wrong_submission_id
stringlengths 10
10
| problem_id
stringlengths 6
6
| user_id
stringlengths 10
10
| time_limit
float64 1k
8k
| memory_limit
float64 131k
1.05M
| wrong_status
stringclasses 2
values | wrong_cpu_time
float64 10
40k
| wrong_memory
float64 2.94k
3.37M
| wrong_code_size
int64 1
15.5k
| problem_description
stringlengths 1
4.75k
| wrong_code
stringlengths 1
6.92k
| acc_submission_id
stringlengths 10
10
| acc_status
stringclasses 1
value | acc_cpu_time
float64 10
27.8k
| acc_memory
float64 2.94k
960k
| acc_code_size
int64 19
14.9k
| acc_code
stringlengths 19
14.9k
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s830081522 | p02240 | u963402991 | 1,000 | 131,072 | Wrong Answer | 40 | 8,584 | 803 | Write a program which reads relations in a SNS (Social Network Service), and judges that given pairs of users are reachable each other through the network. | from queue import deque
n, m = map(int, input().split())
color = [-1 for i in range(n)]
G = deque([[0] * n for i in range(n)])
for i in range(m):
s, t = map(int, input().split())
G[s][t] = 1
G[t][s] = 1
def dfs(r, c):
S = deque([r])
color[r] = c
while (len(S)):
u = S.popleft()
for i in [j for j in range(0, m) if G[u][j] == 1]:
if color[i] == -1:
color[i] = c
S.append(i)
def assginColor():
i_d = 1
for u in range(n):
if color[u] == -1:
dfs(u, i_d)
i_d += 1
assginColor()
q = int(input())
for i in range(q):
s,t = map(int, input().split())
if color[s] == color[t]:
print("yes")
else:
print("no") | s716413462 | Accepted | 970 | 30,280 | 977 | from queue import deque
n, m = map(int, input().split())
G = deque([[] for i in range(n)])
for i in range(m):
s, t = map(int, input().split())
G[s].append(t)
G[t].append(s)
def dfs(r, c):
S = deque()
S.append(r)
color[r] = c
while (len(S) > 0):
u = S.popleft()
for j in G[u]:
if color[j] == -1:
color[j] = c
S.append(j)
def assginColor():
i_d = 1
for u in range(n):
if color[u] == -1:
dfs(u, i_d)
i_d += 1
color = [-1] * n
assginColor()
q = int(input())
s_t = [list(map(int, input().split())) for i in range(q)]
for s, t in s_t:
if color[s] == color[t]:
print("yes")
else:
print("no") |
s415889953 | p03089 | u379234461 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 447 | Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it. | # AGC032 Problem-AGC032
n = int(input())
a = list (map(int, input().split()))
#print (a)
a.insert(0,-1)
result = []
for i in range (n, 0, -1) :
for j in range (i, -1, -1) :
# print (a)
if a[j] == j :
result.append(j)
# print (result)
a = a[:j+1] + a[j+2:]
break
if j == 0 :
result = [-1]
break
result.reverse()
for ans in result :
print (ans)
| s297953200 | Accepted | 19 | 3,060 | 445 | # AGC032 Problem-AGC032
n = int(input())
a = list (map(int, input().split()))
#print (a)
a.insert(0,-1)
result = []
for i in range (n, 0, -1) :
for j in range (i, -1, -1) :
# print (a)
if a[j] == j :
result.append(j)
# print (result)
a = a[:j] + a[j+1:]
break
if j == 0 :
result = [-1]
break
result.reverse()
for ans in result :
print (ans)
|
s698704451 | p03549 | u595952233 | 2,000 | 262,144 | Wrong Answer | 1,428 | 9,096 | 313 | Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds. Now, he goes through the following process: * Submit the code. * Wait until the code finishes execution on all the cases. * If the code fails to correctly solve some of the M cases, submit it again. * Repeat until the code correctly solve all the cases in one submission. Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer). | from math import ceil
n, m = map(int, input().split())
base = (n-m)*100+1900*m
allok = 2**m
def prob(k):
assert k > 0
return (allok-1)**(k-1)
i = 1
temp = 0
for i in range(1, 10000):
temp += i*base*prob(i)/pow(allok, i)
print(ceil(temp))
| s466001250 | Accepted | 28 | 8,896 | 231 | from math import ceil
n, m = map(int, input().split())
base = (n-m)*100+1900*m
allok = pow(2,m)
print(base*allok)
|
s272190115 | p03720 | u935254309 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 279 | There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city? | N,M = map(int,input().split())
glaf = [[] for i in range(N)]
for i in range(M):
a,b = map(int,input().split())
a=a-1
b=b-1
glaf[a].append(b)
glaf[b].append(a)
print(glaf)
for i in glaf:
cnt =0
for j in i:
cnt+=1
print(cnt) | s579839397 | Accepted | 18 | 3,064 | 262 | N,M = map(int,input().split())
glaf = [[] for i in range(N)]
for i in range(M):
a,b = map(int,input().split())
a=a-1
b=b-1
glaf[a].append(b)
glaf[b].append(a)
for i in glaf:
cnt =0
for j in i:
cnt+=1
print(cnt) |
s861230811 | p02558 | u751843916 | 5,000 | 1,048,576 | Wrong Answer | 1,328 | 102,336 | 512 | You are given an undirected graph with N vertices and 0 edges. Process Q queries of the following types. * `0 u v`: Add an edge (u, v). * `1 u v`: Print 1 if u and v are in the same connected component, 0 otherwise. | from networkx.utils.union_find import UnionFind
n,q=map(int,input().split())
a=[[int(i) for i in input().split()] for _ in range(q)]
uf=UnionFind()
for i in range(n):
uf.union(i,i)
for x in a:
if x[0]==0:
uf.union(x[1],x[2])
else:
f=x[1]
while True:
if f==uf.parents[f]:break
f=uf.parents[f]
b=x[2]
while True:
if b==uf.parents[b]:break
b=uf.parents[b]
print(0 if f==b else 1) | s359935219 | Accepted | 1,328 | 102,228 | 512 | from networkx.utils.union_find import UnionFind
n,q=map(int,input().split())
a=[[int(i) for i in input().split()] for _ in range(q)]
uf=UnionFind()
for i in range(n):
uf.union(i,i)
for x in a:
if x[0]==0:
uf.union(x[1],x[2])
else:
f=x[1]
while True:
if f==uf.parents[f]:break
f=uf.parents[f]
b=x[2]
while True:
if b==uf.parents[b]:break
b=uf.parents[b]
print(1 if f==b else 0) |
s268307905 | p03644 | u062306892 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 82 | Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times. | n = int(input())
ans = [1, 2, 4, 8, 16, 32, 64]
max([i for i in ans if n-i >= 0]) | s106941271 | Accepted | 17 | 2,940 | 89 | n = int(input())
ans = [1, 2, 4, 8, 16, 32, 64]
print(max([i for i in ans if n-i >= 0])) |
s856947351 | p03251 | u604398799 | 2,000 | 1,048,576 | Wrong Answer | 23 | 3,064 | 409 | Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out. |
N, M, X, Y = map(int, input().split())
x = map(int, input().split())
y = map(int, input().split())
x_max = max(x)
y_min = min(y)
flag_war = True
for z in range(X+1, Y):
if x_max < z:
if y_min >= z:
print('No War')
print(z)
flag_war = False
else:
continue
else:
continue
if flag_war == True:
print('War')
| s069402121 | Accepted | 17 | 3,064 | 408 |
N, M, X, Y = map(int, input().split())
x = map(int, input().split())
y = map(int, input().split())
x_max = max(x)
y_min = min(y)
flag_war = True
for z in range(X+1, Y+1):
if x_max < z:
if y_min >= z:
print('No War')
flag_war = False
break
else:
continue
else:
continue
if flag_war == True:
print('War') |
s080329688 | p02264 | u650459696 | 1,000 | 131,072 | Wrong Answer | 30 | 5,996 | 400 | _n_ _q_ _name 1 time1_ _name 2 time2_ ... _name n timen_ In the first line the number of processes _n_ and the quantum _q_ are given separated by a single space. In the following _n_ lines, names and times for the _n_ processes are given. _name i_ and _time i_ are separated by a single space. | from collections import OrderedDict
n, q = map(int, input().split())
od = OrderedDict()
for i in range(n):
name, time = input().split()
od.update({name: int(time)})
total = 0
while len(od) > 0:
name, time = od.popitem(False)
if time < q:
total += time
print("{} {}".format(name, total))
else:
total += q
time -= q
od.update({name: time}) | s033509592 | Accepted | 480 | 14,932 | 401 | from collections import OrderedDict
n, q = map(int, input().split())
od = OrderedDict()
for i in range(n):
name, time = input().split()
od.update({name: int(time)})
total = 0
while len(od) > 0:
name, time = od.popitem(False)
if time <= q:
total += time
print("{} {}".format(name, total))
else:
total += q
time -= q
od.update({name: time}) |
s181000092 | p02237 | u193453446 | 1,000 | 131,072 | Wrong Answer | 30 | 7,704 | 371 | There are two standard ways to represent a graph $G = (V, E)$, where $V$ is a set of vertices and $E$ is a set of edges; Adjacency list representation and Adjacency matrix representation. An adjacency-list representation consists of an array $Adj[|V|]$ of $|V|$ lists, one for each vertex in $V$. For each $u \in V$, the adjacency list $Adj[u]$ contains all vertices $v$ such that there is an edge $(u, v) \in E$. That is, $Adj[u]$ consists of all vertices adjacent to $u$ in $G$. An adjacency-matrix representation consists of $|V| \times |V|$ matrix $A = a_{ij}$ such that $a_{ij} = 1$ if $(i, j) \in E$, $a_{ij} = 0$ otherwise. Write a program which reads a directed graph $G$ represented by the adjacency list, and prints its adjacency-matrix representation. $G$ consists of $n\; (=|V|)$ vertices identified by their IDs $1, 2,.., n$ respectively. | def main():
