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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s306875050 | p02255 | u398978447 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 246 | Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step. |
n=int(input("n:"))
A=[0 for i in range(n)]
A=input().split()
for i in range(n):
A[i]=int(A[i])
print(A)
for i in range(n):
key=A[i]
j=i-1
while j>=0 and A[j]>key:
A[j+1]=A[j]
j=j-1
A[j+1]=key
print(A)
| s239479861 | Accepted | 30 | 5,976 | 343 | n=int(input())
A=[0 for i in range(n)]
A=input().split()
for i in range(n):
A[i]=(int)(A[i])
for i in range(n):
key=A[i]
j=i-1
while j>=0 and A[j]>key:
A[j+1]=A[j]
j=j-1
A[j+1]=key
for i in range(n):
if i!=n-1:
print(A[i],end=" ")
if i==n-1:
print(A[i])
|
s057692443 | p03598 | u987164499 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 133 | There are N balls in the xy-plane. The coordinates of the i-th of them is (x_i, i). Thus, we have one ball on each of the N lines y = 1, y = 2, ..., y = N. In order to collect these balls, Snuke prepared 2N robots, N of type A and N of type B. Then, he placed the i-th type-A robot at coordinates (0, i), and the i-th type-B robot at coordinates (K, i). Thus, now we have one type-A robot and one type-B robot on each of the N lines y = 1, y = 2, ..., y = N. When activated, each type of robot will operate as follows. * When a type-A robot is activated at coordinates (0, a), it will move to the position of the ball on the line y = a, collect the ball, move back to its original position (0, a) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. * When a type-B robot is activated at coordinates (K, b), it will move to the position of the ball on the line y = b, collect the ball, move back to its original position (K, b) and deactivate itself. If there is no such ball, it will just deactivate itself without doing anything. Snuke will activate some of the 2N robots to collect all of the balls. Find the minimum possible total distance covered by robots. | N = int(input())
K = int(input())
x = list(map(int,input().split()))
length = 0
for i in x:
length += min(i,K-i)
print(length) | s855128166 | Accepted | 18 | 2,940 | 135 | N = int(input())
K = int(input())
x = list(map(int,input().split()))
length = 0
for i in x:
length += min(i,K-i)*2
print(length) |
s876549866 | p02612 | u869566286 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,048 | 32 | 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())
print(n % 1000) | s311710014 | Accepted | 24 | 9,076 | 79 | n = int(input())
n = n % 1000
if n == 0:
print(0)
else:
print(1000 - n) |
s340619953 | p03408 | u626337957 | 2,000 | 262,144 | Wrong Answer | 20 | 3,064 | 354 | Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards. | N = int(input())
words = []
for _ in range(N):
words.append(('plus', input()))
M = int(input())
for _ in range(M):
words.append(('minus', input()))
ans = 0
for i in range(N):
point = 1
for j in range(N+M):
if words[i][1] == words[j][1]:
if words[j][0] == 'plus':
point += 1
else:
point -= 1
ans = max(ans, point) | s053092077 | Accepted | 20 | 3,064 | 377 | N = int(input())
words = []
for _ in range(N):
words.append(('plus', input()))
M = int(input())
for _ in range(M):
words.append(('minus', input()))
ans = 0
for i in range(N):
point = 1
for j in range(N+M):
if words[i][1] == words[j][1] and i != j:
if words[j][0] == 'plus':
point += 1
else:
point -= 1
ans = max(ans, point)
print(ans)
|
s747163711 | p03494 | u014139588 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 132 | 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. | A = list(map(int, input().split()))
count = 0
while all(a % 2 == 0 for a in A):
A = [a/2 for a in A]
count += 1
print(count) | s070517222 | Accepted | 20 | 3,060 | 259 | n = int(input())
a = list(map(int,input().split()))
kaisuu = 0
while a[0]%2 == 0:
for i in range(n):
if a[i]%2 == 0:
a[i] /= 2
if i == n-1:
kaisuu += 1
else:
break
else:
print(kaisuu) |
s141085956 | p02795 | u068142202 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 70 | We have a grid with H rows and W columns, where all the squares are initially white. You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions: * Choose one row, then paint all the squares in that row black. * Choose one column, then paint all the squares in that column black. At least how many operations do you need in order to have N or more black squares in the grid? It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations. | h = int(input())
w = int(input())
n = int(input())
print(n % max(h,w)) | s945329368 | Accepted | 17 | 2,940 | 128 | h = int(input())
w = int(input())
n = int(input())
if n % max(h,w) == 0:
print(n // max(h,w))
else:
print(n // max(h,w) + 1) |
s620258730 | p02678 | u920204936 | 2,000 | 1,048,576 | Wrong Answer | 710 | 87,288 | 628 | 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. | from collections import deque
N,M = map(int,input().split())
graph = [deque([]) for _ in range(N+1)]
for _ in range(M):
a,b = map(int,input().split())
graph[a].append(b)
graph[b].append(a)
def bfs(u):
ans = [None] * (N+1)
queue = deque([u])
d = [None] * (N+1)
d[u] = 0
while queue:
v = queue.popleft()
for i in graph[v]:
if d[i] is None:
d[i] = d[v] + 1
ans[i] = v
queue.append(i)
return ans
d = bfs(1)
for i in range(2,N+1):
print(d[i]) | s056930565 | Accepted | 839 | 87,376 | 641 | from collections import deque
N,M = map(int,input().split())
graph = [deque([]) for _ in range(N+1)]
for _ in range(M):
a,b = map(int,input().split())
graph[a].append(b)
graph[b].append(a)
def bfs(u):
ans = [None] * (N+1)
queue = deque([u])
d = [None] * (N+1)
d[u] = 0
while queue:
v = queue.popleft()
for i in graph[v]:
if d[i] is None:
d[i] = d[v] + 1
ans[i] = v
queue.append(i)
return ans
d = bfs(1)
print("Yes")
for i in range(2,N+1):
print(d[i]) |
s734099150 | p03971 | u996564551 | 2,000 | 262,144 | Wrong Answer | 189 | 4,988 | 393 | There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass. | N, A, B = input().split(' ')
N = int(N)
A = int(A)
B = int(B)
S = []
S = list(input())
JS = 0
AS = 0
OT = 0
for i in range(N):
print(S[i])
if S[i] == 'a':
JS += 1
elif S[i] == 'b':
AS += 1
elif S[i] == 'c':
OT += 1
if S[i] == 'a' and A + B >= (JS + AS):
print('Yes')
elif S[i] == 'b' and A + B >= (JS + AS) and AS <= B:
print('Yes')
else:
print('No') | s663703454 | Accepted | 111 | 4,712 | 291 | N, A, B = input().split(' ')
N = int(N)
A = int(A)
B = int(B)
S = []
S = list(input())
JS = 0
AS = 0
OT = 0
for i in range(N):
if S[i] == 'a' and A+B > JS+AS:
print('Yes')
JS += 1
elif S[i] == 'b' and A+B > JS+AS and AS < B:
print('Yes')
AS += 1
else:
print('No') |
s180240790 | p03007 | u757446793 | 2,000 | 1,048,576 | Wrong Answer | 273 | 14,016 | 626 | There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer. | # coding: utf-8
# Your code here!
N = int(input())
A = list(map(int,input().split()))
A = sorted(A)
print(sum(A[N//2:]) - sum(A[:N//2]))
l = 0
r = N - 1
if N % 2 == 0:
for i in range(N-1):
if i % 2 == 0:
print(A[r],A[l])
A[r] = A[r] - A[l]
l += 1
else:
print(A[l],A[r])
A[l] = A[l] - A[r]
r -= 1
else:
for i in range(N-1):
if i % 2 == 1:
print(A[r],A[l])
A[r] = A[r] - A[l]
l += 1
else:
print(A[l],A[r])
A[l] = A[l] - A[r]
