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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s967806993 | p03457 | u962819039 | 2,000 | 262,144 | Wrong Answer | 1,014 | 4,340 | 329 | 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())
x = 0
y = 0
t = 0
for i in range(n):
t_new, x_new, y_new = map(int, input().split())
print(t_new, x_new, y_new)
if ((abs(x_new - x) + abs(y_new - y)) <= (t_new - t)) and ((abs(x_new - x) + abs(y_new - y) - (t_new - t)) % 2 == 0):
x = x_new
y = y_new
t = t_new
else:
print('No')
exit()
print('YES')
| s242614677 | Accepted | 376 | 3,060 | 292 | n = int(input())
x = 0
y = 0
t = 0
for i in range(n):
t_new, x_new, y_new = map(int, input().split())
l = abs(x_new - x) + abs(y_new - y)
dt = t_new - t
if (l <= dt) and (l % 2 == dt % 2 ):
x = x_new
y = y_new
t = t_new
else:
#print(t, x, y)
print('No')
exit()
print('Yes')
|
s260364581 | p03575 | u817328209 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 853 | 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())
gragh = [[] for _ in range(n)]
used_v = [False]*n
used_e = [[False]*n for _ in range(n)]
Ord = [0]*n
lowlink = [0]*n
k = 0
for i in range(m) :
a, b = map(int, input().split())
gragh[a-1].append(b-1)
gragh[b-1].append(a-1)
def dfs(v) :
global k
used_v[v] = True
Ord[v] = lowlink[v] = k
k += 1
for node in gragh[v] :
if not used_v[node] :
used_e[v][node] = True
dfs(node)
lowlink[v] = min(lowlink[v], lowlink[node])
elif not used_e[node][v] :
lowlink[v] = min(lowlink[v], Ord[node])
dfs(0)
ans = 0
used_e = [[False]*n for _ in range(n)]
for i in range(n) :
for j in gragh[i] :
if not used_e[i][j] and Ord[i] < lowlink[j] :
ans += 1
used_e[i][j] = True
used_e[j][i] = True
print(ans) | s418201589 | Accepted | 22 | 3,316 | 610 | from collections import defaultdict
def dfs(i, used, g, ng):
for n in g[i]:
if n not in used and (i, n) != ng and (n, i) != ng:
used.add(n)
dfs(n, used, g, ng)
def main():
N, M = map(int, input().split())
g = defaultdict(list)
k = []
for _ in range(M):
a, b = map(int, input().split())
g[a].append(b)
g[b].append(a)
k.append((a, b))
ans = 0
for a, b in k:
used = set()
dfs(1, used, g, (a, b))
if len(used) != N:
ans += 1
print(ans)
if __name__ == '__main__':
main() |
s779912813 | p03993 | u193264896 | 2,000 | 262,144 | Wrong Answer | 68 | 12,392 | 417 | There are N rabbits, numbered 1 through N. The i-th (1≤i≤N) rabbit likes rabbit a_i. Note that no rabbit can like itself, that is, a_i≠i. For a pair of rabbits i and j (i＜j), we call the pair (i，j) a _friendly pair_ if the following condition is met. * Rabbit i likes rabbit j and rabbit j likes rabbit i. Calculate the number of the friendly pairs. | import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 8)
INF = float('inf')
MOD = 10 ** 9 + 7
def main():
N = int(readline())
A = list(map(int, readline().split()))
ans = 0
print(A)
for i in range(N):
if i==A[A[i]-1]-1:
ans += 1
print(ans//2)
if __name__ == '__main__':
main() | s544369223 | Accepted | 58 | 12,392 | 405 | import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 8)
INF = float('inf')
MOD = 10 ** 9 + 7
def main():
N = int(readline())
A = list(map(int, readline().split()))
ans = 0
for i in range(N):
if i==A[A[i]-1]-1:
ans += 1
print(ans//2)
if __name__ == '__main__':
main()
|
s622709753 | p03456 | u301624971 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 305 | AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number. | def myAnswer(a:int,b:int) -> str:
ab=int(str(a)+str(b))
number = 2
while number**2 < ab:
if(number**2 ==ab): return "Yes"
number += 1
return "No"
def modelAnswer():
tmp=1
def main():
a,b=map(int,input().split())
print(myAnswer(a,b))
if __name__ == '__main__':
main() | s744837555 | Accepted | 19 | 3,060 | 335 | def myAnswer(a:int,b:int) -> str:
ab=int(str(a)+str(b))
number = 2
while True:
if(number**2 == ab): return "Yes"
if(number**2 > ab): break
number += 1
return "No"
def modelAnswer():
tmp=1
def main():
a,b=map(int,input().split())
print(myAnswer(a,b))