num = int(input())
T = [[0] * num for i in range(num)]
for n in range(num):
a = list(map(int,input().split()))
u = a[0] - 1
if a[1] > 0:
for i in a[2:]:
print(i)
T[u][i-1] = 1
for i in range(num):
print(" ".join(map(str,T[i])))
if __name__ == '__main__':
main() | s677680318 | Accepted | 20 | 7,792 | 346 | def main():
num = int(input())
T = [[0] * num for i in range(num)]
for n in range(num):
a = list(map(int,input().split()))
u = a[0] - 1
if a[1] > 0:
for i in a[2:]:
T[u][i-1] = 1
for i in range(num):
print(" ".join(map(str,T[i])))
if __name__ == '__main__':
main() |
s533708605 | p03693 | u393881437 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 84 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4? | r, g, b = map(int, input().split())
print('Yes' if r*100+g*10+b % 4 == 0 else 'No')
| s221241783 | Accepted | 17 | 2,940 | 89 | r, g, b = map(int, input().split())
print('YES' if (100*r + 10*g + b) % 4 == 0 else 'NO') |
s946169656 | p03150 | u977193988 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,064 | 249 | A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string. | import sys
s=input()
l=len(s)-7
if s=='keyence':
print('YES')
sys.exit()
if l<=0:
print('NO')
sys.exit()
for i in range(l):
print(s[:i]+s[i+l:])
if s[:i]+s[i+l:]=='keyence':
print('YES')
sys.exit()
print('NO') | s074168964 | Accepted | 18 | 3,060 | 248 | import sys
s=input()
l=len(s)-7
if l<0:
print('NO')
sys.exit()
for i in range(7):
# print(s[:i]+s[i+l:])
if s[:i]+s[i+l:]=='keyence':
print('YES')
sys.exit()
if s[:7]=='keyence':
print('YES')
else:
print('NO') |
s720505235 | p03693 | u291373585 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 108 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4? | a,b,c= list(map(int,input().split()))
N = 100*a + 10*b + c
if N % 4 == 0:
print('Yes')
else:
print('No') | s231370806 | Accepted | 17 | 2,940 | 108 | a,b,c= list(map(int,input().split()))
N = 100*a + 10*b + c
if N % 4 == 0:
print('YES')
else:
print('NO') |
s640060337 | p03386 | u777607830 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 154 | Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers. |
A,B,K = map(int,input().split())
for i in range(K):
if A+i <= B:
print(A+i)
for i in range(K):
if B-K+i >= A + K:
print(B-K+i)
| s211708837 | Accepted | 17 | 3,060 | 158 |
A,B,K = map(int,input().split())
for i in range(K):
if A+i <= B:
print(A+i)
for i in range(K):
if B-K+i+1 >= A + K:
print(B-K+i+1)
|
s700028747 | p03162 | u952081880 | 2,000 | 1,048,576 | Wrong Answer | 1,104 | 31,672 | 296 | Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains. | # C - Vacation
import numpy as np
N = int(input())
value = np.zeros((N, 3))
for i in range(N):
value[i, :] = list(map(int, input().split()))
value = value.tolist()
dp = [0]*3
for a, b, c in value:
dp = [max(dp[1], dp[2])+a, max(dp[0], dp[2])+b, max(dp[0], dp[1])+c]
print(max(dp)) | s662803770 | Accepted | 1,090 | 33,464 | 301 | # C - Vacation
import numpy as np
N = int(input())
value = np.zeros((N, 3))
for i in range(N):
value[i, :] = list(map(int, input().split()))
value = value.tolist()
dp = [0]*3
for a, b, c in value:
dp = [max(dp[1], dp[2])+a, max(dp[0], dp[2])+b, max(dp[0], dp[1])+c]
print(int(max(dp))) |
s026324058 | p03370 | u026155812 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 140 | Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition. | N, X = map(int, input().split())
m = []
cur = 0
for i in range(N):
cur += int(input())
m.append(cur)
print(len(m) + (X-cur)//min(m)) | s304499019 | Accepted | 19 | 2,940 | 148 | N, X = map(int, input().split())
m = []
cur = 0
for i in range(N):
a = int(input())
cur += a
m.append(a)
print(len(m) + (X-cur)//min(m)) |
s830622416 | p03636 | u549002697 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 121 | The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way. | string = input()
cnt = 0
le = len(string)
for i in range(1,le-1):
cnt =cnt +1
print(string[0],cnt,string[le-1])
| s035134869 | Accepted | 17 | 3,060 | 123 | string = input()
cnt = 0
le = len(string)
for i in range(1,le-1):
cnt =cnt +1
print(string[0],cnt,string[le-1],sep ='') |
s473875840 | p03588 | u780709476 | 2,000 | 262,144 | Wrong Answer | 691 | 13,472 | 449 | A group of people played a game. All players had distinct scores, which are positive integers. Takahashi knows N facts on the players' scores. The i-th fact is as follows: the A_i-th highest score among the players is B_i. Find the maximum possible number of players in the game. | n = int(input())
A = []
B = []
for i in range(n):
a, b = map(int, input().split())
A.append(a)
B.append(b)
A = sorted(A)
B = sorted(B, reverse = True)
ans = 0
ans += A[0]
print(ans)
for i in range(1, n):
print(A[i] - A[i - 1], B[i - 1] - B[i])
if(A[i] - A[i - 1] > B[i - 1] - B[i]):
ans += B[i - 1] - B[i]
print("B", ans)
else:
ans += A[i] - A[i - 1]
print("A", ans)
ans += B[-1]
print(ans)
| s757805139 | Accepted | 445 | 12,484 | 346 | n = int(input())
A = []
B = []
for i in range(n):
a, b = map(int, input().split())
A.append(a)
B.append(b)
A = sorted(A)
B = sorted(B, reverse = True)
ans = 0
ans += A[0]
for i in range(1, n):
if(A[i] - A[i - 1] > B[i - 1] - B[i]):
ans += B[i - 1] - B[i]
else:
ans += A[i] - A[i - 1]
ans += B[-1]
print(ans)
|
s254629848 | p03729 | u027675217 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 89 | You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`. | a,b,c = input().split()
if a[-1]==b[0] and b[-1]==c[0]:
print("Yes")
else:
print("No")
| s292756920 | Accepted | 17 | 2,940 | 89 | a,b,c = input().split()
if a[-1]==b[0] and b[-1]==c[0]:
print("YES")
else:
print("NO")
|
s429660964 | p03361 | u306241759 | 2,000 | 262,144 | Wrong Answer | 34 | 9,208 | 506 | We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j). Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`. However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black. Determine if square1001 can achieve his objective. | h, w = map(int,input().split())
grid = []
for _ in range(h):
grid.append(list(input()))
dx=[-1,0,0,1]
dy=[0,-1,1,0]
for i in range(h):
for j in range(w):
flag=False
if grid[i][j]==".":
continue
for x,y in zip(dx,dy):
if i+x<0 or i+x>h-1 or j+y<0 or j+y>w-1:
continue
if grid[i+x][j+y]=="#":
flag=True
break
if not flag:
print("NO")
exit()
print("YES") | s797700902 | Accepted | 27 | 9,140 | 507 | h, w = map(int,input().split())
grid = []
for _ in range(h):
grid.append(list(input()))
dx=[-1,0,0,1]
dy=[0,-1,1,0]
for i in range(h):
for j in range(w):
flag=False
if grid[i][j]==".":
continue
for x,y in zip(dx,dy):
if i+x<0 or i+x>h-1 or j+y<0 or j+y>w-1:
continue
if grid[i+x][j+y]=="#":
flag=True
break
if not flag:
print("No")
exit()
print("Yes")
|
s610360943 | p03457 | u142023109 | 2,000 | 262,144 | Wrong Answer | 400 | 11,816 | 529 | AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan. | N = int(input())
t = [0 for i in range(N)]
x = [0 for i in range(N)]
y = [0 for i in range(N)]
for i in range(N):
t[i], x[i], y[i] = map(int, input().split())
a = "YES"
for i in range(N):
if i == 0:
dt = t[0]
dist = abs(x[0]) + abs(y[0])
else:
dt = t[i]-t[i-1]
dist = abs(x[i]-x[i-1]) + abs(y[i]-y[i-1])
if dist > dt or (dist+dt) % 2 == 1:
a = "NO"
print(a)
| s185361630 | Accepted | 391 | 11,824 | 527 | N = int(input())
t = [0 for i in range(N)]
x = [0 for i in range(N)]
y = [0 for i in range(N)]
for i in range(N):
t[i], x[i], y[i] = map(int, input().split())
a = "Yes"
for i in range(N):
if i == 0:
dt = t[0]
dist = abs(x[0]) + abs(y[0])
else:
dt = t[i]-t[i-1]
dist = abs(x[i]-x[i-1]) + abs(y[i]-y[i-1])
if dist > dt or (dist+dt) % 2 == 1:
a = "No"
print(a) |
s519406111 | p03556 | u702582248 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 51 | Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer. | import math
n=int(input())
print(int(math.sqrt(n))) | s100365395 | Accepted | 34 | 4,372 | 74 | n=int(input())
print(list(filter(lambda x:x*x<=n,range(int(1e5))))[-1]**2) |
s929141294 | p02865 | u029935728 | 2,000 | 1,048,576 | Wrong Answer | 256 | 3,060 | 138 | How many ways are there to choose two distinct positive integers totaling N, disregarding the order? | import math
N1 = input()
N = int(N1)
i = 0
j = 1
r = 0
for j in range (N):
i=N-j
if i!=j :
r+=1
