines describing your construction. The first line contains an integer $$$z$$$ representing the number of different colours used. The $$$i$$$-th of the next $$$n-1$$$ lines contains three integers $$$p$$$, $$$q$$$, and $$$... | null | The following is the given graph.
Chaneka can direct all undirected edges and colour each edge as follows.
As an example, consider a stackable walk $$$3→1→5→2→5→4→5$$$. Let's show that that walk is ball-stackable.
1. Chaneka starts in vertex $$$3$$$. The stack is $$$[]$$$.
2. Chaneka moves to vertex $$$1$$$. She put... | [
{
"input": "5\n2 5 1\n1 3 0\n5 4 0\n1 5 1",
"output": "3\n1 5 2\n5 4 1\n2 5 2\n3 1 3"
}
] | Let's try to solve the problem if all edges are directed.
First, assign some arbitrary vertex $x$ as the root of the tree. Consider all walks that start from $x$. Notice that for a fixed vertex $y$, no matter how Chaneka walks from $x$, when she reaches $y$, the number of balls in the stack will always be the same. Mor... | The key observation is that the condition for all stackable walks reduces to constraints only on the simple paths from a single, carefully chosen root. The problem becomes about finding a vertex that minimizes a certain imbalance, then performing a DFS from it where edge colors are determined by a stack of colors you p... | {
"exploration": "Okay so we have a tree with n vertices and n-1 edges, some already directed, some undirected. The constraint is n ≤ 10^5. We need to direct all undirected edges and color each edge such that every stackable walk is also ball-stackable — meaning whenever you traverse an edge backwards and pop from th... | {
"exploration": "I'll model this as a flow problem: each edge is a pipe, colors are flow types, and we need to assign directions and colors so that any walk's stack operations (push forward, pop backward) respect color matching. The condition \"all stackable walks are ball-stackable\" means whenever you traverse an ... | {
"exploration": "I'll randomly orient the undirected edges and assign random colors, then verify by checking all (x,y) pairs on their unique tree path.\n\nLet me trace on Example 1: n=5, edges: (2,5,1) directed 2→5, (1,3,0) undirected, (5,4,0) undirected, (1,5,1) directed 1→5.\n\nFirst, random orientation and colors... | null | null | null | None of those got me anywhere useful. The greedy couldn't see global reachability, the flow blew up in a star, and the randomized verification couldn't even finish one check.
Hmm. But that nagging thread — reducing all stackable walks to constraints on simple paths from a root — feels like the only way out. The tree s... | function solve():
READ n
adj = array of size n+1, each an empty list
directed_edges = empty list
for i from 1 to n-1:
READ u, v, t
if t == 0:
add (v, 0) to adj[u]
add (u, 0) to adj[v]
else:
add (v, 1) to adj[u]
add (u, -1) to adj[v]... |
1491/H | Yuezheng Ling and Dynamic Tree | 3,400 | data structures, trees | 1.5 | 256 | Yuezheng Ling gives Luo Tianyi a tree which has $$$n$$$ nodes, rooted at $$$1$$$.
Luo Tianyi will tell you that the parent of the $$$i$$$-th node is $$$a_i$$$ ($$$1 \leq a_i<i$$$ for $$$2 \le i \le n$$$), and she will ask you to perform $$$q$$$ queries of $$$2$$$ types:
1. She'll give you three integers $$$l$$$, $$$r... | The first line contains two integers $$$n$$$ and $$$q$$$ ($$$2\leq n,q \leq 10^5$$$) — the number of nodes and the number of queries, respectively.
The second line contains $$$n-1$$$ integers $$$a_2, a_3,\dots, a_n$$$ ($$$1 \le a_i < i$$$), where $$$a_i$$$ is the parent of the node $$$i$$$.
Next $$$q$$$ lines contain... | For each query of the second type output answer on a new line. | null | The tree in example is shown below.
After the query of the first type, the tree changes and is looking as shown below. | [
{
"input": "6 4\n1 2 3 3 4\n2 3 4\n1 2 3 1\n2 5 6\n2 2 3",
"output": "3\n3\n1"
}
] | Divide the nodes into $\sqrt{n}$ blocks. The $i$-th block will contain the nodes in $[(i-1) \sqrt{n}+1,i \sqrt{n}]$.
Let's define $f_x$ as an ancestor of $x$ such that $f_x$ is in the same block as $x$ and $a_{f_x}$ is not in the same block as $x$.
Notice that for a given block, if all $a_x$ is not in the same block as... | The key observation is that the parent updates are monotone (always moving nodes closer to the root) and bounded. This suggests a data structure that can handle range decrements while maintaining ancestor relationships efficiently. What if you could group nodes into blocks and maintain a compressed representation of th... | {
"exploration": "Okay so we have a rooted tree at node 1, parent array a_i where a_i < i, and two query types: range decrement of parents with floor at 1, and LCA queries. n and q up to 10^5, time 1.5s. The parent pointers change over time, so any LCA structure needs to stay current.\n\nI'll maintain a segment tree ... | {
"exploration": "I'll build the HLD on the initial tree, maintain an Euler tour array, and use a Fenwick tree to track subtree sizes. For each type-1 update on range [l, r], I'll iterate over each i in the range, compute new_parent = max(1, a_i - x), then remove i's subtree from its current position in the Euler tou... | {
"exploration": "I'll maintain a link-cut tree where each node's parent pointer is stored as a value parent[i]. For a type-1 query on [l, r] with x, I iterate i from l to r, cut the edge from i to its current parent, compute new_parent = max(1, parent[i] - x), update parent[i] = new_parent, and link i to new_parent.... | null | null | null | So none of those panned out. The lifting tables were too heavy to rebuild, naive iteration over ranges was too slow, and even fancy dynamic trees couldn't avoid touching every node in the update interval.
Hmm. The monotonicity is the thing — each update only moves parents upward, toward 1, and they can't go below 1. S... | READ n, q
B = floor(sqrt(n))
block_id = array of size n+1
for i from 1 to n:
block_id[i] = (i - 1) / B // 0-indexed blocks
READ a[2..n] // a[i] is parent of i
// Initialize a for node 1 (unused, but set for completeness)
a[1] = 0
// For each block, store lazy value and jump pointers
lazy = array of size (n/B ... |
1266/H | Red-Blue Graph | 3,400 | dp, graphs, math, matrices, meet-in-the-middle | 4 | 512 | There is a directed graph on $$$n$$$ vertices numbered $$$1$$$ through $$$n$$$ where each vertex (except $$$n$$$) has two outgoing arcs, red and blue. At any point in time, exactly one of the arcs is active for each vertex. Initially, all blue arcs are active and there is a token located at vertex $$$1$$$. In one secon... | The first line contains a single integer $$$n$$$ ($$$2 \leq n \leq 58$$$) — the number of vertices.
$$$n-1$$$ lines follow, $$$i$$$-th of contains two space separated integers $$$b_i$$$ and $$$r_i$$$ ($$$1 \leq b_i, r_i \leq n$$$) representing a blue arc $$$(i, b_i)$$$ and red arc $$$(i, r_i)$$$, respectively. It is g... | Output $$$q$$$ lines, each containing answer to a single query.
If the state in the $$$i$$$-th query is unreachable, output the integer $$$-1$$$. Otherwise, output $$$t_i$$$ — the first time when the state appears (measured in seconds, starting from the initial state of the graph which appears in time $$$0$$$). | null | The graph in the first example is depticed in the figure below.
The first $$$19$$$ queries denote the journey of the token. On the $$$19$$$-th move the token would reach the vertex $$$6$$$. The last two queries show states that are unreachable. | [
{
"input": "6\n2 1\n5 5\n2 1\n6 3\n4 3\n21\n1 BBBBB\n1 RBBBB\n2 BBBBB\n5 BRBBB\n3 BRBBR\n1 BRRBR\n1 RRRBR\n2 BRRBR\n5 BBRBR\n4 BBRBB\n3 BBRRB\n2 BBBRB\n5 BRBRB\n3 BRBRR\n1 BRRRR\n1 RRRRR\n2 BRRRR\n5 BBRRR\n4 BBRRB\n2 BRBBB\n4 BRBBR",
"output": "0\n1\n2\n3\n4\n5\n6\n7\n8\n9\n10\n11\n12\n13\n14\n15\n16\n17\n1... | Consider a fixed query $\{s_i\}_{i=1}^{n-1}$ and $v$.
Let $x_i$ be the number of times the red arc from $i$ was traversed, and $y_i$ the number of times the blue arc was traversed. The answer for a given query, should the $x_i$ and $y_i$ exist, is the sum of all of them. Let's try to find these values.
When the blue ar... | The key observation is that the process defines a precise flow of the token through the graph. What if you treat the number of times each edge is traversed as unknowns? The structure yields a linear system that, due to the graph's properties, has a unique solution for any query. The challenge is ensuring the solution c... | {
"exploration": "Okay so we have a directed graph on n vertices, each vertex i (except n) has two outgoing arcs: blue to b_i and red to r_i. Initially all blue arcs are active, token at vertex 1. Each second: the vertex with the token flips its active arc (blue↔red), then the token moves along the newly active arc. ... | {
"exploration": "I'll maintain a current configuration string C of length n-1 (initially all 'B') and token position pos (initially 1). For each query (v_target, s_target), I'll repeatedly: find all vertices i where C[i] != s_target[i], compute cost estimate = shortest_path_distance(pos, i) + number of other mismatc... | {
"exploration": "I'll simulate the process from the initial state, storing each visited configuration in a hash map mapping the full state (vertex + active-arc string) to the first time it was seen. For each query, I just look up the target state in the map. If it's not there, I keep simulating until I either hit th... | null | null | null | Alright, back to square one. The graph isn't a tree, so bottom-up DP fails, and the state space is astronomically large, so simulation or memoization are impossible — the trajectory length could be up to 2^(n-1) steps, which for n=58 is about 10^17.
Hmm. But wait — the process is deterministic and each vertex (except ... | function solve():
READ n
// 1-indexed arrays for convenience
b = array of size n+1
r = array of size n+1
for i from 1 to n-1:
READ b[i], r[i]
// Build matrix A of size (n-1) x (n-1) for system A * y = z
// A[i][j] = coefficient of y_j in equation for vertex i
// Equation: 2*y_i -... |
1572/F | Stations | 3,400 | data structures | 2.5 | 256 | There are $$$n$$$ cities in a row numbered from $$$1$$$ to $$$n$$$.
The cities will be building broadcasting stations. The station in the $$$i$$$-th city has height $$$h_i$$$ and range $$$w_i$$$. It can broadcast information to city $$$j$$$ if the following constraints are met:
- $$$i \le j \le w_i$$$, and
- for each... | The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 2\cdot10^5$$$) — number of cities and number of events.
Then $$$q$$$ lines follow. The $$$i$$$-th line begins with an integer $$$p_i$$$ ($$$p_i = 1$$$ or $$$p_i = 2$$$).
If $$$p_i = 1$$$ a station will be rebuilt. Then two integers $$$c_i$$$ ... | For each query, print in a single line the sum of $$$b_j$$$ over the given interval. | null | In the first test case, only station $$$1$$$ reaches city $$$1$$$ before and after it is rebuilt.
In the second test case, after each rebuild, the array $$$b$$$ looks as follows:
1. $$$[1, 1, 1, 2, 1]$$$;
2. $$$[1, 2, 2, 3, 2]$$$;
3. $$$[1, 2, 2, 1, 2]$$$;
4. $$$[1, 1, 2, 1, 2]$$$;
5. $$$[1, 1, 2, 1, 1]$$$. | [
{
"input": "1 3\n2 1 1\n1 1 1\n2 1 1",
"output": "1\n1"
},
{
"input": "5 10\n1 3 4\n2 3 5\n1 1 5\n2 1 5\n1 4 5\n2 2 4\n1 2 3\n2 1 3\n1 5 5\n2 2 5",
"output": "4\n10\n5\n4\n5"
}
] | We will maintain the array $b$ on a range add/sum segment tree. Queries are done then in $O(\log n)$ per query.
Now lets focus on the station rebuilds. Lets maintain an array $w$, which means how far a station can broadcast information including the fact that some stations might block the signal. When a station is rebu... | The key observation is that the condition about heights being strictly decreasing to the right forces a monotonic structure on the w array. What if you maintain the broadcast ranges w_i directly and propagate their changes efficiently to the answer array b? This requires a data structure that can track which w values a... | {
"exploration": "Okay so we have n cities in a row, each with a station of height h_i and range w_i. A station at i can broadcast to city j if i ≤ j ≤ w_i and all cities strictly between i and j have height less than h_i. Initially all h_i = 0, w_i = i. Then q events: type 1 rebuilds a station at c_i with new range ... | {
"exploration": "I'll build a persistent segment tree over heights. Each version corresponds to processing cities in some order — when a city i rebuilds, its new height becomes the maximum among all stations (strictly highest), so I can treat rebuild events as creating a new \"time\" where that city's height dominat... | null | null | null | null | Alright, every approach I tried hit something. Brute force was impossible, and trying to force a persistent segment tree for the heights couldn't capture the cascading effect a single station rebuild has on the entire answer array.