r -= 1 | s011229508 | Accepted | 230 | 14,016 | 528 | # coding: utf-8
# Your code here!
N = int(input())
A = list(map(int,input().split()))
A = sorted(A)
ans = A[-1] - A[0]
A_nashi = A[1:-1]
plus = len(A_nashi)
for i in range(len(A_nashi)):
if A_nashi[i] >= 0:
plus = i
break
ans += sum(A_nashi[plus:]) - sum(A_nashi[:plus])
print(ans)
res = A[0]
for i in range(len(A_nashi) - plus):
print(res,A_nashi[-i-1])
res -= A_nashi[-i-1]
print(A[-1], res)
res = A[-1] - res
for i in range(plus):
print(res,A_nashi[i])
res -= A_nashi[i]
|
s079724460 | p02842 | u055519694 | 2,000 | 1,048,576 | Wrong Answer | 27 | 9,100 | 177 | 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. | import math
n = int(input())
o = n / 1.08
o_int = math.floor(o)
print(math.floor(o_int * 1.08))
if (math.floor(o_int * 1.08) == n):
print(o_int)
else:
print(":(") | s477238945 | Accepted | 28 | 9,168 | 226 | import math
n = int(input())
o = n / 1.080
o_floor = math.floor(o)
o_ceil = math.ceil(o)
if math.floor(o_floor * 1.08) == n:
print(o_floor)
elif math.floor(o_ceil * 1.08) == n:
print(o_ceil)
else:
print(':(') |
s074897592 | p02613 | u957486750 | 2,000 | 1,048,576 | Wrong Answer | 153 | 9,224 | 306 | 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=str("AC X")
wa=str("WA X")
tle=str("TLE X")
re=str("RE X")
a=0;b=0;c=0;d=0
for i in range(n):
p=input()
if(p=="AC"):
a+=1
if(p=="WA"):
b+=1
if (p == "TLE"):
c += 1
if (p == "RE"):
d += 1
print(ac,a)
print(wa,b)
print(tle,c)
print(re,d) | s788907292 | Accepted | 151 | 9,224 | 306 | n= int(input())
ac=str("AC x")
wa=str("WA x")
tle=str("TLE x")
re=str("RE x")
a=0;b=0;c=0;d=0
for i in range(n):
p=input()
if(p=="AC"):
a+=1
if(p=="WA"):
b+=1
if (p == "TLE"):
c += 1
if (p == "RE"):
d += 1
print(ac,a)
print(wa,b)
print(tle,c)
print(re,d) |
s318313492 | p03385 | u047816928 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 48 | You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`. | print('Yes' if sorted(input())=='abc' else 'No') | s151163947 | Accepted | 17 | 2,940 | 57 | print('Yes' if ''.join(sorted(input()))=='abc' else 'No') |
s168890894 | p03494 | u027403702 | 2,000 | 262,144 | Wrong Answer | 19 | 3,060 | 165 | 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. | N = int(input())
List = list(map(int,input().split()))
count = 1
while all(li % 2 == 0 for li in List):
List = [i // 2 for i in List]
count += 1
print(count) | s365339397 | Accepted | 18 | 3,060 | 167 | N = int(input())
List = list(map(int,input().split()))
count = 0
while all(li % 2 == 0 for li in List):
List = [li // 2 for li in List]
count += 1
print(count) |
s717009275 | p02678 | u945335181 | 2,000 | 1,048,576 | Wrong Answer | 894 | 35,260 | 611 | 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. | from queue import Queue
n,m = map(int,input().split())
grafo = [[] for i in range(n)]
for x in range(m):
a,b = map(int,input().split())
grafo[a-1].append(b)
grafo[b-1].append(a)
placas = [0] * (n)
def bfs(a):
q = Queue(maxsize = n)
placas[1] = 1
q.put(a)
while not q.empty():
popado = q.get()
for adj in grafo[popado-1]:
if not placas[adj-1]:
placas[adj-1] = popado
q.put(adj)
if placas.count(0) == 0:
print("Yes")
for x in placas[1::]:
print(x)
else:
print("No")
bfs(1)
| s901605839 | Accepted | 918 | 35,124 | 611 | from queue import Queue
n,m = map(int,input().split())
grafo = [[] for i in range(n)]
for x in range(m):
a,b = map(int,input().split())
grafo[a-1].append(b)
grafo[b-1].append(a)
placas = [0] * (n)
def bfs(a):
q = Queue(maxsize = n)
placas[0] = 1
q.put(a)
while not q.empty():
popado = q.get()
for adj in grafo[popado-1]:
if not placas[adj-1]:
placas[adj-1] = popado
q.put(adj)
if placas.count(0) == 0:
print("Yes")
for x in placas[1::]:
print(x)
else:
print("No")
bfs(1)
|
s580995222 | p02392 | u899891332 | 1,000 | 131,072 | Wrong Answer | 20 | 7,348 | 77 | Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No". | a = input().split()
if a[0] < a[1] < a[0]:
print('Yes')
else:
print('No') | s953377624 | Accepted | 20 | 7,512 | 77 | a = input().split()
if a[0] < a[1] < a[2]:
print('Yes')
else:
print('No') |
s414527719 | p00007 | u868716420 | 1,000 | 131,072 | Wrong Answer | 20 | 7,572 | 148 | Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week. Write a program which computes the amount of the debt in n weeks. | n = 10 ** 2
for i in range(int(input())) :
n = float(n) * 1.05
if n - int(n) > 0 : n = int(n) + 1
else : n = int(n)
print(n * (10 ** 4)) | s562831918 | Accepted | 40 | 7,572 | 148 | n = 10 ** 2
for i in range(int(input())) :
n = float(n) * 1.05
if n - int(n) > 0 : n = int(n) + 1
else : n = int(n)
print(n * (10 ** 3)) |
s783840291 | p03836 | u517447467 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 443 | Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him. |
Z = list(map(int, input().split()))
S, T = Z[:2], Z[2:]
dy, dx = [abs(T[i] - S[i]) for i in range(2)]
# first
result = ""
result += "U"*dy
result += "R"*dx
result += "D"*dy
result += "L"*dx
result += "L"
result += "U"*(dy + 1)
result += "R"*(dx + 1)
result += "D"
result += "R"
result += "D"*(dy + 1)
result += "L"*(dx + 1)
result += "U"
print(result) | s535893209 | Accepted | 18 | 3,064 | 443 |
Z = list(map(int, input().split()))
S, T = Z[:2], Z[2:]
dx, dy = [abs(T[i] - S[i]) for i in range(2)]
# first
result = ""
result += "U"*dy
result += "R"*dx
result += "D"*dy
result += "L"*dx
result += "L"
result += "U"*(dy + 1)
result += "R"*(dx + 1)
result += "D"
result += "R"
result += "D"*(dy + 1)
result += "L"*(dx + 1)
result += "U"
print(result) |
s660690392 | p03495 | u843032026 | 2,000 | 262,144 | Wrong Answer | 117 | 39,504 | 135 | Takahashi has N balls. Initially, an integer A_i is written on the i-th ball. He would like to rewrite the integer on some balls so that there are at most K different integers written on the N balls. Find the minimum number of balls that Takahashi needs to rewrite the integers on them. | N,K=map(int,input().split())
A={}
for i in input().split():
A[i]=A.get(i,0)+1
A=sorted(A.values())
print(A)
print(sum(A[0:len(A)-K])) | s153654918 | Accepted | 99 | 39,608 | 126 | N,K=map(int,input().split())
A={}
for i in input().split():
A[i]=A.get(i,0)+1
A=sorted(A.values())
print(sum(A[0:len(A)-K])) |
s345643808 | p03943 | u236592202 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 88 | Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students. | a,b,c=map(int,input().split())
s=a+b+c
if s%2==0:
print("YES")
else:
print("NO") | s121790393 | Accepted | 17 | 3,060 | 214 | a,b,c=map(int,input().split())
s=a+b+c
if s%2==0:
if a+b==c:
print("Yes")
elif a+c==b:
print("Yes")
elif c+b==a:
print("Yes")
else:
print("No")
else:
print("No") |
s989479692 | p03795 | u386089355 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 48 | Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y. | n = int(input())
print(800 * n - 200 * n // 15) | s092163203 | Accepted | 17 | 2,940 | 50 | n = int(input())
print(800 * n - 200 * (n // 15)) |
s995356990 | p00017 | u546285759 | 1,000 | 131,072 | Wrong Answer | 30 | 7,384 | 293 | In cryptography, Caesar cipher is one of the simplest and most widely known encryption method. Caesar cipher is a type of substitution cipher in which each letter in the text is replaced by a letter some fixed number of positions down the alphabet. For example, with a shift of 1, 'a' would be replaced by 'b', 'b' would become 'c', 'y' would become 'z', 'z' would become 'a', and so on. In that case, a text: this is a pen is would become: uijt jt b qfo Write a program which reads a text encrypted by Caesar Chipher and prints the corresponding decoded text. The number of shift is secret and it depends on datasets, but you can assume that the decoded text includes any of the following words: "the", "this", or "that". | d= dict(zip([chr(i) for i in range(97, 123)], [chr(i) for i in range(119,123)]+[chr(i) for i in range(97, 119)]))
while 1:
try:
text= input()
for i in input():
t= d.get(i)
print(i if t== None else t, end='')
print()
except:
break | s267610159 | Accepted | 40 | 6,684 | 768 | import string
strings = string.ascii_lowercase
clues = [(19, 7, 8, 18), (19, 7, 0, 19), (19, 7, 4)]
while True:
try:
data = input()
except:
break
for word in data.split():
if len(word) == 4 or 3:
dif = 19 - (ord(word[0]) - 97)
enc = ["" for _ in range(26)]
for k, v in zip([i for i in range(dif, dif+26)], strings):
enc[k%26] = v
candidate = tuple(enc.index(c) for c in word)
try:
clues.index(candidate)
except:
continue
break
ans = ""
for c in data:
try:
ans += strings[enc.index(c)]
except:
ans = ans + "." if c == "." else ans + " "
print(ans)
|
s455187123 | p04012 | u863723142 | 2,000 | 262,144 | Wrong Answer | 29 | 9,056 | 405 | 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. | str_line = input()
str_line = list(str_line)
str_line.sort()
#print(str_line)
temp = 0
stnum_list = []
for i in range(1,len(str_line)-1):
if str_line[i] == str_line[i-1]:
temp += 1
else:
stnum_list.append(temp)
temp = 0
rsult = 1
for i in range(len(stnum_list)):
rsult *= stnum_list[i]
if rsult%2 == 1:
print("YES")
else:
print("NO") | s404265002 | Accepted | 29 | 9,076 | 472 | str_line = input()
str_line = list(str_line)
str_line.sort()
#print(str_line)
temp = 0
stnum_list = []
for i in range(1,len(str_line)-1):
if str_line[i] == str_line[i-1]:
temp += 1
else:
stnum_list.append(temp)
temp = 0
rsult = 1
#print(len(str_line))
if len(str_line) == 1:
rsult = 0
else:
for i in range(len(stnum_list)):
rsult *= stnum_list[i]
if rsult%2 == 1:
print("Yes")
else:
print("No") |