math.floor (r/2) | s904099019 | Accepted | 260 | 3,060 | 157 | import math
N1 = input()
N = int(N1)
i = 0
j = 1
r = 0
a = 0
for j in range (N):
i=N-j
if i!=j :
r+=1
a = math.floor (r/2)
print(a) |
s778774457 | p03377 | u765590009 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 105 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | a, b, x = map(int, input().split())
if (x >= a) and (a+b >= x) :
print("Yes")
else :
print("No") | s766956212 | Accepted | 17 | 2,940 | 105 | a, b, x = map(int, input().split())
if (x >= a) and (a+b >= x) :
print("YES")
else :
print("NO") |
s220825082 | p03494 | u813993459 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 215 | There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform. | s=map(int,input().split())
ans = []
for i in s:
tmp=i
count=0
while(1):
if tmp%2==0:
tmp=tmp/2
count=count+1
else:
break
ans.append(count)
min(ans) | s632055737 | Accepted | 19 | 3,060 | 249 | s=map(int,input().split())
s=map(int,input().split())
ans = []
for i in s:
tmp=i
count=0
while(1):
if tmp%2==0:
tmp=tmp/2
count=count+1
else:
break
ans.append(count)
print(min(ans)) |
s624934523 | p02646 | u619672182 | 2,000 | 1,048,576 | Wrong Answer | 22 | 9,208 | 202 | Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally. | AV = input().split()
BW = input().split()
T = input().split()
A = int(AV[0])
V = int(AV[1])
B = int(BW[0])
W = int(BW[1])
T = int(T[0])
if abs(A-B) < (V - W) * T:
print("Yes")
else:
print("No")
| s055121778 | Accepted | 23 | 9,188 | 203 | AV = input().split()
BW = input().split()
T = input().split()
A = int(AV[0])
V = int(AV[1])
B = int(BW[0])
W = int(BW[1])
T = int(T[0])
if abs(A-B) <= (V - W) * T:
print("YES")
else:
print("NO")
|
s170554728 | p03251 | u722189950 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,060 | 368 | Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out. | #ABC110 B
N, M, X, Y = list(map(int, input().split()))
x = list(map(int, input().split()))
y = list(map(int, input().split()))
isWar = True
if not X < Y:
isWar = False
else:
for Z in range(X+1,Y+1):
if max(x) < Z and min(y) >= Z:
isWar = True
if isWar:
print("War")
else:
print("NO War") | s784534990 | Accepted | 18 | 3,064 | 350 | #ABC110 B
N, M, X, Y = list(map(int, input().split()))
x = list(map(int, input().split()))
y = list(map(int, input().split()))
isnotWar = False
if X < Y:
for Z in range(X+1,Y+1):
if max(x) < Z and min(y) >= Z:
isnotWar = True
if isnotWar:
print("No War")
else:
print("War") |
s385809240 | p02612 | u455381597 | 2,000 | 1,048,576 | Wrong Answer | 31 | 9,148 | 95 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. |
def main():
N = int(input())
print(int(N%1000))
if __name__ == "__main__":
main() | s166796588 | Accepted | 29 | 9,176 | 145 |
def main():
N = int(input())
if N%1000==0:
print(0)
else :
print(1000-N%1000)
if __name__ == "__main__":
main() |
s255484041 | p02612 | u504348975 | 2,000 | 1,048,576 | Wrong Answer | 2,206 | 9,140 | 67 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | num = int(input())
while num < 1000:
num = num % 1000
print(num) | s376449132 | Accepted | 30 | 9,088 | 85 | num = int(input())
num = num % 1000
if num == 0:
print(num)
else:
print(1000-num) |
s443908255 | p03999 | u543954314 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 189 | You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion. All strings that can be obtained in this way can be evaluated as formulas. Evaluate all possible formulas, and print the sum of the results. | s = input()
le = len(s)
ans = 0
for i in range(le):
amini = 10**(le-i-1)
for j in range(le-i-1):
amini += 10**j * 2**(le-i-j-1)
amini *= int(s[i]) * 2**i
ans += amini
print(ans) | s263620679 | Accepted | 17 | 2,940 | 145 | s = list(map(int,list(input())))
n = len(s)
m = 0
for i in range(n):
for j in range(n-i):
m += s[i]*2**i*10**j*2**(max(n-i-j-2,0))
print(m) |
s002553312 | p04000 | u884982181 | 3,000 | 262,144 | Wrong Answer | 2,314 | 198,116 | 483 | We have a grid with H rows and W columns. At first, all cells were painted white. Snuke painted N of these cells. The i-th ( 1 \leq i \leq N ) cell he painted is the cell at the a_i-th row and b_i-th column. Compute the following: * For each integer j ( 0 \leq j \leq 9 ), how many subrectangles of size 3×3 of the grid contains exactly j black cells, after Snuke painted N cells? | from collections import defaultdict
h,w,n = map(int,input().split())
x = []
for i in range(n):
x.append(list(map(int,input().split())))
d = defaultdict(int)
for i in x:
a = []
for j in range(3):
for k in range(3):
if (i[0]-j-1>=0 and i[1]-k-1 >= 0 and i[0]-j-1<=h-3 and i[1]-k-1 <= w-3):
a.append((i[0]-j-1,i[1]-k-1))
for j in a:
d[j] += 1
a = list(d.values())
ans = [0]*9
ans[0] = (h-2)*(w-2)-len(a)
for i in a:
ans[i-1] += 1
for i in ans:
print(i) | s086522703 | Accepted | 2,206 | 198,116 | 482 | from collections import defaultdict
h,w,n = map(int,input().split())
x = []
for i in range(n):
x.append(list(map(int,input().split())))
d = defaultdict(int)
for i in x:
a = []
for j in range(3):
for k in range(3):
if (i[0]-j-1>=0 and i[1]-k-1 >= 0 and i[0]-j-1<=h-3 and i[1]-k-1 <= w-3):
a.append((i[0]-j-1,i[1]-k-1))
for j in a:
d[j] += 1
a = list(d.values())
ans = [0]*10
ans[0] = (h-2)*(w-2)-len(a)
for i in a:
ans[i] += 1
for i in ans:
print(i) |
s137481553 | p03636 | u735069283 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 49 | The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way. | s = str(input())
print(s[0]+str(len(s)) + s[-1]) | s514863995 | Accepted | 17 | 2,940 | 42 | n=input()
print(n[0]+str(len(n)-2)+n[-1])
|
s044107960 | p03815 | u758815106 | 2,000 | 262,144 | Wrong Answer | 29 | 9,088 | 47 | Snuke has decided to play with a six-sided die. Each of its six sides shows an integer 1 through 6, and two numbers on opposite sides always add up to 7. Snuke will first put the die on the table with an arbitrary side facing upward, then repeatedly perform the following operation: * Operation: Rotate the die 90° toward one of the following directions: left, right, front (the die will come closer) and back (the die will go farther). Then, obtain y points where y is the number written in the side facing upward. For example, let us consider the situation where the side showing 1 faces upward, the near side shows 5 and the right side shows 4, as illustrated in the figure. If the die is rotated toward the right as shown in the figure, the side showing 3 will face upward. Besides, the side showing 4 will face upward if the die is rotated toward the left, the side showing 2 will face upward if the die is rotated toward the front, and the side showing 5 will face upward if the die is rotated toward the back. Find the minimum number of operation Snuke needs to perform in order to score at least x points in total. | x = int(input())
s = x // 11 * 2 + 1
print(s) | s044915645 | Accepted | 27 | 9,136 | 127 | x = int(input())
s = x // 11
t = x % 11
if t > 6:
print(s * 2 + 2)
elif t > 0:
print(s * 2 + 1)
else:
print(s * 2) |
s746860400 | p03693 | u726439578 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 93 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4? | a,b,c=map(int,input().split())
n=int(a+b+c)
if n%4==0:
print("YES")
else:
print("NO") | s918732962 | Accepted | 17 | 2,940 | 98 | a,b,c=map(str,input().split())
n=int(a+b+c)
if n%4==0:
print("YES")
else:
print("NO")
|
s148599051 | p03860 | u114687417 | 2,000 | 262,144 | Wrong Answer | 24 | 3,068 | 41 | Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name. | word = input()
print("A"+word[1][0]+"C") | s480561268 | Accepted | 23 | 3,064 | 70 | word = input() .split()
print( word[0][0] + word[1][0] + word[2][0] ) |
s144253651 | p03861 | u513793759 | 2,000 | 262,144 | Wrong Answer | 2,132 | 447,084 | 126 | You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x? | a,b,x = map(int, input().split())
T = []
for i in range(a,b+1):
if i%x==0:
T.append(i)
else:
quit
print(T)
print(len(T)) | s234228852 | Accepted | 18 | 2,940 | 70 | import math
a,b,x = map(int, input().split())
print(b//x-(a+x-1)//x+1) |
s452363112 | p02608 | u394731058 | 2,000 | 1,048,576 | Wrong Answer | 2,205 | 9,280 | 520 | Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N). | import sys
input = sys.stdin.readline
def calc(n):
for i in range(1,int(n**0.5)):
for j in range(1,int(n**0.5)):
for k in range(1, int(n**0.5)):
if (i+j+k)**2 - i*j - j*k - k*i == n:
if i == j == k:
return 1
else:
return 3
return 0
def main():
ans = 0
n = int(input())
for i in range(1,n+1):
print(calc(i))
print(ans)
if __name__ == '__main__':
main() | s329259086 | Accepted | 379 | 9,192 | 382 | import sys