Hmm. Wait — the condition about heights being strictly decreasing to the right. That me... | function solve():
READ n, q
// Segment tree for b: range add, range sum
seg_b = SegmentTree(n) // supports add(l, r, val) and query_sum(l, r)
// Segment tree beats for w: range chmin, point set, with ability to collect changed values
// Each node stores: max_val, second_max, count_max, sum, lazy_c... |
865/E | Hex Dyslexia | 3,300 | bitmasks, brute force, dp, graphs | 3 | 256 | Copying large hexadecimal (base 16) strings by hand can be error prone, but that doesn't stop people from doing it. You've discovered a bug in the code that was likely caused by someone making a mistake when copying such a string. You suspect that whoever copied the string did not change any of the digits in the string... | Input will contain a hexadecimal string S consisting only of digits 0 to 9 and lowercase English letters from a to f, with length at most 14. At least one of the characters is non-zero. | If it is not possible, print "NO" (without quotes).
Otherwise, print the lowercase hexadecimal string corresponding to the smallest possible numerical value, including any necessary leading zeros for the length to be correct. | null | The numerical value of a hexadecimal string is computed by multiplying each digit by successive powers of 16, starting with the rightmost digit, which is multiplied by 160. Hexadecimal digits representing values greater than 9 are represented by letters: a = 10, b = 11, c = 12, d = 13, e = 14, f = 15.
For example, the... | [
{
"input": "f1e",
"output": "NO"
},
{
"input": "0f1e",
"output": "00f1"
},
{
"input": "12d2c",
"output": "00314"
}
] | First, observe that for a solution to exist, the sum of the digits in the input must be divisible by 15. This is because of the Casting out Nines rule, but applied in base 16. Furthermore, the sum of digits, when divided by 15, tells us how many carries must be performed when adding the answer to the input. We can try ... | The key observation is that the problem reduces to a combinatorial search over carry positions and then constructing a valid permutation. Think about the base-16 analogue of digit sums and how carries affect the digit-wise differences. The minimal solution has a special structure related to cycles in the permutation. | {
"exploration": "I need to brute-force all permutations of digits. The length n is at most 14. The original string and the copied string are permutations of each other, meaning they have exactly the same multiset of digits. Let the original string be A (digits a_0...a_{n-1}) and the copied string be B (digits b_0...... | {
"exploration": "Okay so we have a hexadecimal string S of length n ≤ 14, zero-extended to match the original and copied strings. The copied string is a permutation of the original string's digits — same multiset of hex digits, same length. Their numerical absolute difference equals S. We need the smallest possible ... | null | null | null | null | So that's a few failed attempts now. The brute-force over multisets was astronomically too slow, and trying to model it as a flow network couldn't capture the global permutation constraint — the digits used in the original string must exactly match the digits in the permuted string, which is a perfect matching across t... | function solve():
READ S as string
n = length(S)
// Convert S to list of integers, least significant digit first
digits = array of size n
for i from 0 to n-1:
hex_char = S[n-1-i]
if '0' ≤ hex_char ≤ '9':
digits[i] = int(hex_char)
else: // 'a' to 'f'
di... |
2154/F2 | Bombing (Hard Version) | 3,300 | combinatorics,dp | null | null | This is the hard version of the problem. The difference between the versions is that in this version, $$$n \le 10^6$$$. You can hack only if you solved all versions of this problem.
A permutation$$$^{\text{∗}}$$$ $$$b$$$ is considered a riffle shuffle of a permutation $$$a$$$ if $$$|a| = |b|$$$ and there exists $$$k$$... | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.
The first line of each test case contains an integer $$$n$$$ ($$$2 \le n \le 10^6$$$) — the length of the permutation.
The second line of each test case... | For each testcase, output the number of ways to fill $$$p$$$ so that it is a riffle shuffle of $$$[1,2,\ldots,n]$$$ modulo $$$998\,244\,353$$$. | null | null | [] | **Step 1**
Solve the easy version.
**Step 2**
If $p_i \lt i$ we know that $p_i$ must belong to the left side of the split, similarly if $p_i \gt i$ we know that $p_i$ must belong to the right side of the split.
**Step 3**
If a permutation p is a riffle shuffle of $[1,2,\ldots,n]$, then it will have three regions d... | The key observation is that the condition for being a riffle shuffle imposes strict constraints on which side (left or right subsequence) each known element must belong to, based on its value relative to its index. This turns the problem into a combinatorial counting of valid assignments for the unknown positions, wher... | {
"exploration": "Okay so we have a permutation p of length n with some -1s, and we need to count ways to fill them so that p becomes a riffle shuffle of [1,2,...,n]. A riffle shuffle means there exists a split point k where the first k elements of the sorted permutation appear in order as a subsequence, and the rema... | {
"exploration": "I'll set up dp[i][l][r] as described: number of ways to fill first i positions where the last element taken from the left subsequence is l and the last from the right subsequence is r. Initially, before any positions are filled, l=0 and r=0 (no elements taken yet). dp[0][0][0]=1. For each position i... | {
"exploration": "I'll model each position i as a node that must be assigned a unique value from 1..n, and also assigned a side: left subsequence (L) or right subsequence (R). The riffle shuffle condition means: if we take all positions assigned L in order, their values must be strictly increasing; similarly for R. A... | null | null | null | Okay, so I've burned through a few ideas and none stuck. Backtracking is impossible, a naive DP can't track which values are used, and flow can't capture the global structure that the left subsequence must be exactly the first k numbers in sorted order. The core difficulty is that the riffle shuffle condition couples t... | MOD = 998244353
MAXN = 1000000
Precompute factorial fac[0..MAXN] and inverse factorial ifac[0..MAXN]
fac[0] = 1
for i from 1 to MAXN:
fac[i] = fac[i-1] * i % MOD
ifac[MAXN] = pow(fac[MAXN], MOD-2, MOD)
for i from MAXN-1 down to 0:
ifac[i] = ifac[i+1] * (i+1) % MOD
function nCr(n, k):
if k < 0 or k > n: re... |
1530/G | What a Reversal | 3,300 | constructive algorithms | 2 | 512 | You have two strings $$$a$$$ and $$$b$$$ of equal length $$$n$$$ consisting of characters 0 and 1, and an integer $$$k$$$.
You need to make strings $$$a$$$ and $$$b$$$ equal.
In one step, you can choose any substring of $$$a$$$ containing exactly $$$k$$$ characters 1 (and arbitrary number of characters 0) and reverse... | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 2000$$$). Description of the test cases follows.
Each test case consists of three lines. The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 2000$$$; $$$0 \le k \le n$... | For each test case, if it's impossible to make $$$a$$$ equal to $$$b$$$ in at most $$$4n$$$ reversals, print a single integer $$$-1$$$.
Otherwise, print an integer $$$m$$$ ($$$0 \le m \le 4n$$$), denoting the number of reversals in your sequence of steps, followed by $$$m$$$ pairs of integers $$$l_i, r_i$$$ ($$$1 \le ... | null | In the first test case, after the first reversal $$$a = $$$ 011010, after the second reversal $$$a = $$$ 010110, after the third reversal $$$a = $$$ 010101. | [
{
"input": "6\n6 1\n101010\n010101\n6 3\n101010\n010101\n6 0\n101010\n010101\n6 6\n101010\n010101\n4 2\n0000\n1111\n9 2\n011100101\n101001011",
"output": "3\n1 2\n3 4\n5 6\n1\n1 6\n-1\n-1\n-1\n5\n4 8\n8 9\n3 6\n1 4\n3 6"
}
] | First of all, the number of 1's in $a$ and $b$ must match. Let $c$ be the number of 1's in $a$.
If $k = 0$ or $k > c$, we can not do a meaningful reversal, so we just check if $a = b$.
If $k = c$, we can not change the contents of $a$ between the leftmost and the rightmost 1's, we can only reverse it and shift with reg... | The key observation is that the operation is reversible and allows you to manipulate the gaps of zeros between the ones in a structured way. Think about representing the string as a sequence of zero-block lengths and using the operation to systematically move zeros to designated positions, with the parity of k determin... | {
"exploration": "Okay so we have two binary strings a and b of length n, and an integer k. We can reverse any substring of a that contains exactly k ones, and we need to make a equal to b in at most 4n steps. The sum of n over all test cases is at most 2000, so we can afford O(n²) per test case. The key constraint i... | {
"exploration": "I'll model the problem as a graph where nodes are binary strings of length n, edges are valid reversals of substrings with exactly k ones. BFS from a to b, depth limit 4n. For each state, I need to generate all (l, r) pairs where the substring has exactly k ones. I'll precompute prefix sums of ones ... | {
"exploration": "Let me define dp[l][r] as the minimum reversals to transform a[l..r] into b[l..r] using only operations fully inside [l,r]. I'll precompute prefix sums of ones in a so I can check if any substring has exactly k ones in O(1). Then for each interval [l,r], I try every possible reversal [i,j] ⊆ [l,r] w... | {
"exploration": "I'll try simulating adjacent swaps using reversals of substrings containing exactly k ones. The idea: if I can swap any two adjacent characters using a bounded number of reversals, then I can bubble-sort a into b, and the total number of operations would be O(n²) swaps, each costing O(1) reversals, ... | null | null | Alright, back to square one. Greedy undid itself, BFS exploded, DP was impossible, and adjacent swaps were too slow — all because they tried to fix positions locally without a global invariant that the operation preserves.
hmm... The operation reverses a substring with exactly k ones. So the total number of ones in th... | function solve():
READ t
for each test case:
READ n, k
READ string a of length n
READ string b of length n
// Count ones
cnt_a = count of '1' in a
cnt_b = count of '1' in b
if cnt_a != cnt_b:
PRINT -1
continue
m = cnt_a
... |
1098/E | Fedya the Potter | 3,400 | binary search, implementation, math, number theory | 2.5 | 256 | Fedya loves problems involving data structures. Especially ones about different queries on subsegments. Fedya had a nice array $$$a_1, a_2, \ldots a_n$$$ and a beautiful data structure. This data structure, given $$$l$$$ and $$$r$$$, $$$1 \le l \le r \le n$$$, could find the greatest integer $$$d$$$, such that $$$d$$$ ... | First line contains a single integer $$$n$$$ — number of elements in array $$$a$$$ ($$$1 \le n \le 50\,000$$$).
Second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ — elements of the array ($$$1 \le a_i \le 100\,000$$$). | Print a single integer — the lower median of array $$$c$$$. | null | In the first sample array $$$b$$$ is equal to $$${3, 3, 6}$$$, then array $$$c$$$ is equal to $$${3, 3, 6, 6, 9, 12}$$$, so the lower median is $$$6$$$.
In the second sample $$$b$$$ is $$${8, 8, 8}$$$, $$$c$$$ is $$${8, 8, 8, 16, 16, 24}$$$, so the lower median is $$$8$$$. | [
{
"input": "2\n6 3",
"output": "6"
},
{
"input": "2\n8 8",
"output": "8"
},
{
"input": "5\n19 16 2 12 15",
"output": "12"
}
] | There are $O(N\cdot logN)$ different values of gcd, because if we fix the left bound of a segment and iterate right bound from left bound to the end, the gcd stays unchanged or decreases in two or more times. We can get a compressed version of array $b$ if we compress equal consecutive elements in to pair $(value, coun... | The key observation is that the number of distinct gcd values across all subarrays is only O(n log max a). This means array b has long runs of equal values. To count subarray sums of b below a threshold, you can treat each run as a block and reduce the problem to counting integer lattice points under a line. | {
"exploration": "Okay so we have an array a of length n, and for every subarray we compute the GCD of that subarray. That gives us array b with n(n+1)/2 elements. Then we compute the sum of every contiguous subarray of b, giving array c, and we need the lower median of c. The constraint is n ≤ 50,000, a_i ≤ 100,000.... | {
"exploration": "I'll build array b explicitly by enumerating every subarray of a and computing its gcd. For each i from 1 to n, for each j from i to n, I'll query a segment tree for range gcd on [i,j], push the result into b. Then sort b. The size of b is n(n+1)/2.\n\nLet me trace Example 1: n=2, a=[6,3].\nSubarray... | {
"exploration": "For each possible gcd value d from 1 to max(a)=100,000, I'll find all maximal contiguous segments where every element is divisible by d. For each such segment of length len, it contributes len*(len+1)/2 subarrays whose gcd is at least d. Then the exact count of subarrays with gcd exactly d is: count... | null | null | null | Okay, all of those went nowhere. Trying to materialize b is impossible—it's 1.25 billion elements—and c is astronomically larger. Even the DP on gcd segments collapsed because tracking multiplicities for sums of b's subarrays seemed to require enumerating all of them.
hmm. But that nagging thought—distinct gcd values ... | function generate_runs(a, n):
runs = empty list of (value, count)
for i from 1 to n:
stack = empty list of (gcd, rightmost_index)
// For each i, compute gcd segments ending at i
for j from i to n:
g = gcd(stack.empty ? a[j] : gcd(stack.top.g, a[j]))
// Merge with ... |
2135/E1 | Beyond the Palindrome (Easy Version) | 3,300 | combinatorics,dp,math | null | null | This is the easy version of the problem. The difference between the versions is that in this version, $$$n\le 10^6$$$, and the sum of $$$n$$$ over all test cases does not exceed $$$10^6$$$. You can hack only if you solved all versions of this problem.
For a binary string$$$^{\text{∗}}$$$ $$$r$$$, we define $$$f(r)$$$ ... | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^4$$$). The description of the test cases follows.