s966376199 | p03674 | u923270446 | 2,000 | 262,144 | Wrong Answer | 879 | 42,432 | 762 | You are given an integer sequence of length n+1, a_1,a_2,...,a_{n+1}, which consists of the n integers 1,...,n. It is known that each of the n integers 1,...,n appears at least once in this sequence. For each integer k=1,...,n+1, find the number of the different subsequences (not necessarily contiguous) of the given sequence with length k, modulo 10^9+7. | from collections import Counter
n = int(input())
a = list(map(int, input().split()))
mod = 10 ** 9 + 7
class nCr():
def __init__(self, n):
self.n = n
self.fa = [1] * (self.n + 1)
self.fi = [1] * (self.n + 1)
for i in range(1, self.n + 1):
self.fa[i] = self.fa[i - 1] * i % mod
self.fi[i] = pow(self.fa[i], mod - 2, mod)
def comb(self, n, r):
if n < r:return 0
if n < 0 or r < 0:return 0
return self.fa[n] * self.fi[r] % mod * self.fi[n - r] % mod
comb = nCr(2 * n)
c = Counter(a)
num = c.most_common()[0][0]
l, r = a.index(num) + 1, n - list(reversed(a)).index(num) + 1
print(l, r)
for i in range(1, n + 2):
print(comb.comb(n + 1, i) - comb.comb(n - (r - l) - 1, i - 1)) | s696729371 | Accepted | 879 | 42,480 | 746 | from collections import Counter
n = int(input())
a = list(map(int, input().split()))
mod = 10 ** 9 + 7
class nCr():
def __init__(self, n):
self.n = n
self.fa = [1] * (self.n + 1)
self.fi = [1] * (self.n + 1)
for i in range(1, self.n + 1):
self.fa[i] = self.fa[i - 1] * i % mod
self.fi[i] = pow(self.fa[i], mod - 2, mod)
def comb(self, n, r):
if n < r:return 0
if n < 0 or r < 0:return 0
return self.fa[n] * self.fi[r] % mod * self.fi[n - r] % mod
comb = nCr(2 * n)
c = Counter(a)
num = c.most_common()[0][0]
l, r = a.index(num), n - list(reversed(a)).index(num)
for i in range(1, n + 2):
print((comb.comb(n + 1, i) - comb.comb(n - (r - l), i - 1)) % mod) |
s339618812 | p03192 | u553070631 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 105 | You are given an integer N that has exactly four digits in base ten. How many times does `2` occur in the base-ten representation of N? | a,b,c,d = map(int,input())
if a==2:
a=1
if b==2:
b=1
if c==2:
c=1
if d==2:
d=1
print(a+b+c+d) | s549004712 | Accepted | 17 | 2,940 | 177 | a,b,c,d = map(int,input())
if a==2:
e=1
else:
e=0
if b==2:
f=1
else:
f=0
if c==2:
g=1
else:
g=0
if d==2:
h=1
else:
h=0
print(e+f+g+h)
|
s715320555 | p03545 | u496821919 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 262 | 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. | X = list(input())
for i in range(2**3):
W = ["+"]*3
for j in range(3):
if (i >> j) & 1:
W[j] = "-"
formula = ""
for k,l in zip(X,W+[""]):
formula += k+l
if eval(formula) == 7:
print(formula)
break
| s672225515 | Accepted | 17 | 3,064 | 267 | X = list(input())
for i in range(2**3):
W = ["+"]*3
for j in range(3):
if (i >> j) & 1:
W[j] = "-"
formula = ""
for k,l in zip(X,W+[""]):
formula += k+l
if eval(formula) == 7:
print(formula+"=7")
break
|
s493632489 | p03386 | u543894008 | 2,000 | 262,144 | Wrong Answer | 2,246 | 2,071,944 | 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())
numList = [x for x in range(a, b+1)]
ansList = numList[:k] + numList[-k:]
for num in list(set(ansList)):
print(num) | s570452385 | Accepted | 17 | 3,060 | 186 | a, b, k = map(int, input().split())
if (b - a) / 2 >= k:
for i in range(a, a+k):
print(i)
for i in range(b-k+1, b+1):
print(i)
else:
for i in range(a, b+1):
print(i) |
s395188113 | p03565 | u231189826 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 571 | E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`. | S = list(input())
T = list(input())
s_reversed = list(reversed(S))
T_reversed = list(reversed(T))
F = 0
for i in range(len(S)-len(T)+1):
s = s_reversed[i:i+len(T)]
flag = 0
for j in range(len(T)):
if s[j] == T_reversed[j] or s[j] == '?':
continue
else:
flag += 1
break
if flag:
F += 1
for j in range(len(T)):
s_reversed[i+j-1] = T_reversed[j]
break
if F:
print('UNRESTORABLE')
else:
R = ''.join(list(reversed(s_reversed)))
print(R.replace('?','a'))
| s456382222 | Accepted | 19 | 3,188 | 588 | S = list(input())
T = list(input())
S.reverse()
T.reverse()
flag = False
for i in range(len(S)-len(T)+1):
for j in range(len(T)):
if S[j+i] == T[j] or S[j+i] == '?':
pass
else:
break
if j == len(T)-1:
flag = True
for k in range(len(T)):
if S[k+i] == '?':
S[k+i] = T[k]
break
if flag:
break
if flag:
for i in range(len(S)):
if S[i] == '?':
S[i] = 'a'
S.reverse()
print(''.join(S))
else:
print('UNRESTORABLE')
|
s135075395 | p03760 | u337626942 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 157 | Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password. | o=input()
e=input()
ans=[]
for i in range(2*len(o) -1):
if i%2==0:
ans.append(o[i//2])
else:
ans.append(e[i//2])
print(*ans, sep='') | s273005112 | Accepted | 17 | 3,060 | 204 | o=input()
e=input()
ans=[]
cnt=0
for i in range(2*len(e)):
if i%2==0:
ans.append(o[i//2])
else:
ans.append(e[i//2])
if len(o)!=len(e):
ans.append(o[-1])
print(*ans, sep='') |
s782607947 | p03945 | u853586331 | 2,000 | 262,144 | Wrong Answer | 45 | 3,188 | 101 | Two foxes Jiro and Saburo are playing a game called _1D Reversi_. This game is played on a board, using black and white stones. On the board, stones are placed in a row, and each player places a new stone to either end of the row. Similarly to the original game of Reversi, when a white stone is placed, all black stones between the new white stone and another white stone, turn into white stones, and vice versa. In the middle of a game, something came up and Saburo has to leave the game. The state of the board at this point is described by a string S. There are |S| (the length of S) stones on the board, and each character in S represents the color of the i-th (1 ≦ i ≦ |S|) stone from the left. If the i-th character in S is `B`, it means that the color of the corresponding stone on the board is black. Similarly, if the i-th character in S is `W`, it means that the color of the corresponding stone is white. Jiro wants all stones on the board to be of the same color. For this purpose, he will place new stones on the board according to the rules. Find the minimum number of new stones that he needs to place. | S=input()
t=len(S)
ans=0
for i in range(t-1):
if S[i-1]!=S[i]:
ans+=1
else:
ans+=0
print(ans)
| s418156380 | Accepted | 43 | 3,188 | 101 | S=input()
t=len(S)
ans=0
for i in range(t-1):
if S[i]!=S[i+1]:
ans+=1
else:
ans+=0
print(ans)
|
s682677728 | p02406 | u821624310 | 1,000 | 131,072 | Wrong Answer | 30 | 7,448 | 122 | In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; } | n = int(input())
result = 0
i = 1
while result < n:
result = i * 3
print(" {0}".format(result), end="")
i += 1 | s374421811 | Accepted | 20 | 7,904 | 167 | n = int(input())
for i in range(3, n+1):
if i % 3 == 0:
print(" " + str(i), end = "")
elif "3" in str(i):
print(" " + str(i), end = "")
print() |
s046271792 | p03779 | u580316060 | 2,000 | 262,144 | Time Limit Exceeded | 2,105 | 28,348 | 201 | There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X. | X = map(int,input().split())
#X=7
P = {0}
i=1
while True:
Q = {0}
for p in P:
Q.add(p+i)
Q.add(p-i)
P = P.union(Q)
if X in P:
print(i)
break
i = i+1
| s811405921 | Accepted | 27 | 3,060 | 66 | X = int(input())
a = 0
while X >0:
a += 1
X -= a
print (a) |
s250443187 | p03693 | u258073778 | 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? | r, g, b = map(int, input().split())
if (10*g + b)%4 == 0:
print('Yes')
else:
print('No') | s392866609 | Accepted | 17 | 2,940 | 43 | print('YES'*(int(input()[::2])%4==0)or'NO') |
s900743758 | p03157 | u588794534 | 2,000 | 1,048,576 | Wrong Answer | 533 | 27,680 | 1,060 | There is a grid with H rows and W columns, where each square is painted black or white. You are given H strings S_1, S_2, ..., S_H, each of length W. If the square at the i-th row from the top and the j-th column from the left is painted black, the j-th character in the string S_i is `#`; if that square is painted white, the j-th character in the string S_i is `.`. Find the number of pairs of a black square c_1 and a white square c_2 that satisfy the following condition: * There is a path from the square c_1 to the square c_2 where we repeatedly move to a vertically or horizontally adjacent square through an alternating sequence of black and white squares: black, white, black, white... | h,w=map(int,input().split())
maze=[list(input()) for _ in range(h)]
group=[[0]*w for _ in range(h)]
group_no=0
color_cnt=[]
for y in range(h):
for x in range(w):
if group[y][x]==0:
stack=[(y,x)]
group_no+=1
color_cnt_tmp=[0,0]
while len(stack)>0:
tmp=stack.pop()
group[tmp[0]][tmp[1]]=group_no
print(tmp)
if maze[tmp[0]][tmp[1]]=="#":
color_cnt_tmp[0]+=1
else:
color_cnt_tmp[1]+=1
for (ny,nx) in [(0,1),(1,0),(-1,0),(0,-1)]:
my=ny+tmp[0]
mx=nx+tmp[1]
if 0<=my<h and 0<=mx<w:
if group[my][mx]==0:
if maze[tmp[0]][tmp[1]]!=maze[my][mx]:
stack.append((my,mx))
group[my][mx]=group_no
color_cnt.append(color_cnt_tmp)
ans=0
for (b,w) in color_cnt:
ans+=b*w
print(ans)
| s566801843 | Accepted | 468 | 27,792 | 1,061 | h,w=map(int,input().split())
maze=[list(input()) for _ in range(h)]
group=[[0]*w for _ in range(h)]
group_no=0
color_cnt=[]
for y in range(h):
for x in range(w):
if group[y][x]==0:
stack=[(y,x)]
group_no+=1
color_cnt_tmp=[0,0]
while len(stack)>0:
tmp=stack.pop()
group[tmp[0]][tmp[1]]=group_no
#print(tmp)
if maze[tmp[0]][tmp[1]]=="#":
color_cnt_tmp[0]+=1
else:
color_cnt_tmp[1]+=1
for (ny,nx) in [(0,1),(1,0),(-1,0),(0,-1)]:
my=ny+tmp[0]
mx=nx+tmp[1]
if 0<=my<h and 0<=mx<w:
if group[my][mx]==0:
if maze[tmp[0]][tmp[1]]!=maze[my][mx]:
stack.append((my,mx))
group[my][mx]=group_no
color_cnt.append(color_cnt_tmp)
ans=0
for (b,w) in color_cnt:
ans+=b*w
print(ans)
|
s351480044 | p02663 | u384124931 | 2,000 | 1,048,576 | Wrong Answer | 22 | 9,152 | 80 | In this problem, we use the 24-hour clock. Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying? | h_1, m_1, h_2, m_2, k = map(int, input().split())