input = sys.stdin.readline
def main():
ans = [0]* 10010
n = int(input())
for x in range(1,101):
for y in range(1,101):
for z in range(1,101):
v = (x+y+z)**2 - x*y - y*z - z*x
if v <= n:
ans[v] += 1
for i in range(1,n+1):
print(ans[i])
if __name__ == '__main__':
main() |
s220837524 | p02257 | u424457654 | 1,000 | 131,072 | Wrong Answer | 30 | 7,688 | 160 | A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list. | n = int(input())
count = 0
for i in range(2, n):
x = int(input())
if x == 2:
count += 1
elif x % i == 0:
break
else:
count += 1
print(count) | s934076803 | Accepted | 50 | 7,700 | 193 | n = int(input())
count = 0
for i in range(n):
x = int(input())
if x == 2:
count += 1
elif x % 2 == 0:
continue
else:
if pow(2, x - 1, x) == 1:
count += 1
print(count) |
s336933859 | p02266 | u742013327 | 1,000 | 131,072 | Wrong Answer | 30 | 7,576 | 1,329 | Your task is to simulate a flood damage. For a given cross-section diagram, reports areas of flooded sections. Assume that rain is falling endlessly in the region and the water overflowing from the region is falling in the sea at the both sides. For example, for the above cross-section diagram, the rain will create floods which have areas of 4, 2, 1, 19 and 9 respectively. | #http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_3_D
import re
def cal_puddle(puddle_list):
depth = 0
score = 0
for symbol in puddle_list:
if symbol == "\\":
score += depth + 0.5
depth += 1
if symbol == "/":
depth -= 1
score += depth + 0.5
if symbol == "_":
score += depth
return int(score)
def get_puddle_index(diagram_list):
depth = 0
for i, symbol in enumerate(diagram_list):
if symbol == "\\":
depth += 1
elif symbol == "/":
depth = depth - 1
if depth == 0:
return i
return False
def cal(diagram):
puddle_areas = []
diagram_list = []
puddle_list = []
i = 0
while i < len(diagram):
if diagram[i] == "\\":
puddle_end = get_puddle_index(diagram[i:])
if puddle_end:
puddle_list.append(diagram[i: i + puddle_end + 1])
i = i + puddle_end
i += 1
for puddle in puddle_list:
puddle_areas.append(cal_puddle(puddle))
return puddle_areas
if __name__ == "__main__":
a = input()
result = cal(a)
print(sum(result))
print(" ".join([str(a) for a in result])) | s361584012 | Accepted | 3,000 | 8,284 | 1,582 | def cal_puddle(puddle):
depth = 0
score = 0
for s in puddle:
if s == "\\":
score += depth + 0.5
depth += 1
elif s == "/":
depth -= 1
score += depth + 0.5
elif s == "_":
score += depth
if depth == 0:
return int(score)
return int(score)
def get_puddles(diagram):
hight = 0
top_list = []
in_puddle = True
for index, s in enumerate(diagram + "\\"):
if s == "/":
in_puddle = True
hight += 1
if s == "\\":
if in_puddle:
in_puddle = False
top_list.append([hight, index])
hight -= 1
puddles = []
prev_hight = 0
hight_list = [h for h, i in top_list]
i = 0
while i < len(top_list) - 1:
cur_top = top_list[i]
next_tops = list(filter(lambda top:cur_top[0] <= top[0], top_list[i + 1:]))
#print(next_tops)
if next_tops:
next_top = next_tops[0]
puddles.append((cur_top[1], next_top[1]))
i = top_list.index(next_top)
else:
cur_top[0] -= 1
cur_top[1] += diagram[cur_top[1] + 1:].index("\\") + 1
#print(hight_list)
#print(top_list)
#print(puddles)
return puddles
def main():
diagram = input()
result = []
for s,e in get_puddles(diagram):
result.append(cal_puddle(diagram[s:e]))
print(sum(result))
print(str(len(result)) + "".join([" " + str(i) for i in result]))
if __name__ == "__main__":
main() |
s553091963 | p03448 | u941753895 | 2,000 | 262,144 | Wrong Answer | 53 | 3,060 | 210 | You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different. |
A=int(input())
B=int(input())
C=int(input())
X=int(input())
cnt=0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
if 500*A+100*B+50*C==X:
cnt+=1
print(cnt) | s172110746 | Accepted | 52 | 3,060 | 210 |
A=int(input())
B=int(input())
C=int(input())
X=int(input())
cnt=0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
if 500*i+100*j+50*k==X:
cnt+=1
print(cnt) |
s407291424 | p02612 | u345389118 | 2,000 | 1,048,576 | Wrong Answer | 28 | 9,144 | 88 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | N = int(input())
if N % 1000 == 0:
print(N // 1000)
else:
print(N // 1000 + 1)
| s471336155 | Accepted | 30 | 9,144 | 93 | N = int(input())
if N % 1000 == 0:
print(0)
else:
print((N // 1000 + 1) * 1000 - N)
|
s557254289 | p03860 | u089376182 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 45 | Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name. | print('A', input().split()[1], 'C', sep = '') | s074478170 | Accepted | 17 | 2,940 | 69 | temp = list(input().split())
print(temp[0][0]+temp[1][0]+temp[2][0])
|
s474567820 | p03025 | u111365362 | 2,000 | 1,048,576 | Time Limit Exceeded | 2,104 | 3,540 | 672 | Takahashi and Aoki will play a game. They will repeatedly play it until one of them have N wins in total. When they play the game once, Takahashi wins with probability A %, Aoki wins with probability B %, and the game ends in a draw (that is, nobody wins) with probability C %. Find the expected number of games that will be played, and print it as follows. We can represent the expected value as P/Q with coprime integers P and Q. Print the integer R between 0 and 10^9+6 (inclusive) such that R \times Q \equiv P\pmod {10^9+7}. (Such an integer R always uniquely exists under the constraints of this problem.) | m = 10**9+7
def inv(x):
m = 10**9+7
t = x
y = 1
while True:
if t == 1:
break
y *= m // t
t = m % t
y *= -1
y %= m
return y
def f(x):
nn = x
y = 1
while True:
if nn == 1:
break
y *= nn
nn -= 1
return y
def cc(x,y):
return f(x) // f(y) // f(x-y)
n,a,b,c = map(int,input().split())
d = 100 - c
awin = a * inv(d)
bwin = b * inv(d)
asc = 0
bsc = 0
for i in range(n):
p = cc(n-1,i) * awin**(n-1) * bwin**i
asc += p
asc %= m
for i in range(n):
p = cc(n-1,i) * awin**i * bwin**(n-1)
bsc += p
bsc %= m
sc0 = asc * awin + bsc * bwin
sc0 %= m
sc1 = sc0 * 100 * inv(d)
sc1 %= m
print(sc1) | s735098801 | Accepted | 927 | 3,064 | 596 | #18:04
n,a,b,c = map(int,input().split())
mod = 10 ** 9 + 7
def inv(x):
ans = 1
now = x
while now != 1:
ans *= mod // now + 1
ans %= mod
now = now - mod % now
return ans
p = (a * inv(a+b)) % mod
q = (b * inv(a+b)) % mod
taka = 0
aoki = 0
tcom = 1
acom = 1
for i in range(n):
taka += tcom * (n+i)
tcom *= n+i
tcom *= inv(i+1)
tcom *= q
tcom %= mod
aoki += acom * (n+i)
acom *= n+i
acom *= inv(i+1)
acom *= p
acom %= mod
for _ in range(n):
taka *= p
taka %= mod
aoki *= q
aoki %= mod
ans = taka + aoki
ans *= 100
ans *= inv(a+b)
ans %= mod
print(ans) |
s993497167 | p02393 | u162598098 | 1,000 | 131,072 | Wrong Answer | 20 | 7,528 | 39 | Write a program which reads three integers, and prints them in ascending order. | print(sorted(map(int,input().split()))) | s634150899 | Accepted | 20 | 7,668 | 67 | a,b,c=map(int,input().split())
[x,y,z]=sorted([a,b,c])
print(x,y,z) |
s189080511 | p02603 | u860002137 | 2,000 | 1,048,576 | Wrong Answer | 34 | 9,004 | 328 | To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally? | n = int(input())
tmp = list(map(int, input().split()))
arr = [tmp[0]]
for i in range(1, n):
if arr[-1] != tmp[i]:
arr.append(tmp[i])
m, s = 1000, 0
for i in range(1, len(arr)):
if arr[i - 1] < arr[i]:
s += m // arr[i - 1]
m -= m // arr[i - 1] * arr[i - 1]
m += s * arr[i]
s = 0 | s186296666 | Accepted | 32 | 9,192 | 554 | n = int(input())
tmp = list(map(int, input().split()))
arr = [tmp[0]]
for i in range(1, n):
if arr[-1] != tmp[i]:
arr.append(tmp[i])
money, stock = 1000, 0
for i in range(1, len(arr)):
if arr[i - 1] < arr[i]:
buy, money = divmod(money, arr[i - 1])
stock += buy
money += stock * arr[i]
stock = 0
print(money) |
s173995844 | p03730 | u483640741 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 130 | We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`. | a,b,c=map(int,input().split())
s=a
for i in range(b+1):
if s%b==c:
print("YES")
else:
s+=a
print("NO")
| s776319382 | Accepted | 17 | 2,940 | 144 | a,b,c=map(int,input().split())
s=a
for i in range(b+1):
if s%b==c:
print("YES")
exit()
else:
s+=a
print("NO")
|
s304186152 | p04030 | u653005308 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 110 | Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now? | string=input()
ans=""
for x in string:
if x!="b":
ans+=x
else:
ans=ans[:-1]
print(ans) | s455367147 | Accepted | 17 | 2,940 | 110 | string=input()
ans=""
for x in string:
if x!="B":
ans+=x
else:
ans=ans[:-1]
print(ans) |
s567986324 | p03448 | u599547273 | 2,000 | 262,144 | Wrong Answer | 47 | 3,060 | 303 | You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different. | a, b, c, x = [int(input()) for i in range(4)]