The only line of each test case contains a single integer $$$n$$$ ($$$1\le n\le 10^6$$$) — the length of the binary strings.
It is guaranteed that the s... | For each test case, output a single integer — the number of distinct almost-palindrome binary strings of length $$$n$$$, modulo $$$998\,244\,353$$$. | null | null | [] | **Hint 1**
Replace $\tt 0$ with $1$ and $\tt 1$ with $-1$. Try to find a sufficient and necessary condition of $s$ almost-palindrome.
**Hint 2**
Denote the $\pm 1$-s you get in Hint 1 as array $a$, and $b_i = \sum_{j=1}^n a_j$. ($s$ is almost-palindrome) $\Longleftrightarrow$ ($\min b_i + \max b_i = b_n$),
**Hint 3... | What if you think of the string as a walk on integers? The key observation is that the deletion process simplifies to a condition on the extreme values of this walk. The counting reduces to a classic combinatorial problem of paths constrained between two parallel boundaries. | {
"exploration": "I'll brute-force simulate the deletion process for every binary string of length n, compute f(s) and f(rev(s)), and count those where they match. That's 2^n strings per n. For each string, I'll simulate the deletion process: while there exists a \"10\" substring, delete all such substrings simultane... | {
"exploration": "I need to understand what f(r) does. The process repeatedly deletes all \"10\" substrings simultaneously until no \"10\" remains. Let me simulate on a few strings to see the final result. Take r=\"100110001\". They gave the steps: start with \"100110001\". Simultaneously delete all \"10\" substrings... | {
"exploration": "I'll model each binary string as a sequence of runs, because the process deletes \"10\" substrings, which are essentially transitions from 1 to 0. Let me think about what f(r) does: repeatedly delete all \"10\" substrings simultaneously. This means any '1' immediately followed by a '0' gets removed,... | null | null | null | So all those paths failed — brute-force enumeration is exponential, and trying to characterize the final strings directly got stuck in messy equivalence classes without a clean recurrence.
hmm. That half-formed thought about a walk on integers — maybe that's the key. The deletion process removes "10" substrings, which... | MOD = 998244353
function pow2(exp):
result = 1
base = 2
while exp > 0:
if exp % 2 == 1:
result = (result * base) % MOD
base = (base * base) % MOD
exp = exp // 2
return result
READ t
for each test case:
READ n
k = (n + 1) // 2 // ceil(n/2)
ans = pow2(k)... |
1307/F | Cow and Vacation | 3,300 | dfs and similar, dsu, trees | 2 | 256 | Bessie is planning a vacation! In Cow-lifornia, there are $$$n$$$ cities, with $$$n-1$$$ bidirectional roads connecting them. It is guaranteed that one can reach any city from any other city.
Bessie is considering $$$v$$$ possible vacation plans, with the $$$i$$$-th one consisting of a start city $$$a_i$$$ and destina... | The first line contains three integers $$$n$$$, $$$k$$$, and $$$r$$$ ($$$2 \le n \le 2 \cdot 10^5$$$, $$$1 \le k,r \le n$$$) — the number of cities, the maximum number of roads Bessie is willing to travel through in a row without resting, and the number of rest stops.
Each of the following $$$n-1$$$ lines contain two... | If Bessie can reach her destination without traveling across more than $$$k$$$ roads without resting for the $$$i$$$-th vacation plan, print YES. Otherwise, print NO. | null | The graph for the first example is shown below. The rest stop is denoted by red.
For the first query, Bessie can visit these cities in order: $$$1, 2, 3$$$.
For the second query, Bessie can visit these cities in order: $$$3, 2, 4, 5$$$.
For the third query, Bessie cannot travel to her destination. For example, if sh... | [
{
"input": "6 2 1\n1 2\n2 3\n2 4\n4 5\n5 6\n2\n3\n1 3\n3 5\n3 6",
"output": "YES\nYES\nNO"
},
{
"input": "8 3 3\n1 2\n2 3\n3 4\n4 5\n4 6\n6 7\n7 8\n2 5 8\n2\n7 1\n8 1",
"output": "YES\nNO"
}
] | We will run a BFS from all the rest stops in parallel and use union-find to determine which rest stops can reach each other directly. We will split each edge into two to simplify this process. Note that this means Bessie can now travel at most $2k$ roads before needing a rest. While we perform the BFS, we also color al... | The key observation is that the problem reduces to connectivity among rest stops within a doubled travel radius. What if you think about expanding reachable regions from each rest stop and merging them when they intersect? This suggests a BFS-based preprocessing step combined with union-find to cluster rest stops, afte... | {
"exploration": "I'll try a DP on the tree where for each node u I store the minimum distance to the nearest rest stop, call it dist[u], via a BFS from all rest stops. Then for each query (a,b), I'll walk along the unique path from a to b and check if any segment of length > k exists where the distance to a rest sto... | {
"exploration": "I'll split each city into an in-node and an out-node. For a non-rest city, the edge from in to out has capacity 1 — meaning Bessie can pass through it at most once without resting. For a rest city, that edge has infinite capacity — she can pass through it as many times as needed. Then I connect out-... | null | null | null | null | So that's a few failed attempts now. The pairwise distance checks exploded quadratically, and the flow model blew up in size — both trying to capture the global constraint of "never exceed k steps without a rest" with local, per-plan computations.
hmm. But that constraint is really about reachability through a network... | READ n, k, r
adj = adjacency list of size n+1
for i from 1 to n-1:
READ x, y
adj[x].append(y)
adj[y].append(x)
rest = list of size r
READ rest[1..r]
READ v
queries = list of v pairs (a, b)
dist = array of size n+1, initialized to -1
src = array of size n+1, initialized to 0
queue = empty queue
dsu = Unio... |
1012/F | Passports | 3,400 | dp, implementation | 2 | 512 | Gleb is a famous competitive programming teacher from Innopolis. He is planning a trip to N programming camps in the nearest future. Each camp will be held in a different country. For each of them, Gleb needs to apply for a visa.
For each of these trips Gleb knows three integers: the number of the first day of the tri... | In the first line of the input there are two integers N (1 ≤ N ≤ 22) and P (1 ≤ P ≤ 2)—the number of trips and the number of passports Gleb has, respectively.
The next N lines describe Gleb's trips. Each line contains three positive integers si, leni, ti (1 ≤ si, leni, ti ≤ 109)—the first day of the trip, the length o... | If it is impossible to get all visas on time, just print "NO" (quotes for clarity). Otherwise, print "YES" and N lines describing trips. For each trip, first print number of the passport Gleb should put this country's visa in, and then print number of the day he should apply for it. Print trips in the same order as the... | null | Examples with answer "YES" are depicted below.
Each cell of the stripe represents a single day. Rectangles represent trips, each trip starts in the morning and ends in the evening. Rectangles with angled corners represent visa applications. Each application starts in the middle of a day and ends ti days after. The tri... | [
{
"input": "2 1\n3 1 1\n6 1 1",
"output": "YES\n1 1\n1 4"
},
{
"input": "3 1\n13 2 2\n7 3 1\n19 3 4",
"output": "YES\n1 10\n1 1\n1 2"
},
{
"input": "7 2\n15 1 1\n14 1 1\n18 1 1\n21 1 1\n9 4 6\n22 2 5\n5 4 3",
"output": "YES\n2 13\n1 1\n1 16\n1 19\n1 2\n2 16\n2 1"
},
{
"input"... | Let's solve the P = 1 case first. We'll use dynamic programming on subsets. Let's try to add visas to subset in order of application. Notice that if we only have one passport, every visa processing segment should lie between some two consecutive trips. For convenience, let's find all these segments beforehand.
Define d... | What if you think about assigning trips to passports as a partition problem? The key observation is that for a fixed assignment, the schedule for each passport becomes a sequence of non-overlapping processing intervals that must fit into the gaps between trips. This suggests a state-space approach where you track which... | {
"exploration": "Okay so we have N trips (N ≤ 22), each with start day s_i, length len_i, and visa processing time t_i. Gleb has P passports (P ≤ 2). He can apply for a visa on day d only if he's in Innopolis that day — meaning d cannot fall inside any trip interval [s_i, s_i+len_i-1], and also cannot be the day aft... | {
"exploration": "I'll sort the trips by their start day s_i and then try to schedule visa applications using backtracking with interval scheduling heuristics. For each trip, I need to assign it to one of the P passports and pick an application day d such that: d is not during any trip (including the trip itself), d ... | null | null | null | null | Okay, so both the flow network and the naive backtracking failed because they couldn't properly enforce the core constraint: a passport is a single physical object that can't be in two consulates at once, meaning its visa processing intervals [apply_day, apply_day + t_i) must be completely non-overlapping, and they mus... | function solve():
READ N, P
trips = array of size N
for i from 0 to N-1:
READ trips[i].s, trips[i].len, trips[i].t
// Sort trips by start day s
order = indices 0..N-1 sorted by trips[].s
sorted_trips = array of size N
for i from 0 to N-1:
sorted_trips[i] = trips[order[i]]
... |
2147/I1 | Longest Increasing Path (Easy Version) | 3,300 | constructive algorithms,math | null | null | This is the easy version of the problem. The difference between the versions is that in this version the second test is $$$n=100\,000$$$. Hacks are not allowed in this problem.
Let's call an integer sequence $$$a_1, a_2, \ldots, a_n$$$ distance-convex if $$$|a_{i} - a_{i-1}| \lt |a_{i+1} - a_{i}|$$$ for every $$$1 ... | Input consists of two integers $$$n$$$ and $$$m$$$ on a single line.
There are only 2 tests for this problem.
• $$$n = 8$$$, $$$m = 6$$$;
• $$$n = 100\,000$$$, $$$m = 15\,000$$$.
It is guaranteed that an answer exists for both tests.Hacks are not allowed in this problem. | Print a distance-convex sequence $$$a_1, a_2, \ldots, a_n$$$ that contains at most $$$m$$$ distinct values. $$$-10^{18} \le a_i \le 10^{18}$$$ should hold for all $$$1 \le i \le n$$$.
If there are multiple solutions, print any of them. | null | null | [] | I1. Longest Increasing Path (Easy Version)
**Hint1**
The expected is $n = m\log m - O(m)$ steps using harmonic sum.
**Hint2**
Think of a solution in 2D (not needed but good intuition).
**Hint3**
Put points as an almost regular polygon and keep doing circles around it jumping with step $1$, then $2$, then $3$...
... | The key observation is that you can generate many jumps of a fixed length by cycling through a fixed set of points. The technique is to treat the m distinct values as positions on a circle and use modular arithmetic to produce jumps of exact, increasing sizes. The structural property to exploit is the harmonic sum rela... | {
"exploration": "Okay so we need a sequence of n jumps with strictly increasing lengths, visiting at most m distinct points. The jumps are the absolute differences between consecutive terms. We have n=8,m=6 for the first test and n=100000,m=15000 for the second. Values can be anywhere in [-10^18, 10^18].\n\nI'll sta... | {
"exploration": "I'll fix a maximum allowed final jump length L and try to construct a sequence with strictly increasing integer jumps from 1 to n-1, all bounded by L. Start at position 0. For each jump length d from 1 to n-1, I need to move exactly d away from the current position. I maintain a set of used position... | {
"exploration": "I'll try to make DP work by tracking the last value and the last jump length, plus a count of distinct values used so far. Let dp[i][v][d][c] = possible to have sequence of length i ending at value v, last jump = d, using c distinct values. But v can be huge (up to 10^18), so I'll compress by only s... | {
"exploration": "I'll place m=6 points on a circle in 2D, then project to 1D by taking the x-coordinate. The idea is that if the points are arranged around a circle and I visit them in order around the circle, the Euclidean distances between consecutive points in 2D will be roughly equal — but I need them to be stri... | null | null | Hmm, all those ideas fell apart. The arithmetic progression couldn't generate jumps long enough without blowing up the value range, the greedy construction used too many distinct points, and DP or geometry just collapsed under the constraints.
wait — that thread about cycling through a fixed set of points. If I have m... | function generate_sequence(n, m):
// Step 1: Generate m distinct points as 2^i - 1
points = array of size m
for i from 0 to m-1:
points[i] = (1 << i) - 1 // 2^i - 1
// Step 2: Greedy walk
cur = 0
prev = 0
answer = array of size n
answer[0] = cur
// For each step, find the... |
2180/H1 | Bug Is Feature (Unconditional Version) | 3,400 | games | null | null | This is the unconditional version of the problem. The difference between the versions is that in this version, there is no "non-decreasing common difference" condition. You can hack only if you solved all versions of this problem.
Note that neither version is necessarily easier than the other, and they can be solved i... | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). The description of the test cases follows.
The first line of each test case consists of a single integer $$$n$$$ ($$$1 \le n \le 2 \cdot 10^5$$$) — the number of series of games that Bug and Featu... | For each test, output $$$\mathtt{Bug}$$$ if Bug wins, and $$$\mathtt{Feature}$$$ if Feature wins. | null | null | [] | **Solution**
Step 1: reduce to independent series.
First, we can split the whole match into independent series $(a_i,b_i,c_i,[l_i,r_i])$. Since moves never mix two different games, by the Sprague-Grundy theorem we can compute the Grundy number contributed by each series and then solve the question.