print(h_1*60+m_1-h_2*60-m_2-k) | s477431505 | Accepted | 20 | 9,156 | 81 | h_1, m_1, h_2, m_2, k = map(int, input().split())
print(-h_1*60-m_1+h_2*60+m_2-k) |
s822811618 | p02744 | u843135954 | 2,000 | 1,048,576 | Wrong Answer | 160 | 12,496 | 436 | In this problem, we only consider strings consisting of lowercase English letters. Strings s and t are said to be **isomorphic** when the following conditions are satisfied: * |s| = |t| holds. * For every pair i, j, one of the following holds: * s_i = s_j and t_i = t_j. * s_i \neq s_j and t_i \neq t_j. For example, `abcac` and `zyxzx` are isomorphic, while `abcac` and `ppppp` are not. A string s is said to be in **normal form** when the following condition is satisfied: * For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison. For example, `abcac` is in normal form, but `zyxzx` is not since it is isomorphic to `abcac`, which is lexicographically smaller than `zyxzx`. You are given an integer N. Print all strings of length N that are in normal form, in lexicographically ascending order. | import sys
stdin = sys.stdin
sys.setrecursionlimit(10**9)
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
nn = lambda: list(stdin.readline().split())
ns = lambda: stdin.readline().rstrip()
al = ['a','b','c','d','e','f','g','h','i','j']
n = ni()
import numpy as np
for i in range(2**(n-1)):
k = [0]+list(map(int,format(i, 'b')))
p = np.cumsum(k)
ans = [str(al[j]) for j in p]
print(' '.join(ans)) | s916591850 | Accepted | 413 | 4,916 | 464 | import sys
stdin = sys.stdin
sys.setrecursionlimit(10**9)
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
nn = lambda: list(stdin.readline().split())
ns = lambda: stdin.readline().rstrip()
al = ['a','b','c','d','e','f','g','h','i','j']
import copy
n = ni()
def f(a,b,m):
if a == n:
print(''.join([al[i] for i in b]))
else:
for i in range(m+2):
c = copy.copy(b)
c.append(i)
f(a+1,c,max(m,i))
f(1,[0],0) |
s795295139 | p03556 | u408620326 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 29 | 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. | print(int(int(input())**0.5)) | s084280797 | Accepted | 17 | 2,940 | 32 | print(int(int(input())**0.5)**2) |
s627164672 | p03455 | u350997995 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 92 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a,b = map(int,input().split())
if a*b%2==0:
print("Odd")
else:
print("Even") | s603329148 | Accepted | 17 | 2,940 | 73 | a,b = map(int,input().split())
c = a*b
print("Odd" if c%2==1 else "Even") |
s199100132 | p02578 | u416123847 | 2,000 | 1,048,576 | Wrong Answer | 101 | 32,208 | 191 | N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a person includes the stool. Find the minimum total height of the stools needed to meet this goal. | n = int(input())
h = list(map(int, input().split()))
left_max = h[0]
steps = 0
for i in range(1, n):
if(left_max > h[i]):
steps += h[i]-left_max
left_max = steps + h[i]
print(steps) | s463137985 | Accepted | 128 | 32,128 | 191 | n = int(input())
h = list(map(int, input().split()))
left_max = h[0]
steps = 0
for i in range(1, n):
if(left_max > h[i]):
steps += left_max-h[i]
else:
left_max = h[i]
print(steps) |
s161117223 | p04012 | u135616177 | 2,000 | 262,144 | Wrong Answer | 193 | 3,572 | 170 | 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 collections
w = input()
count_dict = collections.Counter(w)
for k, v in count_dict.items():
if v % 2 != 0:
print('No')
break
print('yes')
| s640525710 | Accepted | 540 | 3,572 | 209 | import collections
w = input()
flg = 0
count_dict = collections.Counter(w)
for k, v in count_dict.items():
if v % 2 != 0:
flg = 1
break
if flg == 1:
print('No')
else:
print('Yes')
|
s295694338 | p03448 | u476048753 | 2,000 | 262,144 | Wrong Answer | 44 | 3,064 | 317 | 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()) # 500
B = int(input()) # 100
C = int(input()) # 50
X = int(input()) # target
ans = 0
for i in range(A+1):
amount = 500* (i - 1)
for j in range(B+1):
amount += 100*(j - 1)
for h in range(C+1):
amount += 50*(h - 1)
if amount == X:
ans += 1
print(ans) | s692004615 | Accepted | 57 | 3,060 | 265 | A = int(input()) # 500
B = int(input()) # 100
C = int(input()) # 50
X = int(input()) # target
ans = 0
for i in range(A+1):
for j in range(B+1):
for h in range(C+1):
amount = 500*i + 100 * j + 50 * h
if amount == X:
ans += 1
print(ans) |
s219938648 | p00118 | u940389926 | 1,000 | 131,072 | Wrong Answer | 140 | 19,080 | 973 | タナカ氏が HW アールの果樹園を残して亡くなりました。果樹園は東西南北方向に H × W の区画に分けられ、区画ごとにリンゴ、カキ、ミカンが植えられています。タナカ氏はこんな遺言を残していました。 果樹園は区画単位でできるだけ多くの血縁者に分けること。ただし、ある区画の東西南北どれかの方向にとなりあう区画に同じ種類の果物が植えられていた場合は、区画の境界が分からないのでそれらは 1 つの大きな区画として扱うこと。 例えば次のような 3 × 10 の区画であれば ('リ'はリンゴ、'カ'はカキ、'ミ'はミカンを表す) 同じ樹がある区画の間の境界を消すと次のようになり、 結局 10 個の区画、つまり 10 人で分けられることになります。 雪が降って区画の境界が見えなくなる前に分配を終えなくてはなりません。あなたの仕事は果樹園の地図をもとに分配する区画の数を決めることです。 果樹園の地図を読み込み、分配を受けられる血縁者の人数を出力するプログラムを作成してください。 | import sys
sys.setrecursionlimit(100000)
fields = []
i = 0
H, W = 0, 0
field = []
for line in sys.stdin:
line = line.rstrip()
if line[0] not in ["#", "@", "*"]:
line = line.split(" ")
tmpH = H
tmpW = W
H = int(line[0])
W = int(line[1])
if (len(field) != 0):
fields.append([field, tmpH, tmpW])
field = []
if (H == 0 and W == 0):
break
else:
field.append(list(line))
i += 1
def dfs(x, y, fruit):
field[x][y] = "0"
for dx in [-1, 0, 1]:
if dx == -1: width = [0]
if dx == 0: width = [-1, 0, 1]
if dx == 1: width = [0]
for dy in width:
nx = x+dx
ny = y+dy
inField = (0 <= nx) and (0 <= ny) and (nx < H) and (ny < W)
if inField and field[nx][ny] == fruit:
dfs(nx, ny, fruit)
print(fields)
count = 0
for field_info in fields:
field = field_info[0]
H = field_info[1]
W = field_info[2]
for x in range(H):
for y in range(W):
if (field[x][y] != "0"):
dfs(x, y, field[x][y])
count += 1
print(count)
count = 0 | s451058469 | Accepted | 150 | 19,028 | 949 | import sys
sys.setrecursionlimit(100000)
def load_fields():
fields = []
i = 0
H, W = 0, 0
field = []
for line in sys.stdin:
line = line.rstrip()
if line[0] not in ["#", "@", "*"]:
line = line.split(" ")
H, W = int(line[0]), int(line[1])
if (len(field) != 0):
fields.append(field)
field = []
if (H == 0 and W == 0):
break
else:
field.append(list(line))
i += 1
return fields
def dfs(x, y, fruit):
field[x][y] = "0"
for dx in [-1, 0, 1]:
if dx == -1: width = [0]
if dx == 0: width = [-1, 0, 1]
if dx == 1: width = [0]
for dy in width:
nx = x+dx
ny = y+dy
inField = (0 <= nx < H) and (0 <= ny < W)
if inField and (field[nx][ny] == fruit):
dfs(nx, ny, fruit)
fields = load_fields()
count = 0
for field in fields:
H = len(field)
W = len(field[0])
for x in range(H):
for y in range(W):
if (field[x][y] != "0"):
dfs(x, y, field[x][y])
count += 1
print(count)
count = 0 |
s092047611 | p03852 | u846150137 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 58 | Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`. | print("vowel" if "aiueo".find(input())>0 else "consonant") | s681972261 | Accepted | 17 | 2,940 | 59 | print("vowel" if "aiueo".find(input())>=0 else "consonant") |
s687774138 | p03337 | u104003430 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 128 | You are given two integers A and B. Find the largest value among A+B, A-B and A \times B. | from sys import stdin
A, B = stdin.readline().rstrip().split()
a = int(A)
b = int(B)
print(type(A))
print(max(a+b, a-b, a*b)) | s407774945 | Accepted | 18 | 2,940 | 112 | from sys import stdin
A, B = stdin.readline().rstrip().split()
a = int(A)
b = int(B)
print(max(a+b, a-b, a*b)) |
s181379773 | p03400 | u094565093 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 204 | Some number of chocolate pieces were prepared for a training camp. The camp had N participants and lasted for D days. The i-th participant (1 \leq i \leq N) ate one chocolate piece on each of the following days in the camp: the 1-st day, the (A_i + 1)-th day, the (2A_i + 1)-th day, and so on. As a result, there were X chocolate pieces remaining at the end of the camp. During the camp, nobody except the participants ate chocolate pieces. Find the number of chocolate pieces prepared at the beginning of the camp. | N=int(input())
D,X=map(int,input().split())
A=[int(input()) for i in range(N)]
sum=0
for i in range(len(A)):
n=1
while 1+(n-1)*A[i]<=D:
sum+=1
n+=1
print(sum)
print(sum+X)
| s758632013 | Accepted | 18 | 3,060 | 191 | N=int(input())
D,X=map(int,input().split())
A=[int(input()) for i in range(N)]
sum=0
for i in range(len(A)):
n=1
while 1+(n-1)*A[i]<=D:
sum+=1
n+=1
print(sum+X)
|
s994663715 | p03360 | u061732150 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 175 | There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations? | def main():
A,B,C = list(map(int, input().split()))
K = int(input())
return A + B + C + max(A,B,C) **K - max(A,B,C)
if __name__ == '__main__':
print(main())
| s753093191 | Accepted | 17 | 3,060 | 296 | def main():
A,B,C = list(map(int, input().split()))
K = int(input())
for i in range(K):
if A == max(A,B,C):
A = A*2
elif B == max(A,B,C):
B = B*2
else:
C = C*2
return A+B+C
if __name__ == '__main__':
print(main())
|
s237514907 | p03644 | u209619667 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 129 | 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. | A = int(input())
s = 2
for i in range(1,A):
if A == 1:
print(1)
elif s <= A:
s = 2**i
else:
print(s//2)
break | s999135377 | Accepted | 17 | 3,060 | 185 | A = int(input())
s = 2
for i in range(1,A):
if A == 1:
print(1)
break;
elif A == 2:
print(2)
break;
elif s <= A:
s = 2**i
else:
break;
print(s//2) |
s188856184 | p03110 | u780698286 | 2,000 | 1,048,576 | Wrong Answer | 33 | 9,020 | 191 | 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())
a = [input().split() for i in range(n)]
yen = 0
bit = 0
for f in a:
if f[1] == "JPY":
yen += int(f[0])
else:
bit += float(f[0])
print(yen + int(380000 * bit)) | s099425746 | Accepted | 25 | 9,088 | 182 | n = int(input())
xu = [input().split() for i in range(n)]
y = 0
b = 0
for j in xu:
if j[1] == "JPY":
y += int(j[0])
else:
b += float(j[0])
ans = y + 380000 * b
print(ans) |
s719696356 | p03371 | u131881594 | 2,000 | 262,144 | Wrong Answer | 28 | 9,128 | 320 | "Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas. | a,b,c,x,y=map(int,input().split())
if a+b<=2*c: print(a*x+b*y)
else:
ans=min(x,y)*2*c
print(ans)
dif=abs(x-y)
if x>y:
if a>=2*c:
ans+=2*dif*c
else:
ans+=a*dif
else:
if b>=2*c:
ans+=2*dif*c
else:
ans+=b*dif
print(ans) | s293733678 | Accepted | 29 | 9,084 | 305 | a,b,c,x,y=map(int,input().split())
if a+b<=2*c: print(a*x+b*y)
else:
ans=min(x,y)*2*c
dif=abs(x-y)
if x>y:
if a>=2*c:
ans+=2*dif*c
else:
ans+=a*dif
else:
if b>=2*c:
ans+=2*dif*c
else:
ans+=b*dif
print(ans) |
s320154327 | p03455 | u716649090 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 66 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | a, b = map(int, input().split())
print("Even" if a*b%2 else "Odd") | s176833195 | Accepted | 17 | 2,940 | 67 | a, b = map(int, input().split())
print("Odd" if a*b%2 else "Even")
|
s024750215 | p03545 | u413165887 | 2,000 | 262,144 | Wrong Answer | 18 | 3,188 | 629 | 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. | import sys
df = list(map(int, list(str(input()))))
print(df)
counter = 0
for i in range(2**3):
bin_i = format(i, '03b')
counter += df[0]
for j in range(3):
print(j, counter)
if bin_i[j] == '1':
counter -= df[j+1]
else:
counter += df[j+1]
if counter == 7:
result = str(df[0])
for j in range(3):
if bin_i[j] =='0':
result += '+' + str(df[j+1])
else:
result += '-' + str(df[j+1])
result += '=7'
print(result)
sys.exit()
else:
print(counter)
counter = 0 | s266248852 | Accepted | 18 | 3,188 | 570 | import sys
df = list(map(int, list(str(input()))))
counter = 0
for i in range(2**3):
bin_i = format(i, '03b')
counter += df[0]
for j in range(3):
if bin_i[j] == '1':
counter -= df[j+1]
else:
counter += df[j+1]
if counter == 7:
result = str(df[0])
for j in range(3):
if bin_i[j] =='0':
result += '+' + str(df[j+1])
else:
result += '-' + str(df[j+1])
result += '=7'
print(result)
sys.exit()
else:
counter = 0 |
s124993369 | p02612 | u055687574 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,000 | 33 | 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())
print(n % 1000) | s976525932 | Accepted | 25 | 9,160 | 81 | n = int(input())
if n % 1000 == 0:
print(0)
else:
print(1000 - n % 1000) |
s707807719 | p03377 | u670180528 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 90 | 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. | n,m,x,*l=map(int,open(0).read().split())
s=sum(i in l for i in range(x))
print(min(s,m-s)) | s119976046 | Accepted | 17 | 2,940 | 60 | a,b,x=map(int,input().split());print("NYOE S"[a<=x<=a+b::2]) |
s407480620 | p03719 | u007550226 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 66 | You are given three integers A, B and C. Determine whether C is not less than A and not greater than B. | A,B,C = map(int,input().split())
print('YES' if A<=C<=B else 'NO') | s258272030 | Accepted | 18 | 2,940 | 66 | A,B,C = map(int,input().split())
print('Yes' if A<=C<=B else 'No') |
s694899124 | p03545 | u371132735 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 403 | 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. | import itertools
status = [(0, 1) for _ in range(3)]
state = list(itertools.product(*status))
num = [int(i) for i in input()]
for ptn in state:
ans = num[0]
ans_str = str(num[0])
for i,ope in enumerate(ptn):
if ope == 0:
ans += num[i+1]
ans_str += ("+" + str(num[i+1]))
else:
ans -= num[i+1]
ans_str += ("-" + str(num[i+1]))
if ans == 7:
break
print(ans_str)
| s547300186 | Accepted | 18 | 3,064 | 408 | import itertools
status = [(0, 1) for _ in range(3)]
state = list(itertools.product(*status))
num = [int(i) for i in input()]
for ptn in state:
ans = num[0]
ans_str = str(num[0])
for i,ope in enumerate(ptn):
if ope == 0:
ans += num[i+1]
ans_str += ("+" + str(num[i+1]))
else:
ans -= num[i+1]
ans_str += ("-" + str(num[i+1]))
if ans == 7:
break
print(ans_str+"=7")
|
s755077072 | p04029 | u564060397 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 129 | There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total? | a=input()
b=""
for i in a:
if i=="0":
b+="0"
if i=="1":
b+="1"
if i=="B":
b=b[:-1]
print(b) | s591527040 | Accepted | 17 | 2,940 | 72 | a=int(input())
answer=0
for i in range(1,a+1):
answer+=i
print(answer) |
s370160485 | p02841 | u571646975 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 127 | In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019. Integers M_1, D_1, M_2, and D_2 will be given as input. It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1. Determine whether the date 2019-M_1-D_1 is the last day of a month. | str = [input() for i in range(2)]
s1 = str[0].split()
s2 = str[1].split()
if s1[0] == s2[1]:
print("0")
else:
print("1")
| s344684065 | Accepted | 17 | 2,940 | 156 | str = [input() for i in range(2)]
s1 = str[0].split()
s2 = str[1].split()
if s1[0] != s2[0] and int(s1[1]) >= int(s2[1]):
print("1")
else:
print("0") |
s122336644 | p02613 | u395010524 | 2,000 | 1,048,576 | Wrong Answer | 160 | 16,324 | 318 | 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 = []
ac = 0
wa = 0
tle = 0
re = 0
for i in range(n):
s.append(input())
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)) | s444703687 | Accepted | 160 | 16,320 | 314 | n = int(input())
s = []
ac = 0
wa = 0
tle = 0
re = 0
for i in range(n):
s.append(input())
if s[i] == "AC":
ac += 1
elif s[i] == "WA":
wa += 1
elif s[i] == "TLE":
tle += 1
else:
re += 1
print("AC x " + str(ac))
print("WA x " + str(wa))
print("TLE x " + str(tle))
print("RE x "+ str(re)) |
s509915729 | p03493 | u981931040 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 62 | Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble. | S = list(input().split())
S = list(map(int, S))
print(sum(S))
| s202627701 | Accepted | 17 | 2,940 | 54 | S = list(input())
S = list(map(int, S))
print(sum(S))
|
s718232404 | p00341 | u019169314 | 1,000 | 262,144 | Wrong Answer | 20 | 5,592 | 202 | アイヅ放送協会の教育番組(ＡＨＫ教育)では、子ども向けの工作番組「あそんでつくろ」を放送しています。今回は棒で箱を作る回ですが、用意した１２本の棒を使って直方体ができるかを確かめたいと思います。ただし、棒は切ったり折ったりしてはいけません。 １２本の棒の長さが与えられるので、それらすべてを辺とする直方体が作れるかどうか判定するプログラムを作成せよ。 | li = list(map(int,input().split()))
so = sorted(li)
a = so[0]
b = so[4]
c = so[8]
res = False
for n in so[:4]:
res = a==n
for n in so[4:8]:
res = b==n
for n in so[8:]:
res = c==n
print(res)
| s567721684 | Accepted | 20 | 5,596 | 240 | li = list(map(int,input().split()))
so = sorted(li)
a = so[0]
b = so[4]
c = so[8]
for n in so[:4]:
res1= a==n
for n in so[4:8]:
res2= b==n
for n in so[8:]:
res3= c==n
res = 'yes' if res1 and res3 and res2 else 'no'
print(res)
|
s771721256 | p03860 | u881100099 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 29 | 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. | s=input()
print('A'+s[1]+'C') | s770762483 | Accepted | 17 | 2,940 | 40 | s=input().split()[1]
print('A'+s[0]+'C') |
s744064473 | p02401 | u981238682 | 1,000 | 131,072 | Wrong Answer | 20 | 5,604 | 261 | 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. | while 1:
a,op,b = input().split()
a = int(a)
b = int(b)
if op == '?' :
break
if op == '+' :
print(a + b)
if op == '-' :
print(a - b)
if op == '*' :
print(a * b)
if op == '/' :
print(a / b) | s787118419 | Accepted | 20 | 5,596 | 262 | while 1:
a,op,b = input().split()
a = int(a)
b = int(b)
if op == '?' :
break
if op == '+' :
print(a + b)
if op == '-' :
print(a - b)
if op == '*' :
print(a * b)
if op == '/' :
print(a // b) |
s254384457 | p03730 | u492605584 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 126 | 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(" "))
for i in range(1, 1000):
if A * i % B == C:
print("Yes")
exit(0)
print("No") | s042488764 | Accepted | 18 | 2,940 | 174 | A, B, C = map(int, input().split(" "))
l = []
for i in range(10000000):
tmp = A*i%B
if tmp in l:
break
l.append(tmp)
if C in l:
print("YES")
else:
print("NO")
|
s714217472 | p03371 | u883203948 | 2,000 | 262,144 | Wrong Answer | 128 | 7,096 | 326 | "Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas. | a,b,c,x,y = list(map(int,input().split()))
cost = []
tmp = 0
i = 0
while i <= max(x,y) :
tmp += i * 2 * c
if i >= x :
tmp += (y-i) * b
elif i >= y:
tmp += (x-i) * a
else:
tmp += (x - i) * a + (y - i) * b
cost.append(tmp)
i += 1
tmp = 0
print(min(cost))
| s307889750 | Accepted | 156 | 7,100 | 256 | a,b,c,x,y = list(map(int,input().split()))
cost = []
tmp = 0
i = 0
while i <= max(x,y) :
tmp += i * 2 * c
tmp += max(x-i,0)*a + max(y-i,0) * b
cost.append(tmp)
i += 1
tmp = 0
print(min(cost))
|
s528633389 | p03545 | u584563392 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 606 | 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. | input_str = input()
A = int(input_str[0])
B = int(input_str[1])
C = int(input_str[2])
D = int(input_str[3])
list = []
list.append(A + B + C + D)
list.append(A + B + C - D)
list.append(A + B - C + D)
list.append(A + B - C - D)
list.append(A - B + C + D)
list.append(A - B + C - D)
list.append(A - B - C + D)
list.append(A - B - C - D)
num = list.index(7)
print(num)
if num <= 3:
op1 = '+'
else:
op1 = '-'
if num == 0 or num == 1 or num == 4 or num == 5:
op2 = '+'
else:
op2 = '-'
if num % 2 == 0:
op3 = '+'
else:
op3 = '-'
print('{}{}{}{}{}{}{}=7'.format(A,op1,B,op2,C,op3,D))
| s242619992 | Accepted | 17 | 3,064 | 594 | input_str = input()
A = int(input_str[0])
B = int(input_str[1])
C = int(input_str[2])
D = int(input_str[3])
list = []
list.append(A + B + C + D)