can_count = 0
for i in range(a):
coins_price_500 = i * 500
for j in range(b):
coins_price_100 = j * 100
for k in range(c):
coins_price_50 = k * 50
if x == coins_price_500 * coins_price_100 * coins_price_50:
can_count += 1
print(can_count) | s570756108 | Accepted | 45 | 3,060 | 309 | a, b, c, x = [int(input()) for i in range(4)]
can_count = 0
for i in range(a+1):
coins_price_500 = i * 500
for j in range(b+1):
coins_price_100 = j * 100
for k in range(c+1):
coins_price_50 = k * 50
if x == coins_price_500 + coins_price_100 + coins_price_50:
can_count += 1
print(can_count) |
s656232281 | p02408 | u299798926 | 1,000 | 131,072 | Wrong Answer | 20 | 7,744 | 607 | Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond. | n=int(input())
S=[int(0) for i in range(14)]
H=[int(0) for i in range(14)]
D=[int(0) for i in range(14)]
C=[int(0) for i in range(14)]
for i in range(n):
K,y=[(i) for i in input().split()]
if K=='S':
S[int(y)]=1
elif K=='H':
H[int(y)]=1
elif K=='D':
D[int(y)]=1
else:
C[int(y)]=1
for i in range(14):
if S==0:
print('S'," {0}".format(i))
for i in range(14):
if H==0:
print('H'," {0}".format(i))
for i in range(14):
if D==0:
print('D'," {0}".format(i))
for i in range(14):
if C==0:
print('C'," {0}".format(i)) | s619759307 | Accepted | 30 | 7,804 | 667 | n=int(input())
S=[int(0) for i in range(1,14)]
H=[int(0) for i in range(1,14)]
D=[int(0) for i in range(1,14)]
C=[int(0) for i in range(1,14)]
for i in range(n):
K,y=[(i) for i in input().split()]
if K=='S':
S[int(y)-1]=1
elif K=='H':
H[int(y)-1]=1
elif K=='D':
D[int(y)-1]=1
else:
C[int(y)-1]=1
for i in range(1,14):
if S[int(i)-1]==0:
print('S',"{0}".format(i))
for i in range(1,14):
if H[int(i)-1]==0:
print('H',"{0}".format(i))
for i in range(1,14):
if C[int(i)-1]==0:
print('C',"{0}".format(i))
for i in range(1,14):
if D[int(i)-1]==0:
print('D',"{0}".format(i)) |
s818309799 | p02401 | u876060624 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 158 | Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part. | a,b,c = map(str,input().split())
a,c =int(a),int(c)
if b =='+':
print(a+c)
elif b == '-' : print(a-c)
elif b == '*' : print(a*c)
elif b == '/' : print(a/c)
| s207461518 | Accepted | 20 | 5,596 | 195 | while True :
a,op,b = input().split()
if op == '?' : break
a,b = int(a),int(b)
if op == '+' : print(a+b)
if op == '-' : print(a-b)
if op == '*' : print(a*b)
if op == '/' : print(int(a/b))
|
s158970218 | p03712 | u588081069 | 2,000 | 262,144 | Wrong Answer | 19 | 3,060 | 204 | You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1. | H, W = list(map(int, input().split()))
result = [["*"] * W] * (H + 2)
for i in range(H):
a = list(map(str, input()))
result[i + 1] = a
for i in range(len(result)):
print("".join(result[i]))
| s563528329 | Accepted | 19 | 3,060 | 196 | H, W = list(map(int, input().split()))
for i in range(H + 2):
if i == 0 or i == (H + 1):
print("#" * (W + 2))
else:
print("#{}#".format("".join(list(map(str, input())))))
|
s363163723 | p03545 | u521866787 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 256 | Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted. | ABCD = input()
for i in range(1,2**3):
X = ABCD[0]
for j in range(3):
if 1 << j & i:
X += '+'
else:
X += '-'
X += ABCD[j+1]
# print('eval: ',X)
if eval(X) == 7:
print(X)
break | s191884498 | Accepted | 17 | 2,940 | 261 | ABCD = input()
for i in range(1,2**3):
X = ABCD[0]
for j in range(3):
if 1 << j & i:
X += '+'
else:
X += '-'
X += ABCD[j+1]
# print('eval: ',X)
if eval(X) == 7:
print(X+'=7')
break |
s093604953 | p03645 | u691018832 | 2,000 | 262,144 | Wrong Answer | 687 | 40,472 | 795 | In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him. | import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
n, m = map(int, readline().split())
graph = [[] for _ in range(n + 1)]
for i in range(m):
a, b = map(int, readline().split())
graph[a].append(b)
graph[b].append(a)
def dfs(s):
from collections import deque
check = [0] * (n + 1)
stack = deque([s])
check[s] = 0
while stack:
now = stack.pop()
for next in graph[now]:
if check[next] != 0 and next != n:
continue
if next == n and check[now] == 1:
print('POISSIBLE')
exit()
check[next] += check[now] + 1
stack.appendleft(next)
dfs(1)
print('IMPOSSIBLE')
| s958614852 | Accepted | 575 | 40,420 | 794 | import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
n, m = map(int, readline().split())
graph = [[] for _ in range(n + 1)]
for i in range(m):
a, b = map(int, readline().split())
graph[a].append(b)
graph[b].append(a)
def dfs(s):
from collections import deque
check = [0] * (n + 1)
stack = deque([s])
check[s] = 0
while stack:
now = stack.pop()
for next in graph[now]:
if check[next] != 0 and next != n:
continue
if next == n and check[now] == 1:
print('POSSIBLE')
exit()
check[next] += check[now] + 1
stack.appendleft(next)
dfs(1)
print('IMPOSSIBLE')
|
s868342406 | p04043 | u252745826 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 148 | Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order. | a, b, c = list(map(int, input().split(" ")))
if a+b+c == 17 and (a == b == 5 or b == c == 5 or c == a == 5):
print("Yes")
else:
print("No")
| s901637940 | Accepted | 17 | 2,940 | 168 | a, b, c = list(map(int, input().split(" ")))
if (a == b == 5 and c == 7) or (b == c == 5 and a == 7) or (c == a == 5 and b ==7):
print("YES")
else:
print("NO")
|
s448434979 | p03110 | u044635590 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 226 | Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total? | n = int(input())
summm = 0.0
for i in range(n):
xu = list(map(str, input().split()))
if xu[1] == "JPY":
summm = summm + float(xu[0])
else:
summm = summm + float(xu[0]) * 380000
print(int(summm))
| s186389765 | Accepted | 17 | 3,060 | 221 | n = int(input())
summm = 0.0
for i in range(n):
xu = list(map(str, input().split()))
if xu[1] == "JPY":
summm = summm + float(xu[0])
else:
summm = summm + float(xu[0]) * 380000.0
print(summm) |
s785631420 | p04012 | u242580186 | 2,000 | 262,144 | Wrong Answer | 34 | 9,396 | 586 | Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful. | import sys
from heapq import heappush, heappop, heapify
import math
from math import gcd
import itertools as it
from collections import deque
input = sys.stdin.readline
def inp():
return int(input())
def inpl():
return list(map(int, input().split()))
INF = 1001001001
# ---------------------------------------
def main():
s = input()
dc = {}
for st in s:
if not st in dc:
dc[st] = 1
else:
dc[st] += 1
for k, v in dc.items():
if v % 2 == 1:
print("No")
return
print("Yes")
main()
| s806258032 | Accepted | 31 | 9,348 | 170 | import collections
w = list(input())
cnt = collections.Counter(w)
ans = "Yes"
for v in cnt.values():
if v%2!=0:
ans = "No"
break
print(ans)
|
s999647747 | p02262 | u938045879 | 6,000 | 131,072 | Wrong Answer | 20 | 5,600 | 548 | Shell Sort is a generalization of [Insertion Sort](http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A) to arrange a list of $n$ elements $A$. 1 insertionSort(A, n, g) 2 for i = g to n-1 3 v = A[i] 4 j = i - g 5 while j >= 0 && A[j] > v 6 A[j+g] = A[j] 7 j = j - g 8 cnt++ 9 A[j+g] = v 10 11 shellSort(A, n) 12 cnt = 0 13 m = ? 14 G[] = {?, ?,..., ?} 15 for i = 0 to m-1 16 insertionSort(A, n, G[i]) A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every $g$-th elements. Beginning with large values of $g$, it repeats the insertion sort with smaller $g$. Your task is to complete the above program by filling ?. Write a program which reads an integer $n$ and a sequence $A$, and prints $m$, $G_i (i = 0, 1, ..., m − 1)$ in the pseudo code and the sequence $A$ in ascending order. The output of your program must meet the following requirements: * $1 \leq m \leq 100$ * $0 \leq G_i \leq n$ * cnt does not exceed $\lceil n^{1.5}\rceil$ | n = int(input())
li = [int(input()) for i in range(n)]
cnt = 0
def shell_sort(li, n):
m = n//2
G = [i**2 for i in range(m, 0, -1)]
for i in range(m):
insertionSort(li, n, G[i])
return m, G
def insertionSort(A, n, g):
global cnt
for i in range(g, n):
v = A[i]
j = i-g
while j >= 0 and A[j] > v:
A[j+g] = A[j]
j = j-g
cnt += 1
A[j+g] = v
m, G = shell_sort(li, n)
print(m)
strG = [str(i) for i in G]
print(' '.join(strG))
for i in li:
print(i)
| s355599983 | Accepted | 19,810 | 45,588 | 609 | import math
n = int(input())
li = [int(input()) for i in range(n)]
cnt = 0
def shell_sort(li, n):
m = math.floor(math.log(2*n+1, 3))
G = [round((3**i-1)/2) for i in range(m, 0, -1)]
for i in range(m):
insertionSort(li, n, G[i])