Step 2: normalize... | The key observation is that each game reduces to a state defined by the remaining space and the step size. The Grundy numbers have a periodic structure when you factor out the odd part of the step. Think about computing XORs over ranges by exploiting this periodicity. | {
"exploration": "I'll define a recursive function f(a,b,c,x) that returns the Grundy number (or win/loss status) for a single game with parameters (a,b,c) and upper bound x. Since the game is impartial and the moves are independent across games, the overall game is the XOR of Grundy numbers of individual games. I'll... | {
"exploration": "For each series i, I'll compute d = b_i - a_i. The game state is determined by t = x - c_i, the amount we can increase c_i (and correspondingly adjust a and b to maintain arithmetic progression). When we increase one of a, b, c by some positive amount, the arithmetic progression condition forces spe... | null | null | null | null | Alright, every approach I tried hit something. Direct recursion was impossible because the number of games per series is astronomical, and brute-force period detection failed because the period itself could be up to 10^18, making simulation hopeless.
hmm... but that nagging thread — the Grundy numbers have a periodic ... | function xor_prefix(n):
// returns XOR of integers from 0 to n inclusive
if n < 0:
return 0
r = n % 4
if r == 0:
return n
else if r == 1:
return 1
else if r == 2:
return n + 1
else:
return 0
function xor_range(L, R):
// returns XOR of integers fro... |
1684/H | Hard Cut | 3,400 | constructive algorithms, dfs and similar, divide and conquer, math | 2 | 256 | You are given a binary string $$$s$$$. You have to cut it into any number of non-intersecting substrings, so that the sum of binary integers denoted by these substrings is a power of 2. Each element of $$$s$$$ should be in exactly one substring. | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 10^5$$$). Description of the test cases follows.
Each test case contains a binary string $$$s$$$ ($$$1 \le |s| \le 10^6$$$).
It is guaranteed that the sum of $$$|s|$$$ over all test cases does not exceed $... | For each test case output the answer to the problem as follows:
- If the answer does not exist, output $$$-1$$$.
- If the answer exists, firstly output an integer $$$k$$$ — the number of substrings in the answer. After that output $$$k$$$ non-intersecting substrings, for $$$i$$$-th substring output two integers $$$l_i... | null | In the first test case it is impossible to cut the string into substrings, so that the sum is a power of 2.
In the second test case such cut is valid:
- $$$011_2 = 3_{10}$$$,
- $$$0_2 = 0_{10}$$$,
- $$$1_2 = 1_{10}$$$. | [
{
"input": "4\n00000\n01101\n0111011001011\n000111100111110",
"output": "-1\n\n3\n1 3\n4 4\n5 5\n\n8\n1 2\n3 3\n4 4\n5 6\n7 7\n8 10\n11 12\n13 13\n\n5\n1 5\n6 7\n8 11\n12 14\n15 15"
}
] | Let's say that there are $k$ ones in the given string.
We will build the answer for all $k > 0$. $k = 0$ is the only case when the answer does not exist.
$k = 1$, $k = 2$, $k = 4$ — cut into single digits.
$k = 3$ — described later.
Let's consider $k = 5$:
If all ones are in a row, then we should cut them into $1111_2 ... | What if you focus on the total number of ones in the string? The key observation is that the problem becomes about constructing a specific power-of-two sum from these ones, and you can use a recursive splitting strategy that treats blocks of ones as building blocks. | {
"exploration": "Okay so we have a binary string s, and we need to partition it into contiguous substrings such that the sum of the binary integers represented by those substrings equals some power of 2. The total length across all test cases is at most 10^6, so any O(|s|) or O(|s| log |s|) per test case should be f... | {
"exploration": "Let me set up the DP table. I'll define dp[l][r][v] as a boolean: whether substring s[l..r] can be partitioned into segments whose total sum equals v, where v is a power of two. The recurrence: for each split point m between l and r, and for each pair of powers of two p1, p2 with p1 + p2 = v, check ... | {
"exploration": "I'll enumerate all ways to partition the ones into contiguous groups. For each group, the binary number is formed by the substring from the first one to the last one in that group, with intermediate zeros included. The total sum is the sum of these group values. I need to find a partition where the ... | {
"exploration": "I'll cut the string into individual digits. For \"01101\", that's five segments: \"0\", \"1\", \"1\", \"0\", \"1\". Their binary values are 0, 1, 1, 0, 1. Total sum = 3. Powers of 2: 1, 2, 4, 8, 16, ... Closest is 4 (diff=1) or 2 (diff=1). I'll randomly merge adjacent segments and check if the new s... | null | null | So that's a few failed attempts now. Greedy picked the wrong target, DP was astronomically impossible, grouping ones exploded combinatorially, and random merging had no guarantee. The core issue is the sum is not fixed—it's a power of two we must construct by grouping digits, and the naive search space is unthinkably l... | function solve_one(s):
n = length(s)
ones = list of indices i where s[i] == '1'
k = length(ones)
if k == 0:
return [-1]
// Helper: cut into single digits
function single_cuts():
segments = []
for i from 0 to n-1:
segments.append([i+1, i+1]) // 1‑based indice... |
1528/F | AmShZ Farm | 3,300 | combinatorics, fft, math | 2 | 256 | To AmShZ, all arrays are equal, but some arrays are more-equal than others. Specifically, the arrays consisting of $$$n$$$ elements from $$$1$$$ to $$$n$$$ that can be turned into permutations of numbers from $$$1$$$ to $$$n$$$ by adding a non-negative integer to each element.
Mashtali who wants to appear in every pro... | The first line of input contains two integers $$$n$$$ and $$$k$$$ $$$(1 \le n \le 10^9 , 1 \le k \le 10^5)$$$. | Print a single integer — the answer to the problem modulo $$$998244353$$$. | null | There are eight possible pairs for the second example:
1. $$$a = \{1, 1\}, b = \{1, 1\}$$$
2. $$$a = \{1, 1\}, b = \{1, 2\}$$$
3. $$$a = \{1, 1\}, b = \{2, 1\}$$$
4. $$$a = \{1, 1\}, b = \{2, 2\}$$$
5. $$$a = \{1, 2\}, b = \{1, 1\}$$$
6. $$$a = \{1, 2\}, b = \{2, 2\}$$$
7. $$$a = \{2, 1\}, b = \{1, 1\}$$$
8. $$$a = \{... | [
{
"input": "1 1",
"output": "1"
},
{
"input": "2 2",
"output": "8"
},
{
"input": "5 4",
"output": "50400"
},
{
"input": "20 100",
"output": "807645526"
},
{
"input": "10000000 10000",
"output": "883232350"
}
] | Consider the following problem: $n$ cars want to enter a parking lot one by one. The parking lot has $n$ slots numbered $1, 2 , \ldots , n$, the $i$-th of the $n$ cars wants to park in the $a_i$-th slot.
When the $i$-th car drives in, it will park in the first empty slot $s$ such that $s \ge a_i$.
An array $a$ is Good ... | The key observation is that counting pairs (a,b) reduces to a combinatorial sum over the frequency of a fixed value in a. This sum can be expressed using Stirling numbers of the second kind, which count ways to partition the k indices of b into nonempty groups. The challenge is efficiently computing this sum given the ... | {
"exploration": "For each possible array a of length n with elements between 1 and n, I need to check if it is \"more-equal\" — meaning there exists some integer t ≥ 0 such that adding t to every element yields a permutation of 1..n. Then for each such a, I need to count all arrays b of length k with values in 1..n ... | {
"exploration": "I'll try to understand what a \"more-equal array\" actually is first. An array a of n elements from 1 to n that can be turned into a permutation of 1..n by adding a non-negative integer to each element. So there exists some non-negative integer x such that a_i + x is a permutation of 1..n. That mean... | {
"exploration": "Burnside's lemma counts orbits of a group action. Here the group is cyclic shifts of the value set? More-equal arrays are those where adding a constant (mod n?) turns it into a permutation. Wait, the definition: array a of length n with elements 1..n such that there exists a non-negative integer t w... | null | null | null | So none of those panned out. Brute force explodes exponentially, DP needs impossible state space, and my initial interpretation of "more-equal arrays" was wrong — they're not just permutations, they're arrays that can be shifted into permutations, which is a much larger set.
hmm. Wait — if array a can be turned into a... | MOD = 998244353
function mod_pow(base, exp):
result = 1
while exp > 0:
if exp % 2 == 1:
result = result * base % MOD
base = base * base % MOD
exp = exp // 2
return result
function stirling_second_kind(k):
// Compute S(k, i) for i = 0..k using FFT in O(k log k)
/... |
1534/G | A New Beginning | 3,300 | data structures, dp, geometry, sortings | 5 | 256 | Annie has gotten bored of winning every coding contest and farming unlimited rating. Today, she is going to farm potatoes instead.
Annie's garden is an infinite 2D plane. She has $$$n$$$ potatoes to plant, and the $$$i$$$-th potato must be planted at $$$(x_i,y_i)$$$. Starting at the point $$$(0, 0)$$$, Annie begins wa... | The first line contains the integer $$$n$$$ ($$$1 \le n \le 800\,000$$$).
The next $$$n$$$ lines contain two integers $$$x_i$$$ and $$$y_i$$$ ($$$0 \le x_i,y_i \le 10^9$$$), representing the location of the $$$i$$$-th potato. It is possible that some potatoes should be planted in the same location. | Print the minimum total energy to plant all potatoes. | null | In example $$$1$$$, Annie can travel to each spot directly and plant a potato with no energy required.
In example $$$2$$$, moving to $$$(1,0)$$$, Annie plants the second potato using $$$1$$$ energy. Next, she travels to $$$(1,1)$$$ and plants the first potato with $$$0$$$ energy. | [
{
"input": "2\n1 1\n2 2",
"output": "0"
},
{
"input": "2\n1 1\n2 0",
"output": "1"
},
{
"input": "3\n5 5\n7 7\n4 9",
"output": "2"
},
{
"input": "10\n5 1\n4 0\n9 6\n0 2\n10 1\n9 10\n3 10\n0 10\n8 9\n1 5",
"output": "19"
},
{
"input": "10\n1 1\n2 2\n2 0\n4 2\n4 0\n... | Observe that for any point $(x,y)$ and some path $A$, the minimum distance from $A$ to $(x,y)$ will occur on the intersection of $A$ and the antidiagonal of $(x,y)$. Here the antidiagonal is defined as a line $y=-x+c$.
Proof:
Assume $(a,b)$ is the intersection of $A$ and the antidiagonal of $(x,y)$. If $(a,b) = (x,y)$,... | The key observation is that the Chebyshev distance cost is minimized when the planting point lies on the same antidiagonal as your current position. This suggests transforming coordinates to make these antidiagonals vertical. After the transformation, the problem reduces to maintaining a convex function as you move bet... | {
"exploration": "Okay so we have an infinite 2D plane, Annie starts at (0,0), can only move right or up one unit per step, and at any point (X,Y) she can plant potatoes using a gun that costs max(|X-x|, |Y-y|) energy per potato. The goal is to minimize total energy to plant all n potatoes. n ≤ 800,000, coordinates u... | {
"exploration": "I'll build a complete graph on n+1 nodes — the n potatoes plus the origin (0,0). Edge weight between two points (x₁,y₁) and (x₂,y₂) is the Chebyshev distance max(|x₁-x₂|, |y₁-y₂|). Then I run Prim's algorithm starting from the origin, maintaining a min-heap of (cost, node) and a visited set. Total c... | {
"exploration": "I'll binary search on E, the maximum allowed planting energy. For a fixed E, each potato i at (x_i, y_i) can be planted from any point (X, Y) on Annie's path such that X ≤ x_i, Y ≤ y_i, and max(|X - x_i|, |Y - y_i|) ≤ E. Since the path only moves right and up, X and Y are non-decreasing along the pa... | null | null | null | Well, that was a dead end after dead end. The greedy scan was computationally impossible, the MST fundamentally misunderstood the cost structure, and any flow-based approach exploded in scale before we even started.
hmm. That half-formed thought about Chebyshev distance and antidiagonals... Chebyshev distance is max(|... | function solve():
READ n
points = list of size n
for i from 0 to n-1:
READ x_i, y_i
u_i = x_i + y_i
v_i = x_i - y_i
points[i] = (u_i, v_i)
sort points by u_i ascending, breaking ties arbitrarily
L = max-heap (priority queue that returns largest element)
R = min-h... |
582/D | Number of Binominal Coefficients | 3,300 | dp, math, number theory | 4 | 256 | For a given prime integer p and integers α, A calculate the number of pairs of integers (n, k), such that 0 ≤ k ≤ n ≤ A and $${\binom {n}{k}}$$ is divisible by pα.
As the answer can be rather large, print the remainder of the answer moduly 109 + 7.
Let us remind you that $${\binom {n}{k}}$$ is the number of ways k ob... | The first line contains two integers, p and α (1 ≤ p, α ≤ 109, p is prime).
The second line contains the decimal record of integer A (0 ≤ A < 101000) without leading zeroes. | In the single line print the answer to the problem. | null | In the first sample three binominal coefficients divisible by 4 are $${ \binom { 4 } { 1 } } = 4$$, $${ \binom { 4 } { 3 } } = 4$$ and $${ \binom { 6 } { 3 } } = 20$$. | [
{
"input": "2 2\n7",
"output": "3"
},
{
"input": "3 1\n9",
"output": "17"
},
{
"input": "3 3\n9",
"output": "0"
},
{
"input": "2 4\n5000",
"output": "8576851"
}
] | It is a common fact that for a prime $p$ and integer $n$ maximum $α$, such that $p^{α}|n!$ is calculated as $[formula]$, where $p^{w} ≤ n < p^{w + 1}$. As soon as $[formula]$, the maximum $α$ for $[formula]$ is calculated as $[formula]$.