list.append(A + B + C - D)
list.append(A + B - C + D)
list.append(A + B - C - D)
list.append(A - B + C + D)
list.append(A - B + C - D)
list.append(A - B - C + D)
list.append(A - B - C - D)
num = list.index(7)
if num <= 3:
op1 = '+'
else:
op1 = '-'
if num == 0 or num == 1 or num == 4 or num == 5:
op2 = '+'
else:
op2 = '-'
if num % 2 == 0:
op3 = '+'
else:
op3 = '-'
print('{}{}{}{}{}{}{}=7'.format(A,op1,B,op2,C,op3,D)) |
s184517358 | p03854 | u404676457 | 2,000 | 262,144 | Wrong Answer | 17 | 3,188 | 613 | You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`. | import sys
s = input()
i = 0
while s[i:5] != '':
if s[i:5] == 'dream':
if s[i+5:i+10] == 'dream':
i += 5
elif s[i+5:i+10] == 'erase':
i += 5
elif s[i+5:i+7] == 'er':
i += 7
else:
print('No')
sys.exit(0)
elif s[i:5] == 'erase':
if s[i+5:i+10] == 'dream':
i += 5
elif s[i+5:i+10] == 'erase':
i += 5
elif s[i+5:i+6] == 'r':
i += 6
else:
print('No')
sys.exit(0)
else:
print('No')
sys.exit(0)
print('Yes') | s358008221 | Accepted | 36 | 3,188 | 560 | import sys
s = input()
i = 0
while s[i:i+5] != '':
if s[i:i+5] == 'dream':
if s[i+5:i+10] == 'dream':
i += 5
elif s[i+5:i+10] == 'erase':
i += 5
elif s[i+5:i+7] == 'er':
i += 7
else:
i += 5
elif s[i:i+5] == 'erase':
if s[i+5:i+10] == 'dream':
i += 5
elif s[i+5:i+10] == 'erase':
i += 5
elif s[i+5:i+6] == 'r':
i += 6
else:
i += 5
else:
print('NO')
sys.exit(0)
print('YES') |
s079849994 | p02399 | u580227385 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 87 | Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number) | a, b = map(int, input().split())
print("{} {} {}".format(a // b, a % b, float(a / b)))
| s701967367 | Accepted | 20 | 5,600 | 91 | a, b = map(int, input().split())
print("{} {} {:.5f}".format(a // b, a % b, float(a / b)))
|
s287336249 | p03435 | u321035578 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 531 | We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j. Determine if he is correct. | def main():
c=[list(map(int,input().split())) for i in range(3)]
a=[0,0,0]
b=[0,0,0]
for i in range(1,101):
a[0]=i
a[1]=-1
a[2]=-1
b[0]=c[0][0]-a[0]
b[1]=c[0][1]-a[0]
b[2]=c[0][2]-a[0]
for j in range(1,3):
if c[j][0]-b[0]!=c[j][1]-b[1] or c[j][1]-b[1]!=c[j][2]-b[2]:
break
a[j]=c[j][0]-b[0]
if j == 2:
print('YES')
return
print('NO')
if __name__=='__main__':
main()
| s985949847 | Accepted | 17 | 3,064 | 702 | def main():
c=[list(map(int,input().split())) for i in range(3)]
a=[0,0,0]
b=[0,0,0]
for i in range(0,101):
a[0]=i
a[1]=-1
a[2]=-1
b[0]=c[0][0]-a[0]
b[1]=c[0][1]-a[0]
b[2]=c[0][2]-a[0]
for j in range(1,3):
if c[j][0]-b[0]!=c[j][1]-b[1] or c[j][1]-b[1]!=c[j][2]-b[2]:
break
a[j]=c[j][0]-b[0]
if j == 2 and check(a,b):
print('Yes')
return
print('No')
def check(a,b):
for i in range(0,3):
if a[i] >= 0 and b[i] >= 0 :
continue
else:
return False
return True
if __name__=='__main__':
main()
|
s232463853 | p03854 | u989892335 | 2,000 | 262,144 | Wrong Answer | 40 | 9,240 | 312 | You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`. | s=input()[::-1]
a="dream"[::-1]
b="dreamer"[::-1]
c="erase"[::-1]
d="eraser"[::-1]
i=0
ans="Yes"
while i<len(s):
if a in s[i:i+5]:
i+=5
elif b in s[i:i+7]:
i+=7
elif c in s[i:i+5]:
i+=5
elif d in s[i:i+6]:
i+=6
else :
ans="No"
break
print(ans) | s341090950 | Accepted | 38 | 9,260 | 312 | s=input()[::-1]
a="dream"[::-1]
b="dreamer"[::-1]
c="erase"[::-1]
d="eraser"[::-1]
i=0
ans="YES"
while i<len(s):
if a in s[i:i+5]:
i+=5
elif b in s[i:i+7]:
i+=7
elif c in s[i:i+5]:
i+=5
elif d in s[i:i+6]:
i+=6
else :
ans="NO"
break
print(ans) |
s499205913 | p03095 | u319818856 | 2,000 | 1,048,576 | Wrong Answer | 40 | 3,188 | 587 | You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings. Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order. | def colorful_subsequence(N: int, S: str)->int:
MOD = (10 ** 9) + 7
cum = {}
for c in S:
cum.setdefault(c, 0)
cum[c] += 1
values = list(cum.values())
count = 0
# for i, v in enumerate(values):
for v in values:
# temp = v
# for v in values[i+1:]:
# temp = (temp * (1 + v)) % MOD
# count = (count + temp) % MOD
count *= (1 + v)
count %= MOD
return count
if __name__ == "__main__":
N = int(input())
S = input()
ans = colorful_subsequence(N, S)
print(ans)
| s724150930 | Accepted | 35 | 3,188 | 409 | def colorful_subsequence(N: int, S: str)->int:
MOD = (10 ** 9) + 7
cum = {}
for c in S:
cum.setdefault(c, 0)
cum[c] += 1
values = list(cum.values())
count = 1
for v in values:
count *= (1 + v)
count %= MOD
return (count-1) % MOD
if __name__ == "__main__":
N = int(input())
S = input()
ans = colorful_subsequence(N, S)
print(ans)
|
s317690769 | p03997 | u706414019 | 2,000 | 262,144 | Wrong Answer | 24 | 8,924 | 49 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. | print((int(input())+int(input()))/2*int(input())) | s822720657 | Accepted | 27 | 9,088 | 54 | print(int((int(input())+int(input()))/2*int(input()))) |
s593928653 | p03547 | u076894102 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 114 | In programming, hexadecimal notation is often used. In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively. In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`. When X and Y are seen as hexadecimal numbers, which is larger? | a , b = input().split()
if ord(a)<ord(b):
print("/")
elif ord(a)>ord(b):
print(">")
else :
print("=") | s866935299 | Accepted | 17 | 2,940 | 114 | a , b = input().split()
if ord(a)<ord(b):
print("<")
elif ord(a)>ord(b):
print(">")
else :
print("=") |
s255000343 | p02613 | u658915215 | 2,000 | 1,048,576 | Wrong Answer | 145 | 9,184 | 155 | 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())
L = {'AC':0, 'WA':0, 'TLE':0, 'RE':0}
for i in range(n):
s = input()
L[s] += 1
for i in L:
print('{} × {}'.format(i, L[i])) | s450711208 | Accepted | 151 | 9,132 | 149 | n = int(input())
L = {'AC':0, 'WA':0, 'TLE':0, 'RE':0}
for i in range(n):
s = input()
L[s] += 1
for l in L.keys():
print(l, 'x', L[l]) |
s694749017 | p03997 | u289162337 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 49 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. | print((int(input())+int(input()))*int(input())/2) | s074512208 | Accepted | 17 | 2,940 | 54 | print(int((int(input())+int(input()))*int(input())/2)) |
s911831645 | p03457 | u635958201 | 2,000 | 262,144 | Wrong Answer | 401 | 3,060 | 225 | 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())
count = 0
for i in range(N):
K = list(map(int,input().split()))
if K[0]>=K[1]+K[2] and K[0]%2==(K[1]+K[2])%2:
count += 1
#print(count)
if count == N:
print('YES')
else:
print('NO')
| s030743039 | Accepted | 390 | 3,060 | 224 | N = int(input())
count = 0
for i in range(N):
K = list(map(int,input().split()))
if K[0]>=K[1]+K[2] and K[0]%2==(K[1]+K[2])%2:
count += 1
#print(count)
if count == N:
print('Yes')
else:
print('No') |
s593764555 | p03080 | u551692187 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 93 | There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat. | N = int(input())
s = input().split()
print('Yes') if s.count('R')*2 > len(s) else print('No') | s327795247 | Accepted | 17 | 2,940 | 87 | N = int(input())
s = input()
print('Yes') if s.count('R')>s.count('B') else print('No') |
s075430443 | p04000 | u994988729 | 3,000 | 262,144 | Wrong Answer | 3,169 | 316,572 | 854 | 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
import sys
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 7)
H, W, N = map(int, input().split())
xy = []
dot = defaultdict(bool)
seen = defaultdict(bool)
for _ in range(N):
a, b = map(int, input().split())
xy.append((a, b))
dot[(a, b)] = True
ans = [0] * 10
for x, y in xy:
for dx in [-2, -1, 0]:
for dy in [-2, -1, 0]:
nx = x + dx
ny = y + dy
if seen[(nx, ny)] or nx <= 0 or nx + 2 > H or ny <= 0 or ny + 2 > W:
continue
seen[(nx, ny)] = True
cnt = 0
for i in range(3):
for j in range(3):
cnt += int(dot[(nx + i, ny + j)])
ans[cnt] += 1
print(nx, ny, cnt)
zero = (H - 2) * (W - 2) - sum(ans)
ans[0] = zero
print(*ans, sep="\n") | s617305347 | Accepted | 1,528 | 166,872 | 521 | from collections import defaultdict, Counter
import sys
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 7)
H, W, N = map(int, input().split())
d = defaultdict(int)
for _ in range(N):
x, y = map(int, input().split())
for i in range(-2, 1):
if 1 <= x + i <= H - 2:
for j in range(-2, 1):
if 1 <= y + j <= W - 2:
d[(x + i, y + j)] += 1
c = Counter(d.values())
zero = (H - 2) * (W - 2) - len(d)
print(zero)
for i in range(1, 10):
print(c[i]) |
s910890709 | p02401 | u327972099 | 1,000 | 131,072 | Wrong Answer | 30 | 6,740 | 77 | 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. | while True:
a = input()
if "?" in a:
break
print(eval(a)) | s311790184 | Accepted | 40 | 6,724 | 96 | while True:
a = input()
if "?" in a:
break
print(eval(a.replace("/", "//"))) |
s144366081 | p03501 | u586577600 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 82 | You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours. | n, a, b = map(int, input().split())
if n*a > b:
print(n*a)
else:
print(b)
| s358718848 | Accepted | 18 | 2,940 | 82 | n, a, b = map(int, input().split())
if n*a < b:
print(n*a)
else:
print(b)
|
s844954357 | p03369 | u698916859 | 2,000 | 262,144 | Wrong Answer | 28 | 9,020 | 78 | In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen. |
s = input()
count = s.count("1")
print(700 + 100 * count) | s804445547 | Accepted | 23 | 9,028 | 78 |
s = input()
count = s.count("o")
print(700 + 100 * count) |
s654036639 | p02259 | u045830275 | 1,000 | 131,072 | Wrong Answer | 20 | 7,804 | 598 | Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode. | def main() :
n = int(input())
nums = [int(i) for i in input().split()]
flag = True
count = 0
i = 0 # ?????????????????????????????????????????????