return m, G
def insertionSort(A, n, g):
global cnt
for i in range(g, n):
v = A[i]
j = i-g
while j >= 0 and A[j] > v:
A[j+g] = A[j]
j = j-g
cnt += 1
A[j+g] = v
m, G = shell_sort(li, n)
print(m)
strG = [str(i) for i in G]
print(' '.join(strG))
print(cnt)
for i in li:
print(i)
|
s223870255 | p03385 | u754022296 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 61 | You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`. | if sorted(input())=="abc":
print("Yes")
else:
print("No") | s578145577 | Accepted | 18 | 3,068 | 70 | if sorted(input())==["a","b","c"]:
print("Yes")
else:
print("No")
|
s095573358 | p03827 | u319690708 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 268 | You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation). | N = int(input())
S = input().split()
X = 0
S1 = list(S[0])
S1_mod = [-1 if i == "I" else i for i in S1]
S1_mod = [1 if i == "D" else i for i in S1_mod]
l = []
for i in S1_mod:
if i == -1:
X += -1
l.append(X)
else:
X+= 1
l.append(X)
print(max(l)) | s824121961 | Accepted | 17 | 3,064 | 305 | N = int(input())
S = input().split()
X = 0
S1 = list(S[0])
S1_mod = [1 if i == "I" else i for i in S1]
S1_mod = [-1 if i == "D" else i for i in S1_mod]
l = []
for i in S1_mod:
if i == -1:
X += -1
l.append(X)
else:
X+= 1
l.append(X)
if max(l) < 0:
print(0)
else:
print(max(l)) |
s960889794 | p03494 | u798557584 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 29 | There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform. | print(input().count("1"))
| s151121222 | Accepted | 19 | 3,060 | 235 |
import math
n = int(input())
a = list(map(int,input().split()))
cnt = 0
while True:
flg = 0
for i in range(n):
if a[i] % 2 == 1:
flg = 1
break
else:
a[i] = a[i] / 2
if flg == 1:
break
cnt += 1
print(cnt)
|
s829384733 | p03548 | u882370611 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 51 | We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat? | x,y,z=map(int,input().split())
print(int(x-z//y+z)) | s770536303 | Accepted | 17 | 2,940 | 51 | x,y,z=map(int,input().split())
print((x-z)//(y+z))
|
s730619305 | p03992 | u094191970 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 31 | This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s. | s=input()
print(s[:4]+''+s[4:]) | s794395917 | Accepted | 17 | 2,940 | 32 | s=input()
print(s[:4]+' '+s[4:]) |
s129463325 | p03693 | u497952650 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 98 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4? | a,b,c = map(int,input().split())
if (100*a+10*b+c)%4 == 0:
print("Yes")
else:
print("No") | s588025433 | Accepted | 17 | 2,940 | 92 | a,b,c = map(int,input().split())
if (10*b+c)%4 == 0:
print("YES")
else:
print("NO") |
s447526412 | p02393 | u586792237 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 182 | Write a program which reads three integers, and prints them in ascending order. | x, y, z = map(int, input().split())
if x > y :
x, y, z = y, x, z
if x > z:
x, y, z = z, y, x
if y > z:
x, y, z = x, y, z
if x > y and y > z:
x, y, z = z, y, x
print(x, y, z)
| s759635769 | Accepted | 20 | 5,588 | 71 | a = list(map(int, input().split()))
a.sort()
print(a[0], a[1], a[2])
|
s552800703 | p03337 | u174523836 | 2,000 | 1,048,576 | Wrong Answer | 19 | 2,940 | 68 | You are given two integers A and B. Find the largest value among A+B, A-B and A \times B. | x, y = [int(s) for s in input().split()]
max([x + y, x - y, x * y])
| s944734676 | Accepted | 17 | 2,940 | 75 | x, y = [int(s) for s in input().split()]
print(max([x + y, x - y, x * y]))
|
s820931307 | p03657 | u600402037 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 96 | Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies. | a,b=map(int,input().split())
print('Possible' if a%3==0 or b%3==0 or a+b%3==0 else 'Impossible') | s629580424 | Accepted | 17 | 2,940 | 98 | a,b=map(int,input().split())
print('Possible' if a%3==0 or b%3==0 or (a+b)%3==0 else 'Impossible') |
s468739068 | p03796 | u323680411 | 2,000 | 262,144 | Wrong Answer | 25 | 3,064 | 487 | Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7. | from sys import stdin
def main() -> int:
n = next_int()
m = 10 ** 9 + 7
ans = 1
for i in range(n):
j = i % m
ans = (ans * j) % m
print(ans)
return 0
def next_int() -> int:
return int(next_str())
def next_str() -> str:
result = ""
while True:
tmp = stdin.read(1)
if tmp.strip() != "":
result += tmp
elif tmp != '\r':
break
return result
if __name__ == '__main__':
main() | s753684260 | Accepted | 31 | 3,064 | 495 | from sys import stdin
def main() -> int:
n = next_int()
m = 10 ** 9 + 7
ans = 1
for i in range(1, n + 1):
j = i % m
ans = (ans * j) % m
print(ans)
return 0
def next_int() -> int:
return int(next_str())
def next_str() -> str:
result = ""
while True:
tmp = stdin.read(1)
if tmp.strip() != "":
result += tmp
elif tmp != '\r':
break
return result
if __name__ == '__main__':
main() |
s459986205 | p03478 | u358791207 | 2,000 | 262,144 | Wrong Answer | 28 | 2,940 | 244 | Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive). | def sumcal(num):
sum = 0
while(num != 0):
sum += (num % 10)
num = num // 10
return sum
N, A, B = map(int, input().split())
count = 0
for i in range(N):
sum = sumcal(N + 1)
if sum >= A and sum <= B:
count += 1
print(count) | s566000872 | Accepted | 26 | 3,060 | 248 | def sumcal(num):
sum = 0
while(num != 0):
sum += (num % 10)
num = num // 10
return sum
N, A, B = map(int, input().split())
count = 0
for i in range(N):
sum = sumcal(i + 1)
if sum >= A and sum <= B:
count += i + 1
print(count) |
s796761170 | p02534 | u840649762 | 2,000 | 1,048,576 | Wrong Answer | 27 | 9,148 | 46 | You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them. For example, if K = 3, print `ACLACLACL`. | k = int(input())
ans = "ACL" * 5
print(ans)
| s230762258 | Accepted | 29 | 9,040 | 46 | k = int(input())
ans = "ACL" * k
print(ans)
|
s923339936 | p03377 | u995062424 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 125 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | num = input().split()
num_i = [int(s) for s in num]
if(num_i[0]+num_i[2] < num_i[1]):
print('YES')
else:
print('NO') | s645814704 | Accepted | 17 | 2,940 | 154 | num = input().split()
num_i = [int(s) for s in num]
if(num_i[0]+num_i[1]-num_i[2] > 0) and (num_i[2] >= num_i[0]):
print('YES')
else:
print('NO') |
s453883937 | p02678 | u802180430 | 2,000 | 1,048,576 | Wrong Answer | 2,229 | 652,600 | 525 | There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists. | n,m = list(map(int,input().split()))
dist = [[float("inf") for _ in range(n+1)]for _ in range(n+1)]
for i in range(1,n+1):
dist[i][i] = 0;
for _ in range(m):
a,b = list(map(int,input().split()))
dist[a][b] = 1; dist[b][a] = 1;
for k in range(1,n+1):
for j in range(1,n+1):
dist[1][j] = min(dist[1][j],dist[1][k] + dist[k][j])
flag = False;
for i in range(2,n+1):
if dist[1][i] == float("inf"):
print("NO")
flag = True
break;
if not flag:
print("YES")
for i in range(2,n+1):
print(dist[1][i]) | s553195456 | Accepted | 798 | 50,996 | 676 | n,m = map(int,input().split())
g = [[] for _ in range(n)]
for _ in range(m):
a, b = [int(x) for x in input().split()]
g[a-1].append(b-1)
g[b-1].append(a-1)
from collections import deque
def bfs(u):
queue = deque([u])
d = [None] * n
d[u] = 0
l = [[] for _ in range(n)]
dic = {}
while queue:
v = queue.popleft()
for i in g[v]:
if d[i] is None:
d[i] = d[v] + 1
queue.append(i)
l[i].append(v)
dic[i] = v
return d,l,dic
d,l,dic = bfs(0)
if None in d:
print("No")
else:
print("Yes")
l.remove(l[0])
[print(i[0]+1) for i in l] |
s261635815 | p02613 | u688203790 | 2,000 | 1,048,576 | Wrong Answer | 155 | 16,244 | 309 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. |
n = int(input())
s = [input() for _ in range(n)]
ac = 0
wa = 0
tle = 0
re = 0
for i in range(n):
if s[i] == 'AC':
ac += 1
elif s[i] == 'WA':
wa += 1
elif s[i] == 'TLE':
tle += 1
else:
re += 1
print("AC ×" + str(ac))
print("WA ×" + str(wa))
print("TLE ×" + str(tle))
print("RE ×" + str(re)) | s807830643 | Accepted | 153 | 16,268 | 322 |
n = int(input())
s = [input() for _ in range(n)]
ac = 0
wa = 0
tle = 0
re = 0
for i in range(n):
if s[i] == 'AC':
ac += 1
elif s[i] == 'WA':
wa += 1
elif s[i] == 'TLE':
tle += 1
elif s[i] == 'RE':
re += 1
print("AC x " + str(ac))
print("WA x " + str(wa))
print("TLE x " + str(tle))
print("RE x " + str(re)) |
s306099417 | p03563 | u675919123 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 631 | Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it. | #coding: utf-8
SS = input()
T = input()
f = 0
if len(SS) < len(T):
print('UNRESTORABLE')
else:
for i in range(0, len(SS)-len(T)+1):
pSS = SS[i:len(T)+i]
Tt = ''
for j in range(0,len(T)):
if pSS[j] == T[j]:
Tt += T[j]
elif pSS[j] == '?':
Tt += T[j]
else:
break
if Tt == T:
f += 1
SS = SS.replace(SS[i:len(T)+i], Tt)
ans = SS.replace('?','a')
break
else:
f += 0
if f == 1:
print(ans)
else:
print('UNRESTORABLE') | s345405706 | Accepted | 18 | 2,940 | 65 | #coding: utf-8
R = int(input())
G = int(input())
print(2*G - R) |
s382228291 | p03545 | u104005543 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 337 | Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted. | abcd = str(input())
a = int(abcd[0])
b = int(abcd[1])
c = int(abcd[2])
d = int(abcd[3])
for i in range(2 ** 3):
op = [1, 1, 1]
for j in range(3):
if ((i >> j) & 1):
op[j] = -1
if (a + b * op[0] + c * op[1] + d * op[2]) == 7:
print(str(a) + str(b * op[0]) + str(c * op[1]) + str(d * op[2]) + '=7') | s251123832 | Accepted | 18 | 3,064 | 436 | abcd = str(input())
a = int(abcd[0])
b = int(abcd[1])
c = int(abcd[2])
d = int(abcd[3])
for i in range(2 ** 3):
op = ['+', '+', '+']
ans = a
for j in range(3):
if ((i >> j) & 1):
op[j] = '-'
ans -= int(abcd[j + 1])
else:
ans += int(abcd[j + 1])
if ans == 7:
print(str(a) + str(op[0]) + str(b) + str(op[1]) + str(c) + str(op[2]) + str(d) + '=7')
exit() |
s360369619 | p03150 | u811000506 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 137 | A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string. | S = str(input())
for i in range(8):
word = S[:i]+S[-7:][i:]
if word == "keyence":
print("Yes")
exit()
print("No") | s773621663 | Accepted | 17 | 2,940 | 137 | S = str(input())
for i in range(8):
word = S[:i]+S[-7:][i:]
if word == "keyence":
print("YES")
exit()
print("NO") |
s799572239 | p04043 | u178963077 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 112 | Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order. | a = list(map(int, input().split()))
if sum(a) == 17 and a.count(5) == 2:
print("Yes")
else:
print("No")
| s269613527 | Accepted | 17 | 2,940 | 112 | a = list(map(int, input().split()))
if sum(a) == 17 and a.count(5) == 2:
print("YES")
else:
print("NO")
|
s547179767 | p03729 | u186838327 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 106 | You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`. | a, b, c = map(str, input().split())
if a[-1] == b[0] and b[-1] == c[0]:
print('Yes')
else:
print('No') | s895547106 | Accepted | 18 | 2,940 | 106 | a, b, c = map(str, input().split())
if a[-1] == b[0] and b[-1] == c[0]:
print('YES')
else:
print('NO') |
s506352413 | p02928 | u602252807 | 2,000 | 1,048,576 | Wrong Answer | 216 | 3,444 | 424 | We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j. | n,k = map(int,input().split())
a = list(map(int,input().split()))
a2 = a + a
def tentou (l):
count = [0]*len(l)
for i in range(len(l)-1):
for j in range(i+1,len(l)):
if l[i] > l[j]:
count[i] += 1
return count
v = tentou(a)
vk = [0]*len(v)
print(v)
print(vk)
for i in range(len(v)):
vk[i] = v[i] * (k*(k+1))//2
print(vk)
S = sum(vk)
result = S % ((10**9) + 7)
print(result) | s619509234 | Accepted | 999 | 3,336 | 635 | n,k = map(int,input().split())
a = list(map(int,input().split()))
a2 = a + a
def tentou (l):
count = [0]*len(l)
for i in range(len(l)-1):
for j in range(i+1,len(l)):
if l[i] > l[j]:
count[i] += 1
return count
if n == 1:
print(0)
else:
a_v = tentou(a)
a2_v = tentou(a2)
mergin = [x-y-y for (x,y) in zip(a2_v,a_v)]
if k == 1:
S = sum(a_v)
result = S % ((10**9) + 7)
else:
for i in range(len(a_v)):
a_v[i] = a_v[i]*(k*(k+1))//2 + mergin[i]*(k*(k-1))//2
S = sum(a_v)
result = S % ((10**9) + 7)
print(result) |
s314365058 | p02534 | u751724075 | 2,000 | 1,048,576 | Wrong Answer | 23 | 9,012 | 24 | You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them. For example, if K = 3, print `ACLACLACL`. | print(input().strip()*3) | s920286532 | Accepted | 23 | 9,072 | 35 | K = int(input())
print("ACL" * K)
|
s406825093 | p03730 | u794603719 | 2,000 | 262,144 | Wrong Answer | 21 | 3,316 | 199 | We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`. | a, b, c = map(int, input().split(" "))
print(a,b,c)
flag = 1
for i in range(101):
temp = a * i
if(temp % b == c):
print("YES")
flag = 0
break
if(flag):
print("NO") | s702185955 | Accepted | 17 | 2,940 | 201 | a, b, c = map(int, input().split(" "))
# print(a,b,c)
flag = 1
for i in range(101):
temp = a * i
if(temp % b == c):
print("YES")
flag = 0
break
if(flag):
print("NO") |
s753303710 | p03635 | u224156735 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 47 | In _K-city_ , there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city? | S=input()
a=str(len(S)-2)
print(S[:1]+a+S[-1:]) | s572270254 | Accepted | 17 | 2,940 | 47 | a,b=map(int,input().split())
print((a-1)*(b-1)) |
s929850128 | p03471 | u414626225 | 2,000 | 262,144 | Wrong Answer | 263 | 9,148 | 378 | The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough. | N, Y = map(int, input().split())
if (Y > N * 10000) or (Y < N * 1000):
print("{} {} {}".format(-1, -1, -1))
found = False
for x in range(Y//10000+2):
for y in range(Y//5000+2):
z = N - x - y
if (x * 10000 + y * 5000 + z * 1000) == Y:
print("{} {} {}".format(x, y, z))
found = True
break
if found:
break
| s268599835 | Accepted | 945 | 9,176 | 541 | N, Y = map(int, input().split())
# if (Y > N * 10000) or (Y < N * 1000):
# print("{} {} {}".format(-1, -1, -1))
found = False
for x in range(N+1):
for y in range(N+1):
if (x + y) > N:
continue
z = N - x - y
if z < 0:
continue
if (x * 10000 + y * 5000 + z * 1000) == Y:
ans_x = x
ans_y = y
ans_z = z
found = True
if found:
print("{} {} {}".format(ans_x, ans_y, ans_z))
else:
print("{} {} {}".format(-1, -1, -1))
|
s222533217 | p03352 | u652081898 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 220 | You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2. | x = int(input())
result = 0
for i in range(2, 32):
for j in range(1,10):
test = i**j
if test <= x and test >= result:
result = test
else:
pass
print(result) | s064233705 | Accepted | 18 | 2,940 | 221 | x = int(input())
result = 1
for i in range(2, 32):
for j in range(2,10):
test = i**j
if test <= x and test >= result:
result = test
else:
pass
print(result)
|
s347186543 | p04012 | u723583932 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 237 | Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful. | #abc044 b
s=input()
moji=dict()
flag=True
for i in range(len(s)):
if i not in moji:
moji[i]=1
else:
moji[i]+=1
for i in moji.values():
if i%2==1:
flag=False
break
print("Yes" if flag else "No") | s530315175 | Accepted | 17 | 2,940 | 226 | #abc044 b
s=input()
moji=dict()
flag=True
for i in s:
if i not in moji:
moji[i]=1
else:
moji[i]+=1
for i in moji.values():
if i%2==1:
flag=False
break
print("Yes" if flag else "No") |
s850165090 | p03563 | u790877102 | 2,000 | 262,144 | Wrong Answer | 23 | 2,940 | 58 | Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it. | R = int(input())
G = int(input())
W = R + R - G
print(W) | s164251043 | Accepted | 17 | 2,940 | 55 | R = int(input())
G = int(input())
W = G+G-R
print(W)
|
s566341134 | p03457 | u903959844 | 2,000 | 262,144 | Wrong Answer | 394 | 11,844 | 446 | AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan. | N = int(input())
t = [0]
x = [0]
y = [0]
for i in range(N):
a = input().split()
t.append(int(a[0]))
x.append(int(a[1]))
y.append(int(a[2]))
bool = False
for i in range(N):
if t[i+1]-t[i] >= abs(x[i+1]-x[i]) + abs(y[i+1]-y[i]):
t_EO = (t[i+1]-t[i]) % 2
xy_EO = (abs(x[i+1]-x[i]) + abs(y[i+1]-y[i])) % 2
if t_EO == xy_EO:
bool = True
if bool == True:
print("YES")
else:
print("NO")
| s413396831 | Accepted | 352 | 11,816 | 409 | N = int(input())
t = [0]
x = [0]
y = [0]
for i in range(N):
a = input().split()
t.append(int(a[0]))
x.append(int(a[1]))
y.append(int(a[2]))
bool = True
for i in range(N):
d_t = abs(t[i+1]-t[i])
d_xy = abs(x[i+1]-x[i]) + abs(y[i+1]-y[i])
if d_t < d_xy:
bool = False
elif d_t % 2 != d_xy % 2:
bool = False
if bool == True:
print("Yes")
else:
print("No") |
s040072075 | p03433 | u354527070 | 2,000 | 262,144 | Wrong Answer | 26 | 9,040 | 107 | E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins. | n = int(input())
a = int(input())
print(n)
mod = n % 500
if mod < a:
print("Yes")
else:
print("No") | s623281048 | Accepted | 26 | 9,100 | 99 | n = int(input())
a = int(input())
mod = n % 500
if mod <= a:
print("Yes")
else:
print("No") |
s070918481 | p03573 | u467307100 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 100 | You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers. | a, b, c = map(int,input().split())