One could see, that if we consider numbers $n$, $k$ and $n - k$ in $p$-th based n... | The key observation is that the exponent of p dividing a binomial coefficient is related to carries when adding k and n-k in base p. This suggests a digit DP over the base-p representation of A, where you track the number of carries generated so far and an equality flag. | {
"exploration": "Okay so we have a prime p, exponent α, and a huge integer A up to 10^1000 digits. We need to count pairs (n,k) with 0 ≤ k ≤ n ≤ A such that the binomial coefficient C(n,k) is divisible by p^α. The answer modulo 1e9+7.\n\nThe hypothesis is to iterate n from 0 to A, and for each n iterate k from 0 to ... | {
"exploration": "I'll represent n in base p. Let the digits of n in base p be n_d ... n_0, where n = sum n_i p^i. Lucas's theorem says binom(n,k) mod p^α depends on the carries when adding k and n-k in base p. Specifically, the exponent of p dividing binom(n,k) equals the number of carries in the base-p addition k +... | null | null | null | null | Alright, back to square one. Direct enumeration is impossible because A is astronomically large, and meet-in-the-middle fails because p itself can be huge, so splitting on base-p digits doesn't create a manageable number of states.
hmm — but that base-p digit thought is the only thread left. The exponent of p in C(n,k... | function convert_to_base_p(A_str, p):
// Convert decimal string A_str to base-p digits (least significant first)
digits = empty list
while A_str != "0":
remainder = 0
new_A_str = ""
for each char ch in A_str:
current = remainder * 10 + (ch - '0')
new_A_str += ... |
1523/F | Favorite Game | 3,300 | bitmasks, dp | 2 | 256 | After William is done with work for the day, he enjoys playing his favorite video game.
The game happens in a 2D world, starting at turn $$$0$$$. William can pick any cell in the game world and spawn in it. Then, each turn, William may remain at his current location or move from the current location (x, y) to one of t... | The first line contains two integers $$$n$$$ and $$$m$$$ ($$$0 \le n \le 14, 1 \le m \le 100$$$), which are the number of towers and the number of quests, respectively.
Each of the next $$$n$$$ lines contains two integers $$$xa_i, ya_i$$$ ($$$1 \le xa_i, ya_i \le 10^6$$$), which are the coordinates of fast travel towe... | Print a single number — the maximal number of quests William will be able to complete. | null | In the first sample test one of the possible sequences of William's actions is as follows:
- Spawn at $$$(3, 2)$$$
- On turn $$$1$$$ move to $$$(4, 2)$$$
- On turn $$$2$$$ move to $$$(5, 2)$$$. By visiting this cell William activates tower number $$$3$$$.
- On turn $$$3$$$ move to $$$(5, 1)$$$, where he waits for $$$1... | [
{
"input": "3 4\n1 1\n2 3\n5 2\n2 2 12\n5 1 4\n6 2 11\n3 5 10",
"output": "3"
}
] | For convenience we will sort quests by time.
Let's make two DP:
$F(mask, done_q)$ — minimum amount of time it takes to visit the set of $mask$ towers and complete $done_q$ quests. William is in one of the towers.
$G(mask, q)$ — maximum number of quests that William can complete if he visited a set of $mask$ towers, and... | The key observation is that you can model the problem as two interleaved dynamic programming tables that track your position in two different ways. Think about representing your ability to teleport via activated towers as a set mask, and the fact that quests have strict time constraints forces you to consider whether y... | {
"exploration": "Okay so we have n towers (n ≤ 14) and m quests (m ≤ 100). Each quest has a location and a turn t_i when it can be completed. We can move one cell per turn in 4 directions, and after visiting a tower's location, that tower becomes permanently active — from then on we can instantly teleport to it from... | {
"exploration": "I'll compute dp[mask] = minimum time to activate exactly the towers in mask, starting from any cell. For a single tower i, dp[1<<i] = 0 since we can spawn at that tower. Then for each mask, to add a new tower j not in mask, the cost is the minimum over all towers k in mask of (dp[mask] + Manhattan d... | null | null | null | null | So none of those panned out. Greedy ordering locked me into a sequence that might miss activating a crucial tower earlier, and the pure tower-mask DP couldn't capture the fact that my physical position between quests matters — I might finish a quest at some location, then have to walk to activate new towers before the ... | function solve():
READ n, m
towers = array of size n
for i from 0 to n-1:
READ towers[i].x, towers[i].y
quests = array of size m
for i from 0 to m-1:
READ quests[i].x, quests[i].y, quests[i].t
sort quests by t in non-decreasing order
// Precompute distances
towerDist = m... |
576/E | Painting Edges | 3,300 | binary search, data structures | 6 | 600 | Note the unusual memory limit for this problem.
You are given an undirected graph consisting of n vertices and m edges. The vertices are numbered with integers from 1 to n, the edges are numbered with integers from 1 to m. Each edge can be unpainted or be painted in one of the k colors, which are numbered with integer... | The first line contains integers n, m, k, q (2 ≤ n ≤ 5·105, 1 ≤ m, q ≤ 5·105, 1 ≤ k ≤ 50) — the number of vertices, the number of edges, the number of colors and the number of queries.
Then follow m edges of the graph in the form ai, bi (1 ≤ ai, bi ≤ n).
Then follow q queries of the form ei, ci (1 ≤ ei ≤ m, 1 ≤ ci ≤ ... | For each query print "YES" (without the quotes), if it is valid, or "NO" (without the quotes), if this query destroys the bipartivity of the graph formed by the edges of some color. | null | null | [
{
"input": "3 3 2 5\n1 2\n2 3\n1 3\n1 1\n2 1\n3 2\n3 1\n2 2",
"output": "YES\nYES\nYES\nNO\nYES"
}
] | Let's solve an easier task first: independent of bipartivity, the color of edge changes.
Then we could write a solution, which is pretty similar to solution of Dynamic Connectivity Offline task in $O(nlog^{2n})$.
Let's consider only cases, where edges are not being deleted from the color graph. Then, we could use DSU... | What if you could treat each color's graph as a separate dynamic connectivity problem? The key observation is that you can precompute, for each edge, the future time intervals it will spend in each color, and then process these intervals offline. This points to a data structure that can handle edge additions and deleti... | {
"exploration": "Okay so we have an undirected graph with n vertices and m edges, k colors (k ≤ 50), and q queries. Each query tries to repaint an edge to a new color, but the subgraph of each color must remain bipartite at all times. If repainting would create an odd cycle in that color, the query is rejected and t... | {
"exploration": "For each color c from 1 to k, I'll maintain an adjacency list adj[c] storing the current edges of that color. I'll also maintain a color array col[c][v] for each vertex v in each color's graph — the bipartite side (0 or 1). When a query (e, new_c) arrives, I first remove edge e from its current colo... | {
"exploration": "For each color, I'll maintain a union-find structure to track connected components. When an edge is added to a color, I union its endpoints. For each component, I build a heavy-light decomposition with a segment tree that stores parity (XOR) values along paths. To check if adding edge (u,v) creates ... | null | null | null | Okay, all of those went nowhere. The online DSU with rollback choked on arbitrary edge deletions, BFS per query was hopelessly slow, and HLD couldn't handle component splits efficiently. The core problem is that checking bipartiteness after a single edge color change seems to require re-evaluating an entire connected c... | READ n, m, k, q
edges = array of size m+1
FOR i FROM 1 TO m:
READ a, b
edges[i] = (a, b)
queries = array of size q+1
FOR i FROM 1 TO q:
READ e, c
queries[i] = (e, c)
// next_time[e] = next query index that touches edge e, or q+1 if none
next_time = array of size m+1, initialized to q+1
FOR i FROM q D... |
1886/F | Diamond Theft | 3,300 | data structures, greedy | 2 | 256 | Monocarp is the most famous thief in Berland. This time, he decided to steal two diamonds. Unfortunately for Monocarp, there are $$$n$$$ cameras monitoring the diamonds. Each camera has two parameters, $$$t_i$$$ and $$$s_i$$$. The first parameter determines whether the camera is monitoring the first diamond only ($$$t_... | The first line contains a single integer $$$n$$$ ($$$0 \le n \le 1500$$$) — the number of cameras.
Then $$$n$$$ lines follow, the $$$i$$$-th of them contains two integers $$$t_i$$$ and $$$s_i$$$ ($$$1 \le t_i \le 3$$$; $$$1 \le s_i \le 2n$$$) — the parameters of the $$$i$$$-th camera. | Print a single integer — the minimum time it will take for Monocarp to steal the first diamond first and then the second diamond. If it is impossible, print -1. | null | null | [
{
"input": "4\n2 6\n1 2\n1 2\n2 1",
"output": "6"
},
{
"input": "4\n2 8\n3 2\n3 2\n3 5",
"output": "9"
},
{
"input": "2\n3 2\n2 3",
"output": "4"
},
{
"input": "1\n3 1",
"output": "4"
},
{
"input": "8\n2 1\n2 2\n3 5\n3 6\n1 2\n1 3\n1 4\n1 5",
"output": "11"
... | In this editorial, we will denote the number of cameras of type $i$ as $k_i$.
Without loss of generality, we can assume that our sequence of actions will look as follows:
• hack all cameras of type $1$, some cameras of type $2$ and all cameras of type $3$ in some order (let's call it the first block of actions);
• stea... | The key observation is that the timeline splits into two consecutive hacking blocks separated by the first diamond theft. Within each block, you need to schedule hacks to keep relevant cameras disabled. This becomes a feasibility check reminiscent of Hall's theorem on intervals, where you can precompute for each prefix... | {
"exploration": "Okay so we have n cameras, each monitoring diamond 1, diamond 2, or both, with a disable duration s_i. We need to steal diamond 1 first, then diamond 2, and we can hack cameras at any second, each hack disables the camera for s_i seconds starting the next second. The goal is minimum time to steal bo... | {
"exploration": "I'll build a time-expanded flow network. Let me set an upper bound T_max — maybe 4n or something like that, since each camera has s_i ≤ 2n and we might need to hack multiple cameras. I'll create nodes for each second from 1 to T_max. Source S, sink T. For each camera i, I create a hack-event node H_... | null | null | null | null | So both attempts failed — the DP over camera states exploded exponentially, and the flow model fundamentally couldn't enforce the AND condition of all cameras being disabled simultaneously.
hmm. But the timeline is sequential: steal diamond 1, then diamond 2. That means there's a moment T1 when the first diamond is st... | function solve():
READ n
cameras = list of size n
for i from 0 to n-1:
READ t_i, s_i
cameras[i] = (t_i, s_i)
// Separate cameras by type
type1 = []
type2 = []
type3 = []
for (t, s) in cameras:
if t == 1:
type1.append(s)
else if t == 2:
... |
1329/E | Dreamoon Loves AA | 3,300 | binary search, greedy | 3.5 | 256 | There is a string of length $$$n+1$$$ of characters 'A' and 'B'. The first character and last character of the string are equal to 'A'.
You are given $$$m$$$ indices $$$p_1, p_2, \ldots, p_m$$$ ($$$0$$$-indexation) denoting the other indices of characters 'A' in the string.
Let's denote the minimum distance between t... | The first line contains one integer $$$t$$$ denoting the number of test cases ($$$1 \leq t \leq 400\,000$$$).
For each test case, the first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$1 \leq n \leq 10^{15}, 0 \leq m \leq 400\,000, 0 \leq k < n - m$$$).
The second line contains $$$m$$$ integers $$$p_... | For each test case, print one integer: the smallest possible value of balance degree after $$$k$$$ changes of 'B' to 'A'. | null | null | [
{
"input": "5\n80 3 5\n11 24 50\n81 7 12\n4 10 17 26 37 48 61\n25 10 14\n3 4 7 12 13 15 17 19 21 23\n1 0 0\n\n10 2 0\n2 4",
"output": "5\n2\n0\n0\n4"
}
] | The idea of this problem comes to my brain when I recall that I'm cooking and I want to cut a carrot into many pieces and make the size of each piece as evenly as possible. I think this problem is very interesting. So that's why I determine to prepare a Codeforces contest after so many years. I hope I can share this pr... | The key observation is to think about cutting each existing 'A'-gap into smaller pieces. You need to distribute your k cuts across the gaps to make the resulting piece lengths as uniform as possible. This becomes a problem of balancing upper and lower bounds on piece lengths, which you can search for. | {
"exploration": "Okay so we have a string of length n+1 with 'A' at positions 0, p_1..p_m, and n. We can add k more 'A's at currently 'B' positions. The balance degree is r - l where l is minimum gap between consecutive 'A's and r is maximum gap. We want to minimize r - l.\n\nThe hypothesis is: maintain a max-heap o... | {
"exploration": "I'll define dp[i][c] as a pair (min_gap, max_gap) — the best possible min and max gap lengths among the first i segments (between original A's at positions p_0=0, p_1, ..., p_i) after inserting exactly c extra A's total across those segments. The goal is to minimize max_gap - min_gap at the end. For... | {
"exploration": "Let me start with a concrete representation. I'll maintain a sorted list of all 'A' positions — the original m+2 positions (0 and n are the endpoints, plus the m given p_i) and the k new positions I'm adding. Initially I'll pick k random integers in [1, n-1] that aren't already 'A' positions, sort t... | null | null | null | So that's a few failed attempts now. Greedy splitting the largest gap blindly worsens the minimum, DP can't handle k up to 10^15, and any explicit simulation of cut positions is impossible.
hmm. But that thread — distributing k cuts across gaps to make piece lengths uniform — feels like the right framing. Each existin... | function solve():
READ t
for each test case:
READ n, m, k
READ p[1..m]
// compute gaps d[1..M] where M = m+1
d = array of size m+2
d[1] = p[1] - 0
for i from 2 to m:
d[i] = p[i] - p[i-1]
d[m+1] = n - p[m]
M = m+1
// helper funct... |
1270/H | Number of Components | 3,300 | data structures | 8 | 256 | Suppose that we have an array of $$$n$$$ distinct numbers $$$a_1, a_2, \dots, a_n$$$. Let's build a graph on $$$n$$$ vertices as follows: for every pair of vertices $$$i < j$$$ let's connect $$$i$$$ and $$$j$$$ with an edge, if $$$a_i < a_j$$$. Let's define weight of the array to be the number of connected components i... | The first line contains two integers $$$n$$$ and $$$q$$$ ($$$1 \le n, q \le 5 \cdot 10^5$$$) — the size of the array and the number of queries.