while flag :
flag = False
for i in reversed(range(i, n)) :
if nums[i-1] > nums[i] :
nums[i-1], nums[i] = nums[i], nums[i-1]
count += 1
flag = True
i += 1
nums_str = [str(i) for i in nums]
print(" ".join(nums_str))
print(count)
if __name__ == '__main__' :
main() | s498736097 | Accepted | 30 | 7,804 | 501 | def main() :
n = int(input())
nums = [int(i) for i in input().split()]
flag = True
count = 0
index = 0
while flag :
flag = False
for i in reversed(range(index+1, n)) :
if nums[i-1] > nums[i] :
nums[i-1], nums[i] = nums[i], nums[i-1]
count += 1
flag = True
index += 1
nums_str = [str(i) for i in nums]
print(" ".join(nums_str))
print(count)
if __name__ == '__main__' :
main() |
s556495938 | p03470 | u952656646 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 78 | An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have? | N = int(input())
d = set()
for i in range(N):
d.add(int(input()))
print(d) | s405768750 | Accepted | 17 | 2,940 | 83 | N = int(input())
d = set()
for i in range(N):
d.add(int(input()))
print(len(d)) |
s444582771 | p02742 | u079182025 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 134 | 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 == 0 or w == 0:
print(1)
elif (h * w)%2 == 0:
print(h * w / 2)
else:
print((h*w+1)/2) | s356651229 | Accepted | 17 | 3,060 | 149 | h, w = map(int, input().split())
if h == 1 or w == 1:
print(int(1))
elif (h * w)%2 == 0:
print(int(h * w / 2))
else:
print(int((h*w+1)/2)) |
s404821336 | p02831 | u780565479 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 104 | Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be divided and distributed to multiple guests. | N,M = map(int,input().split())
n,m=N,M
if n<m:
t=n;n=m;m=t
while n%m:
t=n%m;n=m;m=t
print(N*M/m) | s137877333 | Accepted | 18 | 3,060 | 109 | N,M = map(int,input().split())
n,m=N,M
if n<m:
t=n;n=m;m=t
while n%m:
t=n%m;n=m;m=t
print(int(N*M/m)) |
s352938907 | p03370 | u870297120 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 105 | 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 = sorted(list(int(input()) for _ in range(n)))
print((x+sum(m))//m[0]) | s337542024 | Accepted | 17 | 2,940 | 112 | n,x = map(int, input().split())
m = sorted(list(int(input()) for _ in range(n)))
print((x-sum(m))//m[0]+len(m)) |
s940669111 | p03048 | u970809473 | 2,000 | 1,048,576 | Wrong Answer | 1,497 | 3,060 | 211 | Snuke has come to a store that sells boxes containing balls. The store sells the following three kinds of boxes: * Red boxes, each containing R red balls * Green boxes, each containing G green balls * Blue boxes, each containing B blue balls Snuke wants to get a total of exactly N balls by buying r red boxes, g green boxes and b blue boxes. How many triples of non-negative integers (r,g,b) achieve this? | r,g,b,n = map(int, input().split())
res = 0
for i in range(int(n/r) + 1):
for j in range(int((n - i * r) / g) + 1):
if (n - i * r - j * g) % b == 0 or n - i * r - j * g == n:
res = res + 1
print(res) | s694008273 | Accepted | 1,418 | 3,060 | 211 | r,g,b,n = map(int, input().split())
res = 0
for i in range(int(n/r) + 1):
for j in range(int((n - i * r) / g) + 1):
if (n - i * r - j * g) % b == 0 or n - i * r - j * g == 0:
res = res + 1
print(res) |
s996620071 | p03712 | u553070631 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 137 | 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()))
print('#'*(w+2))
a=[]
for i in range(h):
a='#'
a+=input()
a+='#'
print(a)
print('#'*w) | s710114123 | Accepted | 18 | 3,060 | 141 | h,w=list(map(int,input().split()))
print('#'*(w+2))
a=[]
for i in range(h):
a='#'
a+=input()
a+='#'
print(a)
print('#'*(w+2)) |
s023566154 | p03853 | u016323272 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 100 | There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down). | #ABC049.B
H,W = map(int,input().split())
for _ in range (H):
ans = input()
print(ans)
print(ans) | s925918105 | Accepted | 18 | 3,060 | 107 |
H,W = map(int,input().split())
for _ in range(H):
S =input()
print(S)
print(S) |
s641939783 | p03457 | u298297089 | 2,000 | 262,144 | Wrong Answer | 409 | 3,064 | 301 | 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())
manhat = lambda x,y:abs(x[0]-y[0])+abs(x[1]-y[1])
flag = True
t_,x_,y_ = 0,0,0
for i in range(N):
t,x,y = map(int, input().split())
dt = t-t_
dx = manhat((x,y), (x_,y_))
if dt < dx or (dt - dx) % 2:
flag = False
break
t_,x_,y_ = t,x,y
print('YES' if flag else 'NO') | s784883257 | Accepted | 409 | 3,064 | 301 | N = int(input())
manhat = lambda x,y:abs(x[0]-y[0])+abs(x[1]-y[1])
flag = True
t_,x_,y_ = 0,0,0
for i in range(N):
t,x,y = map(int, input().split())
dt = t-t_
dx = manhat((x,y), (x_,y_))
if dt < dx or (dt - dx) % 2:
flag = False
break
t_,x_,y_ = t,x,y
print('Yes' if flag else 'No') |
s988803913 | p03095 | u295656477 | 2,000 | 1,048,576 | Wrong Answer | 142 | 3,188 | 380 | You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings. Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order. | # coding:utf-8
n = int(input())
s = input()
InWord={i:False for i in 'abcdefghijklmnopqrstuvwxyz'}
answer = 0
marker = 1
length = 1
InWord[s[0]]=True
for i in range(n):
for j in range(marker,n):
if InWord[s[j]] or j==n-1:
length = j-i+1
answer += length
marker = j
InWord[s[i]] = False
break
InWord[s[j]] = True
print(answer%(10**9+7)) | s872568293 | Accepted | 19 | 3,188 | 229 | # coding:utf-8
n = int(input())
s = input()
LetterCount = {i:s.count(i) for i in 'abcdefghijklmnopqrstuvwxyz'}
answer = 1
for i in 'abcdefghijklmnopqrstuvwxyz':
answer = answer * (LetterCount[i]+1) % (10**9+7)
print(answer-1) |
s963705782 | p03477 | u214434454 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 142 | A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right. | a, b, c, d = map(int, input().split())
if a + b > c + d:
print("left")
elif a + b == c + d:
print("Balanced")
else:
print("Right") | s300604893 | Accepted | 17 | 2,940 | 142 | a, b, c, d = map(int, input().split())
if a + b > c + d:
print("Left")
elif a + b == c + d:
print("Balanced")
else:
print("Right") |
s520856371 | p03474 | u593567568 | 2,000 | 262,144 | Wrong Answer | 27 | 9,132 | 295 | The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom. | A,B = map(int,input().split())
S = list(input())
ok = True
for i in range(A+B+1):
s = S[i]
if i == A and s == '-':
continue
else:
ok = False
break
if i != A and s != '-':
continue
else:
ok = False
break
if ok:
print("Yes")
else:
print("No")
| s238198910 | Accepted | 28 | 9,136 | 254 | A,B = map(int,input().split())
S = list(input())
ok = True
for i in range(A+B+1):
s = S[i]
if i == A and s != '-':
ok = False
break
if i != A and s == '-':
ok = False
break
if ok:
print("Yes")
else:
print("No")
|
s767583650 | p03478 | u399721252 | 2,000 | 262,144 | Wrong Answer | 36 | 3,060 | 154 | 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). | n, a, b = [ int(v) for v in input().split() ]
ans = 0
for i in range(1,n+1):
if a <= sum([ int(v) for v in list(str(i)) ]) <= b:
ans += 1
print(ans) | s000010940 | Accepted | 37 | 2,940 | 154 | n, a, b = [ int(v) for v in input().split() ]
ans = 0
for i in range(1,n+1):
if a <= sum([ int(v) for v in list(str(i)) ]) <= b:
ans += i
print(ans) |
s520023369 | p03494 | u102461423 | 2,000 | 262,144 | Wrong Answer | 148 | 12,396 | 136 | 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. | import numpy as np
x = np.array(input().split(),dtype=np.int32)
answer = 0
while (x%2==0).all():
x //= 2
answer += 1
print(answer) | s563943974 | Accepted | 150 | 12,504 | 143 | import numpy as np
input()
x = np.array(input().split(),dtype=np.int32)
answer = 0
while (x%2==0).all():
x //= 2
answer += 1
print(answer) |
s445718401 | p03698 | u527993431 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 66 | You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different. | S=input()