if a == b:
print(a)
if b == c:
print(b)
if a ==c:
print(c) | s084810997 | Accepted | 17 | 2,940 | 100 | a, b, c = map(int,input().split())
if a == b:
print(c)
if b == c:
print(a)
if a ==c:
print(b) |
s458392740 | p02612 | u937238023 | 2,000 | 1,048,576 | Wrong Answer | 28 | 9,152 | 51 | We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required. | n = int(input())
x = (n+999)/1000
print(x*1000 - n) | s408790800 | Accepted | 27 | 9,052 | 53 | n = int(input())
x = (n+999)//1000
print(x*1000 - n)
|
s931151418 | p03998 | u396961814 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 584 | Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game. | SA = input()
SB = input()
SC = input()
WIN = [0, 0, 0]
SA_CNT = 0
SB_CNT = 0
SC_CNT = 0
card = SA[0]
while True:
if card == 'a':
SA_CNT += 1
if SA_CNT == len (SA):
print('a')
break
else:
card = SA[SA_CNT]
if card == 'b':
SB_CNT += 1
if SB_CNT == len(SB):
print('b')
break
else:
card = SB[SA_CNT]
if card == 'c':
SC_CNT += 1
if SC_CNT == len (SC):
print('c')
break
else:
card = SC[SA_CNT] | s809892548 | Accepted | 17 | 3,064 | 577 | SA = input()
SB = input()
SC = input()
card = SA[0]
SA_CNT = 1
SB_CNT = 0
SC_CNT = 0
while True:
if card == 'a':
if SA_CNT < len (SA):
card = SA[SA_CNT]
SA_CNT += 1
else:
print('A')
break
if card == 'b':
if SB_CNT < len (SB):
card = SB[SB_CNT]
SB_CNT += 1
else:
print('B')
break
if card == 'c':
if SC_CNT < len (SC):
card = SC[SC_CNT]
SC_CNT += 1
else:
print('C')
break |
s838318553 | p03228 | u932465688 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,064 | 281 | In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person. Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total. | A,B,K = map(int,input().split())
c = 0
if K%2 == 0:
while c <= K:
B = B+A//2
A = A//2
c += 1
A = A+B//2
B = B//2
c += 1
else:
while c < K-1:
B = B+A//2
A = A//2
c += 1
A = A+B//2
B = B//2
c += 1
B = B+A//2
A = A//2
print(A,B) | s458922355 | Accepted | 18 | 3,064 | 280 | A,B,K = map(int,input().split())
c = 0
if K%2 == 0:
while c < K:
B = B+A//2
A = A//2
c += 1
A = A+B//2
B = B//2
c += 1
else:
while c < K-1:
B = B+A//2
A = A//2
c += 1
A = A+B//2
B = B//2
c += 1
B = B+A//2
A = A//2
print(A,B) |
s857333839 | p03377 | u287431190 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 119 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | a,b,x = map(int, input().split())
if a > x:
print('No')
exit()
if (a+b) > x:
print('No')
exit()
print('Yes') | s675394146 | Accepted | 17 | 2,940 | 119 | a,b,x = map(int, input().split())
if x < a:
print('NO')
exit()
if (a+b) < x:
print('NO')
exit()
print('YES') |
s500792114 | p02936 | u517447467 | 2,000 | 1,048,576 | Wrong Answer | 2,107 | 56,180 | 545 | Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations. | import copy
N, Q = list(map(int, input().split()))
nodes = dict()
for i in range(N-1):
r, e = list(map(int, input().split()))
if r not in nodes:
nodes[r] = [e]
else:
nodes[r].append(e)
#print(nodes)
result = [0] * N
edges = []
for i in range(Q):
r, s = list(map(int, input().split()))
result[r-1] += s
if r in nodes:
edges = copy.copy(nodes[r])
while len(edges) != 0:
r = edges.pop(0)
result[r-1] += s
if r in nodes:
edges.extend(nodes[r])
#print(edges, nodes[r])
#print(result)
print(result) | s962446109 | Accepted | 1,865 | 274,864 | 546 | import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**6)
N, Q = list(map(int, input().split()))
nodes = [[] for i in range(N)]
for i in range(N-1):
r, e = list(map(int, input().split()))
r, e = r - 1, e - 1
nodes[r].append(e)
nodes[e].append(r)
#print(nodes)
result = [0] * N
for i in range(Q):
r, s = list(map(int, input().split()))
result[r-1] += s
def dfs(r, prev=-1):
for i in nodes[r]:
if i == prev:
continue
result[i] += result[r]
dfs(i, r)
if __name__ == "__main__":
dfs(0)
print(*result)
|
s008818401 | p02613 | u637874199 | 2,000 | 1,048,576 | Wrong Answer | 138 | 16,280 | 237 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. | n = int(input())
data = [input() for _ in range(n)]
a = data.count("AC")
b = data.count("WA")
c = data.count("TLE")
d = data.count("RE")
print("AC × " + str(a))
print("WA × " + str(b))
print("TLE × " + str(c))
print("RE × " + str(d)) | s400339084 | Accepted | 138 | 16,208 | 233 | n = int(input())
data = [input() for _ in range(n)]
a = data.count("AC")
b = data.count("WA")
c = data.count("TLE")
d = data.count("RE")
print("AC x " + str(a))
print("WA x " + str(b))
print("TLE x " + str(c))
print("RE x " + str(d)) |
s199477260 | p03338 | u488934106 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,064 | 385 | You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position. | def CutandCount(n , s):
ans = 0
for i in range(1 , n + 1):
count = 0
for w in range(ord('a'), ord('z') + 1):
if s[:i] in chr(w) and s[i:] in chr(w):
count += 1
ans = max(ans , count)
return ans
def main():
n = int(input())
s = str(input())
print(CutandCount(n , s))
if __name__ == '__main__':
main() | s365408365 | Accepted | 18 | 3,064 | 385 | def CutandCount(n , s):
ans = 0
for i in range(1 , n + 1):
count = 0
for w in range(ord('a'), ord('z') + 1):
if chr(w) in s[:i] and chr(w) in s[i:]:
count += 1
ans = max(ans , count)
return ans
def main():
n = int(input())
s = str(input())
print(CutandCount(n , s))
if __name__ == '__main__':
main() |
s969454796 | p03555 | u106297876 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 86 | You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise. | a = list(input())
b = list(input())
if a == b[::-1]:
print('Yes')
else:
print('No') | s476154983 | Accepted | 17 | 2,940 | 86 | a = list(input())
b = list(input())
if a == b[::-1]:
print('YES')
else:
print('NO') |
s198474845 | p02842 | u511457539 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 91 | Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact. | N = int(input())
X = N/1.08
if X*1.08 //1 == N:
print(X)
else:
print(X)
print(":(")
| s341961982 | Accepted | 17 | 2,940 | 104 | import math
N = int(input())
x = math.ceil(N/1.08)
if x*1.08//1 == N:
print(x)
else:
print(":(")
|
s485010444 | p02613 | u882765852 | 2,000 | 1,048,576 | Wrong Answer | 157 | 9,216 | 285 | Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format. | n=int(input())
ac=0
wa=0
tle=0
re=0
for i in range(n) :
s=str(input())
if s=="AC" :
ac=ac+1
elif s=="WA" :
wa=wa+1
elif s=="TLE" :
tle=tle+1
else :
re=re+1
print("AC ×",ac)
print("WA ×",wa)
print("TLE ×",tle)
print("RE ×",re) | s642689967 | Accepted | 156 | 9,208 | 281 | n=int(input())
ac=0
wa=0
tle=0
re=0
for i in range(n) :
s=str(input())
if s=="AC" :
ac=ac+1
elif s=="WA" :
wa=wa+1
elif s=="TLE" :
tle=tle+1
else :
re=re+1
print("AC x",ac)
print("WA x",wa)
print("TLE x",tle)
print("RE x",re) |
s823031175 | p03437 | u010090035 | 2,000 | 262,144 | Wrong Answer | 2,103 | 3,060 | 154 | You are given positive integers X and Y. If there exists a positive integer not greater than 10^{18} that is a multiple of X but not a multiple of Y, choose one such integer and print it. If it does not exist, print -1. | l = list(map(int,input().split()))
A = l[0];
B = l[1];
if(A%B == 0):
print(-1)
else:
xa = A;
while xa%B != 0:
xa += A;
print(xa); | s576131695 | Accepted | 17 | 3,060 | 234 | l = list(map(int,input().split()))
A = l[0];
B = l[1];
if(A%B == 0):
print(-1)
else:
xa = A*2;
while ((xa%B == 0) and (xa < 10**18)):
xa += A;
if(xa > 10**18):
print(-1);
else:
print(xa); |
s144334847 | p03448 | u276115223 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 304 | You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different. | # ABC 087: B – Coins
a, b, c, x = [int(input()) for _ in range(4)]
number_of_ways = 0
for i in range(a + 1):
for j in range(b + 1):
if 0 <= (x - (500 * i + 100 * j)) // 50 <= c:
print(i, j, (x - (500 * i + 100 * j)) // 50)
number_of_ways += 1
print(number_of_ways) | s106827654 | Accepted | 54 | 3,060 | 254 | # ABC 087: B – Coins
a, b, c, x = [int(input()) for _ in range(4)]
comb = 0
for i in range(0, a + 1):
for j in range(0, b + 1):
for k in range(0, c + 1):
comb = comb + 1 if 500 * i + 100 * j + 50 * k == x else comb
print(comb) |
s459081490 | p02742 | u484052148 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 167 | We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move: | h, w = map(int, input().split())
if h==1 or w==1:
print(1)
elif h%2 != w%2 != 0:
print(h*w//2 + 1)
else:
print(h*w//2) | s779604844 | Accepted | 18 | 3,060 | 173 | h, w = map(int, input().split())
if h==1 or w==1:
print(1)
elif h%2 != 0 and w%2 != 0:
print(h*w//2 + 1)
else:
print(h*w//2) |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.