The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^6$$$) — the initial array.
Each of the next $$$q$$$ lines contains two integers $$$p... | After each query, output the weight of the array. | null | After the first query array looks like $$$[25, 40, 30, 20, 10]$$$, the weight is equal to $$$3$$$.
After the second query array looks like $$$[25, 40, 45, 20, 10]$$$, the weight is still equal to $$$3$$$.
After the third query array looks like $$$[48, 40, 45, 20, 10]$$$, the weight is equal to $$$4$$$. | [
{
"input": "5 3\n50 40 30 20 10\n1 25\n3 45\n1 48",
"output": "3\n3\n4"
}
] | Firstly, note that all connected components form segments of sequential indexes. Indeed, let numbers $i$ and $j$ lie in one component and let's consider $i < x < j$. Because $i$ and $j$ lie in one component, there exists a path, which connects them: $v_1 = i, \ldots, v_t = j$, such that $v_k$ and $v_{k+1}$ are connecte... | The key observation is that components correspond to contiguous segments, and a segment boundary occurs exactly when a certain prefix condition holds. This condition can be reinterpreted by looking at the array through many thresholds. The technique involves maintaining a global statistic over all possible thresholds, ... | {
"exploration": "I'll maintain a set of edges that exist because a_i < a_j for i<j. The graph is defined by these edges, and components are determined by connectivity. After each update at position pos, I need to recompute which edges incident to pos change. For each neighbor i<pos, if a_i < a_pos before update and ... | {
"exploration": "I'll try to compute the weight as the number of \"runs\" in the sorted order of positions by value. If I sort indices by a_i, then two consecutive indices in this sorted order are connected if they are adjacent in the original index order? Actually, the graph connects i and j if a_i < a_j, which is ... | null | null | null | null | Alright, every approach I tried hit something — the dynamic connectivity was impossible because the graph is dense, and the BIT for inversions lacked the right invariant to capture components directly. The graph definition is so simple: connect i<j if a_i < a_j. That means a component is just a set of indices where the... | function solve():
READ n, q
READ a[1..n]
// Initialize segment trees
// segA for prefix minima A[i] = min(a[1..i])
// segR for suffix maxima R[i] = max(a[i..n])
// segCnt for counting i where A[i] > R[i+1], over i=1..n-1
// Initialize A and R arrays
A = array of size n
R = array of s... |
1416/F | Showing Off | 3,300 | flows, graph matchings, greedy, implementation | 6 | 256 | Another dull quarantine day was going by when BThero decided to start researching matrices of size $$$n \times m$$$. The rows are numerated $$$1$$$ through $$$n$$$ from top to bottom, and the columns are numerated $$$1$$$ through $$$m$$$ from left to right. The cell in the $$$i$$$-th row and $$$j$$$-th column is denote... | The first line of input file contains a single integer $$$T$$$ ($$$1 \le T \le 100$$$) denoting the number of test cases. The description of $$$T$$$ test cases follows.
First line of a test case contains two integers $$$n$$$ and $$$m$$$ ($$$1 \le n \cdot m \le 10^5$$$).
Each of the following $$$n$$$ lines contain exa... | For each test case, if an answer does not exist, print a single word NO. Otherwise, print YES and both matrices in the same format as in the input.
- The first matrix should be the cost matrix and the second matrix should be the direction matrix.
- All integers in the cost matrix should be positive.
- All characters i... | null | null | [
{
"input": "2\n3 4\n7 6 7 8\n5 5 4 4\n5 7 4 4\n1 1\n5",
"output": "YES\n1 1 1 1\n2 1 1 1\n3 2 1 1\nR D L L\nD R D L\nU L R U\nNO"
}
] | Several unexpected Kuhn solutions passed for D1F. Could you please discuss your solutions in the comments and prove its correctness or provide any counter-examples. Author's solution uses flows with Dinic.
Editorial is not completed yet. Problem D1F will be added later. Hope you enjoyed the problemset!
Editorial was/... | The key observation is that the reachable set from a cell is determined by following a functional graph of directions. This creates a partition into components where values are constant. Within each component, the sum condition forces a relationship between costs and directions. Think about pairing adjacent cells with ... | {
"exploration": "Okay so we have an n×m grid where each cell has a cost and a direction (L,R,U,D) that points to an adjacent cell, never outside. The reachable set S_{i,j} is all cells you eventually visit starting from (i,j) and following directions. The given matrix B[i][j] is the sum of costs of all cells in S_{i... | {
"exploration": "I'll set up a system of equations. For each cell (i,j), the value in the produced matrix is the sum of costs of all cells reachable from (i,j). Since directions form functional edges (each cell points to exactly one neighbor), the graph is a functional graph — each component has exactly one cycle, w... | null | null | null | null | So none of those panned out. The component tree gave zero costs, and the cycle constraints were too rigid — equal-B cells can't just have costs as differences; they need to be in cycles where the costs sum to the budget in a way that keeps everything positive.
hmm. But that nagging thought about equal values being pai... | function solveTestCase():
READ n, m
B = 2D array of size n x m
for i from 0 to n-1:
for j from 0 to m-1:
READ B[i][j]
cost = 2D array of size n x m, initialized to 0
dir = 2D array of size n x m, initialized to -1
visited = 2D array of size n x m, initialized to false
in... |
1010/F | Tree | 3,400 | fft, graphs, trees | 7 | 256 | The Main Martian Tree grows on Mars. It is a binary tree (a rooted tree, with no more than two sons at each vertex) with $$$n$$$ vertices, where the root vertex has the number $$$1$$$. Its fruits are the Main Martian Fruits. It's summer now, so this tree does not have any fruit yet.
Autumn is coming soon, and leaves a... | The first line contains two integers: $$$n$$$ and $$$x$$$ ($$$1 \le n \le 10^5$$$, $$$0 \le x \le 10^{18}$$$) — the size of the tree and the number of fruits in the root.
The $$$i$$$-th of the following $$$(n-1)$$$ lines contains two integers $$$a_i$$$ and $$$b_i$$$ ($$$1 \le a_i, b_i \le n$$$) — vertices connected by... | Print one number — the number of configurations of the resulting tree modulo $$$998244353$$$. | null | Consider the first example.
There are $$$2$$$ fruits at the vertex $$$1$$$. The following $$$13$$$ options are possible:
- there is no vertex $$$2$$$, there is no vertex $$$3$$$;
- there is no vertex $$$2$$$, there are no fruits at the vertex $$$3$$$;
- there is no vertex $$$2$$$, there is $$$1$$$ fruit at the vertex... | [
{
"input": "3 2\n1 2\n1 3",
"output": "13"
},
{
"input": "2 5\n1 2",
"output": "7"
},
{
"input": "4 10\n1 2\n1 3\n3 4",
"output": "441"
}
] | Let $b_v = a_v - \sum{a_{to}}$, (for each $(v, to)$, such that $to$ — is a son of $v$ and tree have both vertices $v$ and $to$), then all that we need is $b_v \geq 0$ and $\sum{b_v} = x$, so for fixed subset of vertices number of ways to arrange weights is just number of ways to partite $x$ into such number of parts.
S... | What if you think of the fruit assignment condition as a non‑negative slack variable at each remaining vertex? The key observation is that the total slack across all remaining vertices must sum to the root's fruit count, turning the problem into counting integer compositions. This reduces the fruit‑counting to a combin... | {
"exploration": "For each vertex, we need to count the number of ways to assign a non-negative integer number of fruits to that vertex, given that the number at a vertex must be at least the sum of the numbers at its remaining children. The root gets exactly x fruits. The tree can lose any subset of vertices (except... | {
"exploration": "I'll decompose the tree into two halves and count configurations for each half independently, then combine via convolution. For a binary tree, I can find a centroid edge or vertex that splits the tree into roughly balanced parts. Let me pick the root's children as the natural split — the root is alw... | null | null | null | null | None of those got me anywhere useful. Both paths tried to brute-force the combinatorial sum term by term, which is O(n²) in the worst case — impossible for n=10⁵.
hmm. The fruit assignment condition: for each remaining vertex, its fruit count ≥ sum of fruits in its remaining children. That's like a flow constraint: th... | MOD = 998244353
PRIMITIVE_ROOT = 3
function ntt(a, invert):
n = length(a)
j = 0
for i from 1 to n-1:
bit = n >> 1
while j & bit:
j ^= bit
bit >>= 1
j ^= bit
if i < j:
swap(a[i], a[j])
len = 2
while len <= n:
wlen = pow(PRIM... |
1290/E | Cartesian Tree | 3,300 | data structures | 5 | 256 | Ildar is the algorithm teacher of William and Harris. Today, Ildar is teaching Cartesian Tree. However, Harris is sick, so Ildar is only teaching William.
A cartesian tree is a rooted tree, that can be constructed from a sequence of distinct integers. We build the cartesian tree as follows:
1. If the sequence is empt... | The first line contains a single integer $$$n$$$ ($$$1 \le n \le 150000$$$).
The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le n$$$). It is guarenteed that each integer from $$$1$$$ to $$$n$$$ appears in the sequence exactly once. | Print $$$n$$$ lines, $$$i$$$-th line should contain a single integer — the answer to the $$$i$$$-th question. | null | After the first round, the sequence is $$$1$$$. The tree is
The answer is $$$1$$$.
After the second round, the sequence is $$$2, 1$$$. The tree is
The answer is $$$2+1=3$$$.
After the third round, the sequence is $$$2, 1, 3$$$. The tree is
The answer is $$$2+1+3=6$$$.
After the fourth round, the sequence is $$$2,... | [
{
"input": "5\n2 4 1 5 3",
"output": "1\n3\n6\n8\n11"
},
{
"input": "6\n1 2 4 5 6 3",
"output": "1\n3\n6\n8\n12\n17"
}
] | Instead of inserting numbers one by one, let's imagine I have $n$ blanks in a row, and I will fill the blanks with integers from $[1,n]$ in order, and I want to know about the cartesian tree after each blank filled.
In the following parts of the solution, I will use positions in the original array instead of positions ... | The key observation is that the total subtree size sum can be expressed as contributions from specific interval endpoints. The structure of the Cartesian tree under insertions causes these endpoints to change via range-min operations on an array. This points to a data structure that can efficiently support many range-c... | {
"exploration": "Okay so we have a sequence that starts empty and we insert values 1 through n one by one, each time inserting at some position in the current sequence. After each insertion we need the sum of subtree sizes of the Cartesian tree built from the current sequence. The Cartesian tree is built by taking t... | {
"exploration": "I'll process the insertions in order 1..n, maintaining a monotonic decreasing stack of (value, position) pairs. When inserting value i at position pos (given by the final array), I pop from the stack while the top value is less than i, then push (i, pos). Each node in the Cartesian tree corresponds ... | {
"exploration": "I'll build a persistent segment tree over the Euler tour of the Cartesian tree as insertions happen. The idea: process insertions in order i=1..n, each time inserting value i at its position. After each insertion, the Cartesian tree changes — I need to capture its structure in a persistent segment t... | null | null | null | Okay, so every approach hit a wall: scanning for the maximum each time is quadratic, union-find can't handle the dynamic parent reassignments when a new maximum splits a subtree, and persistent structures explode in complexity because the entire tree topology can shift with each insertion.
Hmm. Wait — the Cartesian tr... | function solve():
READ n
READ a[1..n] // final positions: a[i] = position of value i in final sequence
// Compute insertion positions pos[i] for i from 1 to n
fenwick = array of size n+1 initialized to 0
pos = array of size n+1
for i from 1 to n:
x = a[i]
// query number of j <... |
1517/G | Starry Night Camping | 3,300 | constructive algorithms, flows, graphs | 2 | 256 | At the foot of Liyushan Mountain, $$$n$$$ tents will be carefully arranged to provide accommodation for those who are willing to experience the joy of approaching nature, the tranquility of the night, and the bright starry sky.