if len(S)==len(set(S)):
print("Yes")
else:
print("No") | s953327671 | Accepted | 18 | 2,940 | 66 | S=input()
if len(S)==len(set(S)):
print("yes")
else:
print("no") |
s834557210 | p00741 | u672443148 | 1,000 | 131,072 | Wrong Answer | 110 | 5,624 | 1,058 | You are given a marine area map that is a mesh of squares, each representing either a land or sea area. Figure B-1 is an example of a map. Figure B-1: A marine area map You can walk from a square land area to another if they are horizontally, vertically, or diagonally adjacent to each other on the map. Two areas are on the same island if and only if you can walk from one to the other possibly through other land areas. The marine area on the map is surrounded by the sea and therefore you cannot go outside of the area on foot. You are requested to write a program that reads the map and counts the number of islands on it. For instance, the map in Figure B-1 includes three islands. | def dfs(w,h,W,H,ID,islands,islandsID):
if islands[h][w]==1 and islandsID[h][w]!=0:
for i in range(-1,2):
for j in range(-1,2):
dw=w+i
dh=h+j
if dw>=0 and dw<W and dh>=0 and dh<H:
if islands[dh][dw]==1:
islandsID[dh][dw]=islandsID[h][w]
if islands[h][w]==1 and islandsID[h][w]==0:
for i in range(-1,2):
for j in range(-1,2):
dw=w+i
dh=h+j
if dw>=0 and dw<W and dh>=0 and dh<H:
if islands[dh][dw]==1:
islandsID[dh][dw]=ID
ID+=1
return ID
while True:
islands=[]
islandsID=[]
ID=1
W,H=map(int,input().split())
if W==0:
break
for i in range(H):
islandsID.append([0 for j in range(W)])
for _ in range(H):
islands.append(list(map(int,input().split())))
for h in range(H):
for w in range(W):
ID=dfs(w,h,W,H,ID,islands,islandsID)
print(ID-1)
| s613369229 | Accepted | 3,190 | 10,428 | 1,101 | while 1:
W, H = list(map(int,input().split()))
if W == 0:break
C = [list(map(int, input().split())) for i in range(H) ]
def work(x,y):
ax = [-1, 0, 1, -1, 1, -1, 0, 1]
ay = [-1, -1, -1, 0, 0, 1, 1, 1]
if C[y][x] != 1:return False
stack = [(x,y)]
while len(stack) > 0:
s = stack.pop(-1)
C[s[1]][s[0]] = -1 #mark
for _x in [s[0] + _x for _x in ax]:
for _y in [s[1] + _y for _y in ay]:
if 0 <= _x < W and 0 <= _y < H and C[_y][_x] == 1:
stack.append((_x, _y))
return True
# ax = [-1, 0, 1, -1, 1, -1, 0, 1]
#
# if C[y][x] != 1:return False
# C[y][x] = -1 #mark
#
# for _ax in ax:
# for _ay in ay:
# if 0 <= _ax + x < W and 0 <= _ay + y < H:
# return True
cnt = 0
for x in range(W):
for y in range(H):
cnt += work(x,y)
print(cnt)
|
s988519741 | p03574 | u162612857 | 2,000 | 262,144 | Wrong Answer | 31 | 3,444 | 514 | You are given an H × W grid. The squares in the grid are described by H strings, S_1,...,S_H. The j-th character in the string S_i corresponds to the square at the i-th row from the top and j-th column from the left (1 \leq i \leq H,1 \leq j \leq W). `.` stands for an empty square, and `#` stands for a square containing a bomb. Dolphin is interested in how many bomb squares are horizontally, vertically or diagonally adjacent to each empty square. (Below, we will simply say "adjacent" for this meaning. For each square, there are at most eight adjacent squares.) He decides to replace each `.` in our H strings with a digit that represents the number of bomb squares adjacent to the corresponding empty square. Print the strings after the process. | h, w=[int(i) for i in input().split()]
field = []
for i in range(h):
field.append( input())
for i in range(h):
for j in range(w):
if field[i][j] == ".":
hmin = max(i-1, 0)
hmax = min(i+1, h-1)
wmin = max(j-1, 0)
wmax = min(j+1, w-1)
count = 0
for m in range(hmin, hmax+1):
for n in range(wmin, wmax+1):
if field[m][n] == "#":
count+=1
print (count)
# field[i][j] = count
field[i] = field[i][0:j]+ str(count) + field[i][j+1:]
for i in range(h):
print (field[i])
| s959660116 | Accepted | 30 | 3,064 | 515 | h, w=[int(i) for i in input().split()]
field = []
for i in range(h):
field.append( input())
for i in range(h):
for j in range(w):
if field[i][j] == ".":
hmin = max(i-1, 0)
hmax = min(i+1, h-1)
wmin = max(j-1, 0)
wmax = min(j+1, w-1)
count = 0
for m in range(hmin, hmax+1):
for n in range(wmin, wmax+1):
if field[m][n] == "#":
count+=1
# print (count)
# field[i][j] = count
field[i] = field[i][0:j]+ str(count) + field[i][j+1:]
for i in range(h):
print (field[i])
|
s591369068 | p03575 | u442877951 | 2,000 | 262,144 | Wrong Answer | 19 | 3,064 | 767 | You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges. | N,M = map(int,input().split())
road = [[] for _ in range(N)]
for _ in range(M):
a,b = map(int,input().split())
road[a-1].append(b-1)
road[b-1].append(a-1)
def dfs(v,w):
l = [0 for _ in range(N)]
l[v] = 1
stack = [v]
while stack:
vv = stack.pop()
for i in road[vv]:
if i == w and v == vv or i == vv and v == w:
continue
else:
if l[i] == 0:
l[i] = 1
stack.append(i)
return sum(l)
ans = 0
check = [[0 for _ in range(N)] for _ in range(N)]
for n in range(N):
for m in range(len(road[n])):
if check[n][road[n][m]] == 0 or check[road[n][m]][n] == 0:
check[n][road[n][m]] = 1
check[road[n][m]][n] = 1
d = dfs(n,road[n][m])
if d != N:
ans += 1
print(ans,road) | s754715116 | Accepted | 19 | 3,064 | 762 | N,M = map(int,input().split())
road = [[] for _ in range(N)]
for _ in range(M):
a,b = map(int,input().split())
road[a-1].append(b-1)
road[b-1].append(a-1)
def dfs(v,w):
l = [0 for _ in range(N)]
l[v] = 1
stack = [v]
while stack:
vv = stack.pop()
for i in road[vv]:
if i == w and v == vv or i == vv and v == w:
continue
else:
if l[i] == 0:
l[i] = 1
stack.append(i)
return sum(l)
ans = 0
check = [[0 for _ in range(N)] for _ in range(N)]
for n in range(N):
for m in range(len(road[n])):
if check[n][road[n][m]] == 0 or check[road[n][m]][n] == 0:
check[n][road[n][m]] = 1
check[road[n][m]][n] = 1
d = dfs(n,road[n][m])
if d != N:
ans += 1
print(ans) |
s611271785 | p03777 | u580316060 | 2,000 | 262,144 | Time Limit Exceeded | 2,104 | 28,268 | 201 | Two deer, AtCoDeer and TopCoDeer, are playing a game called _Honest or Dishonest_. In this game, an honest player always tells the truth, and an dishonest player always tell lies. You are given two characters a and b as the input. Each of them is either `H` or `D`, and carries the following information: If a=`H`, AtCoDeer is honest; if a=`D`, AtCoDeer is dishonest. If b=`H`, AtCoDeer is saying that TopCoDeer is honest; if b=`D`, AtCoDeer is saying that TopCoDeer is dishonest. Given this information, determine whether TopCoDeer is honest. | X = map(int,input().split())
#X=7
P = {0}
i=1
while True:
Q = {0}
for p in P:
Q.add(p+i)
Q.add(p-i)
P = P.union(Q)
if X in P:
print(i)
break
i = i+1
| s370827877 | Accepted | 17 | 2,940 | 71 | a,b = input().split()
if a == b:
print('H')
else:
print('D')
|
s031901170 | p03962 | u418260963 | 2,000 | 262,144 | Wrong Answer | 22 | 3,064 | 143 | AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him. | count = 0
a, b, c = map(int, input().split())
if a == b:
count += 1
if b == c:
count += 1
if b == c:
count += 1
print(count)
| s459029203 | Accepted | 23 | 3,064 | 166 | count = 3
a, b, c = map(int, input().split())
if a == b:
count -= 1
if a == c:
count -= 1
if b == c:
count -= 1
if count == 0:
count += 1
print(count) |
s235760218 | p03379 | u743281086 | 2,000 | 262,144 | Wrong Answer | 147 | 26,772 | 180 | When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N. | n = int(input())
x = map(int, input().split())
temp = sorted(x)
idx = n//2
y = temp[idx-1]
z = temp[idx]
for item in x:
if item < y:
print(y)
else:
print(z) | s522363617 | Accepted | 285 | 25,556 | 187 | n = int(input())
x = list(map(int, input().split()))
temp = sorted(x)
idx = n//2
y = temp[idx-1]
z = temp[idx]
for item in x:
if item >= z:
print(y)
else:
print(z) |
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