The $$$i$$$-th tent is located at the point of $$$(x_i, y_i)$$$ and has a weight of $$$w_i$... | The first line contains a single integer $$$n$$$ ($$$1\leq n\leq 1\,000$$$), representing the number of tents.
Each of the next $$$n$$$ lines contains three integers $$$x_i$$$, $$$y_i$$$ and $$$w_i$$$ ($$$-10^9\leq x_i,y_i \leq 10^9$$$, $$$1\leq w_i\leq 10^9$$$), representing the coordinate of the $$$i$$$-th tent and ... | A single integer — the maximum sum of the weights of the remaining tents. | null | Here is an illustration of the second example. Black triangles indicate the important tents. This example also indicates all $$$8$$$ forbidden patterns. | [
{
"input": "5\n0 0 4\n0 1 5\n1 0 3\n1 1 1\n-1 1 2",
"output": "12"
},
{
"input": "32\n2 2 1\n2 3 1\n3 2 1\n3 3 1\n2 6 1\n2 5 1\n3 6 1\n3 5 1\n2 8 1\n2 9 1\n1 8 1\n1 9 1\n2 12 1\n2 11 1\n1 12 1\n1 11 1\n6 2 1\n7 2 1\n6 3 1\n5 3 1\n6 6 1\n7 6 1\n5 5 1\n6 5 1\n6 8 1\n5 8 1\n6 9 1\n7 9 1\n6 12 1\n5 12 1... | We can label all the integer points in the plane as follow:
...2323...
...1010...
...2323...
...1010...
Where all the good points are labeled with 1. If we draw an edge between every adjacent points, all the forbidden patterns form all paths of length 4 with label 0-1-2-3. So we transform the problem to another problem... | What if you think of the condition in terms of graph paths? The key observation is that the parity constraints on coordinates induce a 4-coloring of the integer grid. The forbidden pattern for an important tent corresponds to a specific short walk through all four colors. This suggests a flow or cut formulation on a bi... | {
"exploration": "I'll enumerate all subsets of the n tents, check each subset for validity against the forbidden pattern, and take the maximum total weight among valid subsets. For n=1000, there are 2^1000 subsets — completely impossible to enumerate. But let me trace the logic on Example 1 first to understand what ... | {
"exploration": "I'll sort all important tents (both x and y even) by weight descending. I'll maintain a set of selected tents, initially empty. For each important tent in sorted order, I'll check whether adding it creates a forbidden parallelogram with any three already-selected tents. If it does, I skip it; otherw... | {
"exploration": "I'll scan each important tent (both coordinates even), and for each one, enumerate all triples of other tents within its 3×3 neighborhood (|dx|≤1, |dy|≤1) to check if they form a forbidden parallelogram with one side parallel to the x-axis. A parallelogram with one side parallel to x-axis means two ... | null | null | null | Okay, all of those went nowhere. Brute force is astronomically impossible, greedy fails because local choices ignore global constraints, and explicitly modeling the hypergraph is computationally intractable.
hmm — that half-formed thought about parity and coloring... wait — let's map it out. An important tent has both... | function solve():
READ n
tents = list of size n
for i from 0 to n-1:
READ x_i, y_i, w_i
tents[i] = (x_i, y_i, w_i)
// Map each tent by its coordinates for quick lookup
coord_to_idx = empty map
for i from 0 to n-1:
(x, y, w) = tents[i]
coord_to_idx[(x, y)] = i
... |
2190/G | Maximize Determinant | 3,300 | graphs | null | null | An $$$n \times n$$$ matrix $$$B$$$ is called an interval matrix if for each row $$$i$$$, there exists a contiguous range of columns $$$[l_i, r_i]$$$ ($$$1 \le l_i \le r_i \le n$$$) filled with ones, while all other elements in the row are zero. Formally, $$$B_{i, j} = 1$$$ if $$$l_i \le j \le r_i$$$, and $$$B_{i, j} = ... | Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 5 \cdot 10^4$$$). The description of the test cases follows.
The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^6$$$) — the size of the matrix.
The next $$$n$$$ lines des... | For each test case, print a single integer — the minimum total cost required to make $$$\operatorname{det}(A) = X$$$. | null | null | [] | First let's find an easy way to calculate the determinant of a fixed $A$. We can easily see that $\operatorname{det}(A) \in [-1, 0, 1]$ for any $A$.
**Proof**
If there's no ones in the first column, the determinant is $0$. Otherwise, let $i$ be the row with $l_i = 1$ and $r_i$ being as small as possible. We can subtr... | The key observation is that the determinant of an interval matrix is directly tied to a graph property. Each row defines a connection between two columns, and the matrix is singular if and only if these connections create a cycle. To maximize the determinant, you need to ensure the underlying graph is a tree, and then ... | {
"exploration": "I'll compute the determinant directly by Gaussian elimination on the interval matrix. For each row i, I have an interval [l_i, r_i] of ones. I'll treat the matrix as a binary matrix over reals and perform elimination to compute the determinant as the product of pivots. I'll iterate rows in order, fi... | {
"exploration": "Let me start by understanding what the maximum determinant X actually is for interval matrices. For an n×n interval matrix, each row is a contiguous block of ones. The determinant over GF(2) is 1 if the rows are linearly independent, 0 otherwise. Over integers, the determinant of a 0-1 interval matr... | {
"exploration": "I'll set up a randomized rank estimation over GF(2). For each trial, I sample a random vector x ∈ {0,1}^n, compute y = A x (mod 2), and collect y vectors to estimate the dimension of the column space. If the matrix is totally unimodular and has full rank n, the determinant is ±1. The goal is to find... | null | null | null | Well, that was a dead end after dead end. Direct computation is astronomically slow, DP over subsets is impossible, and spectral approximations miss the discrete structure. The common failure is treating the determinant as a generic real number — but an interval matrix is a 0-1 matrix with a very specific geometric sha... | function get_cycles(p):
n = length(p)
used = array of size n, initialized to false
cnt = 0
for i from 0 to n-1:
if not used[i]:
cnt += 1
j = i
while not used[j]:
used[j] = true
j = p[j]
return cnt
function solve_zero(a):
... |
1466/H | Finding satisfactory solutions | 3,300 | combinatorics, dp, graphs, greedy, math | 5 | 1,024 | Getting so far in this contest is not an easy feat. By solving all the previous problems, you have impressed the gods greatly. Thus, they decided to spare you the story for this problem and grant a formal statement instead.
Consider $$$n$$$ agents. Each one of them initially has exactly one item, $$$i$$$-th agent has ... | In the first line of input there is an integer $$$n$$$ ($$$1 \leq n \leq 40$$$). The next line contains $$$n$$$ space separated integers, a permutation of numbers from $$$1$$$ to $$$n$$$. The $$$i$$$-th number denotes the item assigned to agent $$$i$$$ in the optimal assignment. | In a single line output one non-negative integer, the number of preference profiles for which the assignment of items given in the input is optimal modulo $$$10^9+7$$$. | null | Assignment from the first test case is optimal only for the following preference profile:
$$$2, 1$$$
$$$1, 2$$$
If any agent wants his initial item the most and is given another item, he would form a dissatisfied set. Hence the allocation is not optimal for any other preference profile. | [
{
"input": "2\n2 1",
"output": "1"
},
{
"input": "3\n1 2 3",
"output": "98"
},
{
"input": "4\n2 1 3 4",
"output": "27408"
}
] | We start with presenting the algorithm for finding the unique optimal good assignment. It treats agents as vertices in a graph and works in the following way:
• For every $i$ from $1$ to $n$ create a vertex (a single vertex will represent both the agent and his good)
• Until there are agents left, perform a round of an... | The key observation is that the optimal assignment is built in rounds where agents point to their favorite available item, forming cycles. The number of profiles depends on how many ways to arrange preferences to generate exactly these cycles. This suggests a combinatorial DP that simulates the round-by-round cycle for... | {
"exploration": "Okay so we have n agents, each with a preference permutation over n items. The input gives an assignment A — a permutation where A(i) is the item agent i gets. We need to count how many preference profiles (n permutations, one per agent) make A optimal. Optimal means no subset S of agents can intern... | {
"exploration": "We need to count preference profiles where the given assignment A is optimal. The hypothesis is to use dynamic programming over item assignments per agent. Let me try to build a DP that processes agents one by one, tracking which items have been assigned so far and which preferences are allowed for ... | {
"exploration": "Let me encode each agent's preference as a permutation of n items, represented by rank variables. For agent i, let r_i(x) be the rank of item x in i's preference (1 = most preferred, n = least). The condition that no subset S is dissatisfied means: for every nonempty S and every possible reassignmen... | null | null | null | Hmm, all those ideas fell apart. Brute force over all profiles is astronomically impossible, DP over subsets is still exponential in n=40, and trying to enumerate all cycle constraints directly explodes combinatorially. The common failure is trying to count profiles by directly checking the optimality condition for eve... | MOD = 10^9 + 7
function solve():
READ n
READ perm[1..n] // permutation A
// Find cycles in the permutation
visited = array of size n+1, initialized to false
cycles = [] // list of cycle lengths
for i from 1 to n:
if not visited[i]:
length = 0
cur = i
... |
1427/G | One Billion Shades of Grey | 3,300 | flows, graphs | 3 | 256 | You have to paint with shades of grey the tiles of an $$$n\times n$$$ wall. The wall has $$$n$$$ rows of tiles, each with $$$n$$$ tiles.
The tiles on the boundary of the wall (i.e., on the first row, last row, first column and last column) are already painted and you shall not change their color. All the other tiles a... | The first line contains $$$n$$$ ($$$3\le n\le 200$$$) – the number of rows and columns.
Then $$$n$$$ lines, each containing $$$n$$$ integers, follow. The $$$i$$$-th of these lines describe the $$$i$$$-th row of tiles. It contains the $$$n$$$ integers $$$a_{ij}$$$ ($$$-1\le a_{ij} \le 10^9)$$$. The value of $$$a_{ij}$$... | Print a single integer – the minimum possible total contrast of the wall. | null | Explanation of the first testcase: The initial configuration of the tiles is (tiles to paint are denoted by ?):
1 7 64 ? 61 1 1
1 7 64 5 61 1 1
Explanation of the second testcase: Since all tiles are either painted or broken, there is nothing to do. The total contrast is $$$396$$$.
Explanation of the third testcase... | [
{
"input": "3\n1 7 6\n4 0 6\n1 1 1",
"output": "26"
},
{
"input": "3\n10 100 1\n1 -1 100\n10 10 10",
"output": "396"
},
{
"input": "5\n6 6 5 4 4\n6 0 0 0 4\n7 0 0 0 3\n8 0 0 0 2\n8 8 1 2 2",
"output": "34"
},
{
"input": "7\n315055237 841510063 581663979 148389224 405375301 24... | We present two solutions: the first solution is mine, the second is by dacin21.
• The first solution reduces the problem to the computation of $O(n)$ min-cuts and then computes the min-cuts with total complexity $O(n^3)$. This solution requires no advanced knowledge and it fits easily in the time-limit.
• The second so... | The key observation is that the total contrast can be expressed as a sum over thresholds. This suggests a flow-based approach where you need to compute a sequence of related cuts efficiently. Think about how the assignment of shades to interior vertices partitions the graph relative to the fixed boundary values. | {
"exploration": "I'll start with all unpainted interior tiles set to some initial value — say the average of their already-painted neighbors — then iteratively smooth each tile to the median or mean of its four neighbors, repeating until convergence or a fixed number of passes.\n\nLet me trace Example 1:\nn=3, grid:... | {
"exploration": "Okay so we have an n×n grid with n up to 200. Boundary tiles are pre-painted with shades in [1, 10^9] and cannot be changed. Interior tiles are either broken (-1), empty (0, must be painted), or pre-painted with a fixed shade. We need to assign shades to all empty interior tiles to minimize the sum ... | {
"exploration": "I'll build a graph where each tile is a node. Boundary tiles and broken tiles have fixed shades (broken tiles are removed). Interior unpainted tiles (a_ij=0) are the ones I need to assign shades to. I'll discretize the possible shades to the set of existing boundary shades — let me collect those fir... | null | null | null | Okay, all of those went nowhere. Greedy smoothing got stuck in local minima, DP exploded exponentially, and MST ignored the fundamental tension: each interior tile's shade must simultaneously satisfy constraints from multiple fixed boundary neighbors, a global trade-off that can't be resolved locally.
hmm. That phrase... | function solve():
READ n
grid = array[1..n][1..n]
for i from 1 to n:
for j from 1 to n:
READ grid[i][j]
// Build graph: each tile is a node, index = (i-1)*n + j
total_nodes = n*n
source = total_nodes + 1
sink = total_nodes + 2
// Adjacency list for the grid edges (u... |
1103/E | Radix sum | 3,400 | fft, math, number theory | 4 | 256 | Let's define radix sum of number $$$a$$$ consisting of digits $$$a_1, \ldots ,a_k$$$ and number $$$b$$$ consisting of digits $$$b_1, \ldots ,b_k$$$(we add leading zeroes to the shorter number to match longer length) as number $$$s(a,b)$$$ consisting of digits $$$(a_1+b_1)\mod 10, \ldots ,(a_k+b_k)\mod 10$$$. The radix ... | The first line contains integer $$$n$$$ — the length of the array($$$1 \leq n \leq 100000$$$).
The second line contains $$$n$$$ integers $$$x_1, \ldots x_n$$$ — array elements($$$0 \leq x_i < 100000$$$). | Output $$$n$$$ integers $$$y_0, \ldots, y_{n-1}$$$ — $$$y_i$$$ should be equal to corresponding number of ways modulo $$$2^{58}$$$. | null | In the first example there exist sequences: sequence $$$(5,5)$$$ with radix sum $$$0$$$, sequence $$$(5,6)$$$ with radix sum $$$1$$$, sequence $$$(6,5)$$$ with radix sum $$$1$$$, sequence $$$(6,6)$$$ with radix sum $$$2$$$. | [
{
"input": "2\n5 6",
"output": "1\n2"
},
{
"input": "4\n5 7 5 7",
"output": "16\n0\n64\n0"
}
] | After reading the problem statement, it is pretty clear that you should apply something like Hadamard transform, but inner transform should have size ten instead of two. There is no $10$-th root of one modulo $2^{58}$ (except $5$-th and $2$-th roots), so it is not possible to solve the problem just calculating all valu... | The key observation is that radix addition modulo 10 behaves like convolution with a cyclic group of order 10. What if you treat each digit as a coefficient in a polynomial and exploit the structure of 10th roots of unity? The challenge is that working directly modulo 2^58 requires handling non-invertible numbers, but ... | {
"exploration": "Okay so we have an array of n numbers, each up to 99999 (5 digits max), and we need to compute the number of ways to pick n numbers with replacement such that their radix sum equals i, for each i from 0 to n-1, modulo 2^58. The radix sum is digit-wise addition mod 10 with no carry between digits — w... | {
"exploration": "I'll represent each input number as a 5-digit vector with leading zeros, since x_i < 100000 means at most 5 decimal digits. For each digit position d (0=units, 1=tens, ..., 4=ten-thousands), I build a polynomial P_d(y) where the coefficient of y^k is the count of input numbers whose digit at positio... | {
"exploration": "I'll split the sequence of n choices into two halves: first ⌊n/2⌋ choices and second ⌈n/2⌉ choices. For each half, I'll enumerate all possible radix sums by iterating over all ways to pick numbers from the array for that many positions. Each radix sum is a 5-digit number (with leading zeros), so I c... | null | null | null | The first two ideas—digit-wise DP with carry and direct NTT convolution—both missed the core structure: radix sum is per-digit modulo 10 with no carry, so each digit position is independent, but the digits of a chosen number are coupled. That’s why the third idea, using roots of unity, clicked: treat each number as a p... | READ n
READ array x[1..n]
// Represent each number as a 5-digit base-10 number (leading zeros)
digits = array of size n x 5
for i from 1 to n:
temp = x[i]
for d from 1 to 5:
digits[i][d] = temp % 10
temp = temp // 10
// Build frequency array freq of size 10 x 10 x 10 x 10 x 10 (5 dimensions, e... |
1381/E | Origami | 3,300 | geometry, math, sortings | 4 | 256 | After being discouraged by 13 time-limit-exceeded verdicts on an ugly geometry problem, you decided to take a relaxing break for arts and crafts.
There is a piece of paper in the shape of a simple polygon with $$$n$$$ vertices. The polygon may be non-convex, but we all know that proper origami paper has the property t... | The first line contains two integers $$$n$$$, $$$q$$$ ($$$3\le n\le 10^5, 1\le q\le 10^5$$$) — the number of polygon vertices and queries, respectively.
Each of the next $$$n$$$ lines contains two integers $$$x_i$$$, $$$y_i$$$ ($$$|x_i|, |y_i|\le 10^5$$$) — the coordinates of the $$$i$$$-th point of the polygon. The... | For each query, output the area $$$A_i$$$ of the paper if you fold it along the line $$$x=f_i$$$.
Your answer is considered correct if its absolute or relative error does not exceed $$$10^{-4}$$$.
Formally, let your answer be $$$a$$$, and the jury's answer be $$$b$$$. Your answer is accepted if and only if $$$\frac{|... | null | The first test, with the fold $$$f=-5$$$:
The second test, with the fold $$$f=1$$$:
The third test, with the fold $$$f=-1$$$: | [
{
"input": "4 7\n0 10\n10 0\n0 -10\n-10 0\n-9\n-5\n-1\n0\n1\n5\n9",
"output": "199.0000000000\n175.0000000000\n119.0000000000\n100.0000000000\n119.0000000000\n175.0000000000\n199.0000000000"
},
{
"input": "4 1\n0 0\n0 2\n2 4\n2 2\n1",
"output": "3.0000000000"
},
{
"input": "9 4\n0 -3\n2 ... | First, let's imagine the problem in one dimension. And let's see how the length of the folded segment changes as we sweep the fold line from left to right.
If the fold line is to the left of the paper, it's just the length of the segment. Then as the fold line enters the left side of the paper, we subtract the length o... | What if you consider the effect of folding on each infinitesimal horizontal strip independently? The key observation is that the folded length along a horizontal line behaves like a piecewise linear function of the fold position, which you can integrate over y to get the total area. This reduces the problem to sweeping... | {
"exploration": "Okay so we have a simple polygon with n vertices, and the key property is that any horizontal line intersects the boundary in at most two points — this means the polygon is y-monotone, essentially a shape where for each y there's a left boundary and a right boundary. The constraint is n ≤ 10^5, q ≤ ... | {
"exploration": "I'll discretize the y-range by collecting all polygon vertex y-coordinates, sorting them, and building a segment tree over the intervals between consecutive y-values. For each vertical slab between consecutive x-sorted polygon vertices, I'll compute the polygon's intersection with that slab as horiz... | {
"exploration": "For each query f, I'll compute the area of the polygon after folding along x=f. The folded shape is the union of the original polygon's part on the right side (x≥f) and the reflection of the left side (x≤f) across the line x=f. The reflection maps a point (x,y) with x≤f to (2f-x, y). So the folded s... | null | null | null | Well, that was a dead end after dead end. Monte Carlo was hopelessly imprecise, the dynamic envelope exploded in complexity, and scanning all segments per query was far too slow. The core issue is that folding creates a reflected shape whose boundary changes in a complicated, non-linear way with the fold line, and tryi... | function solve():
READ n, q
x = array of size n
y = array of size n
for i from 0 to n-1:
READ x[i], y[i]
// Compute original area using shoelace
original_area = 0
for i from 0 to n-1:
j = (i+1) % n
original_area += x[i] * y[j] - x[j] * y[i]
original_area = abs(or... |
1083/F | The Fair Nut and Amusing Xor | 3,300 | data structures | 4 | 256 | The Fair Nut has two arrays $$$a$$$ and $$$b$$$, consisting of $$$n$$$ numbers. He found them so long ago that no one knows when they came to him.
The Fair Nut often changes numbers in his arrays. He also is interested in how similar $$$a$$$ and $$$b$$$ are after every modification.
Let's denote similarity of two arr... | The first line contains three numbers $$$n$$$, $$$k$$$ and $$$q$$$ ($$$1 \le k \le n \le 2 \cdot 10^5$$$, $$$0 \le q \le 2 \cdot 10^5$$$) — the length of the arrays, the length of the subarrays, to which the operations are applied, and the number of queries.
The second line contains $$$n$$$ integers $$$a_1, a_2, \ldot... | On the first line print initial similarity of arrays $$$a$$$ and $$$b$$$.
On the $$$i$$$-th of following $$$q$$$ lines print similarity of $$$a$$$ and $$$b$$$ after applying first $$$i$$$ modifications. | null | In the first sample making arrays $$$[0, 4, 2]$$$ and $$$[1, 2, 3]$$$ is impossible with $$$k=3$$$. After the modification, you can apply the operation with $$$x=1$$$ to the whole first array (its length is equal to $$$k$$$), and it will be equal to the second array.
In order to make arrays equal in the second sample ... | [
{
"input": "3 3 1\n0 4 2\n1 2 3\nb 2 5",
"output": "-1\n1"
},
{
"input": "3 2 2\n1 3 2\n0 0 0\na 1 0\nb 1 1",
"output": "2\n-1\n2"
}
] | Let $c_{i}=a_{i} \oplus b_{i}$.
Let's notice that if there is list of operations to both arrays, which makes them equal, applying these operations to $a$ makes it equal to $b$. Because of this, applying them to $c$ makes all elements equal to $0$.
Now we are processing modifications of $c$. Let's make array $d$ with le... | The key observation is that the operation's effect on the difference array has a periodic structure. This allows you to decompose the problem into k independent subproblems. Within each subproblem, think about tracking prefix XORs and how an operation affects them — the minimal number of moves relates to counting non-z... | {
"exploration": "I'll try directly simulating the greedy operations. The idea: to make arrays equal, I need to make a[i] = b[i] for all i. The operation lets me pick any subarray of length k and XOR every element in it by the same x. So I can think in terms of the difference array d[i] = a[i] ⊕ b[i]. I need to make ... | {
"exploration": "I'll define dp[l][r] as the minimum operations to make a[l..r] equal to b[l..r], considering only operations fully inside [l, r]. The recurrence: for any segment, either we do nothing (if a[l..r] already equals b[l..r], dp=0), or we apply an operation starting at some i with l ≤ i ≤ r-k+1, XORing bo... | null | null | null | null | Alright, back to square one. The greedy simulation was impossibly slow, and any DP over intervals blew up because every update forced a full recomputation — the cost per query was linear in n, and with n and q both up to 2×10^5, that's 4×10^10 operations, which is dead.
hmm — that nagging thread about the operation's ... | function main():
READ n, k, q
READ array a[1..n]
READ array b[1..n]
// compute initial d and f
d = array of size n+2, initialized to 0 // d[0]=0, d[n+1]=0
for i from 1 to n:
d[i] = a[i] XOR b[i]
f = array of size n+2, initialized to 0 // f[1..n+1]
for i from 1 to n+1:
... |
1097/H | Mateusz and an Infinite Sequence | 3,400 | bitmasks, brute force, dp, strings | 7 | 512 | A Thue-Morse-Radecki-Mateusz sequence (Thorse-Radewoosh sequence in short) is an infinite sequence constructed from a finite sequence $$$\mathrm{gen}$$$ of length $$$d$$$ and an integer $$$m$$$, obtained in the following sequence of steps:
- In the beginning, we define the one-element sequence $$$M_0=(0)$$$.
- In the ... | The first line contains two integers $$$d$$$ and $$$m$$$ ($$$2 \leq d \leq 20$$$, $$$2 \leq m \leq 60$$$) — the length of the sequence $$$\mathrm{gen}$$$ and an integer used to perform the modular operations. The second line contains $$$d$$$ integers $$$\mathrm{gen}_i$$$ ($$$0 \leq \mathrm{gen}_i < m$$$). It's guarante... | Print a single integer — the answer to the problem. | null | Thorse-Radewoosh sequence in the first example is the standard Thue-Morse sequence, so the sequence $$$A$$$ is as follows: $$$11010011001011010010$$$. Here are the places where the sequence $$$B$$$ majorizes $$$A$$$: | [
{
"input": "2 2\n0 1\n4\n0 1 1 0\n2 21",
"output": "6"
},
{
"input": "3 4\n0 1 2\n2\n0 2\n6 11",
"output": "1"
}
] | In the beginning, change the query about an interval into two queries about prefixes.
Nextly, let's generalize the problem. Change the condition $A_i \leq B_j$ into $A_i \in X_j$, where $X_j$ is some set of possible values (and will be represented as a bitmask).
The problem is that the length of the sequence from the i... | What if you could turn the condition 'A_i ≤ B_j' into a membership test in a precomputed set for each position? The key observation is that the sequence's recursive generation lets you collapse groups of d positions when you know the index modulo d. This suggests a recursion that splits on the remainder, combined with ... | {
"exploration": "Okay so we have a recursively defined infinite sequence M∞ built from a length-d generator array gen and modulus m. Each step concatenates d copies of the previous sequence, each shifted by gen[i] mod m. The constraint says d ≤ 20, m ≤ 60, n ≤ 30000, and l, r up to 10^18. The time limit is 7 seconds... | {
"exploration": "I'll build a segment tree over the infinite sequence M_∞, where each node represents a contiguous segment at some level k. A node at level k corresponds to a block of size d^k, and stores the maximum value in that block. To answer whether B majorizes a length-n substring starting at position x, I qu... | {
"exploration": "I'll try to binary search the answer: for a candidate x, I need to check if there are at least x starting positions where B majorizes the length-n substring of A. That reduces to counting how many positions i in [l, r-n+1] satisfy B ≥ A[i..i+n-1]. To count efficiently, I can precompute rolling hashe... | null | null | null | None of those got me anywhere useful. Brute-force generation is impossible at 10^18 length, segment trees can't count matches efficiently, and binary search still leaves us with 10^18 positions to check — the sheer scale of the sequence forces us to exploit its recursive structure, not treat it as a flat array.
hmm. T... | function solve():
READ d, m
READ gen[0..d-1]
READ n
READ B[0..n-1]
READ l, r
// Convert to prefix queries: count in [1, r] minus count in [1, l-1]
R = r - n + 1
L = l - 1
if R < 1:
R = 0
if L < 1:
L = 0
// Precompute DP for a given prefix length a: cnt[a][v] =... |
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