if __name__ == '__main__':
main() |
s101053389 | p03407 | u019578976 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 57 | An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan. | a,b,c=map(int,input().split())
print(["No","Yes"][a+b<c]) | s504812327 | Accepted | 19 | 2,940 | 57 | a,b,c=map(int,input().split())
print(["Yes","No"][a+b<c]) |
s886562819 | p02690 | u539692012 | 2,000 | 1,048,576 | Wrong Answer | 44 | 10,704 | 81 | Give a pair of integers (A, B) such that A^5-B^5 = X. It is guaranteed that there exists such a pair for the given integer X. | x=int(input())
r=range(200)
[i**5-x-j**5or exit(print(i,j))for i in r for j in r] | s258269677 | Accepted | 40 | 9,088 | 67 | i=j=x=int(input())
while i**5-j**5-x:i=-~i%127;j=j%257-i
print(i,j) |
s050534647 | p03150 | u388497015 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,116 | 254 | A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string. | import sys
s = input()
target = "keyence"
for i in range(len(target)):
sbst1 = target[0:i]
sbst2 = target[i:]
ind = s.find(sbst1)
indx = s.find(sbst2)
if ind != -1 and ind < indx:
print("Yes")
sys.exit()
print("No") | s286414625 | Accepted | 30 | 9,072 | 258 | import sys
s = input()
target = "keyence"
if s == target:
print("YES")
sys.exit()
for i in range(len(s)):
for j in range(i, len(s)):
st = s[:i] + s[j:]
if st == target:
print("YES")
sys.exit()
print("NO") |
s424771810 | p02612 | u031679985 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,088 | 28 | 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) | s374046422 | Accepted | 27 | 9,160 | 65 | n=int(input())
a=n%1000
if a==0:
print(0)
else:
print(1000-a) |
s392986630 | p02390 | u455935767 | 1,000 | 131,072 | Wrong Answer | 20 | 5,596 | 104 | Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively. | Sec = input()
S = int(Sec)
h = S/60
m = S%60/60
s = S%60%60
st=(str(h)+':'+str(m)+':'+str(s))
print(st)
| s027390583 | Accepted | 30 | 5,600 | 175 | Sec = input()
S = int(Sec)
h = int(S/3600)
m = int(S%3600/60)
s = S%3600%60
st=(str(h)+':'+str(m)+':'+str(s))
print(st)
#1h=3600s
#1m=60s
#1s=1s
'''
8000s
h=2
8000%60=20
'''
|
s139692793 | p03457 | u126823513 | 2,000 | 262,144 | Wrong Answer | 476 | 32,800 | 523 | 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. | int_n = int(input())
result = False
plan_array = [[0, 0, 0]]
for n in range(int_n):
plan_array.append(list(map(int, input().split())))
print(plan_array)
for n in range(int_n):
dt = plan_array[n+1][0] - plan_array[n][0]
dist = plan_array[n+1][1] - plan_array[n][1] + \
plan_array[n+1][2] - plan_array[n][2]
if dist > dt:
result = False
break
if dt & 1 and dist & 1:
result = True
else:
result = False
if result:
print('YES')
else:
print('No')
| s348971750 | Accepted | 274 | 27,124 | 652 | n = int(input())
st_t = 0
st_x = 0
st_y = 0
li_in = list()
for i in range(n):
li_in.append(list(map(int, input().split())))
for t, x, y in li_in:
time = t - st_t
diff_x = abs(x - st_x)
diff_y = abs(y - st_y)
if diff_x + diff_y > time:
print('No')
exit(0)
elif diff_x + diff_y == time:
st_t = t
st_x = x
st_y = y
continue
else:
if (time - (diff_x + diff_y)) % 2 == 0:
st_t = t
st_x = x
st_y = y
continue
else:
print('No')
exit(0)
else:
print('Yes')
|
s064205744 | p03719 | u912862653 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 93 | 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 = list(map(int, input().split()))
if a<=c and c<=b:
print('YES')
else:
print('NO') | s388621885 | Accepted | 18 | 2,940 | 93 | a, b, c = list(map(int, input().split()))
if c>=a and c<=b:
print('Yes')
else:
print('No') |
s913296572 | p03478 | u092278825 | 2,000 | 262,144 | Wrong Answer | 37 | 2,940 | 185 | 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(i) for i in input().split())
s = 0
for i in range(1, N+1):
list_1 = [int(x) for x in list(str(i))]
k = sum(list_1)
if k >= A and k <= B:
s += 1
print(s) | s886673576 | Accepted | 36 | 2,940 | 177 | N,A,B = map(int, input().split())
s = 0
for i in range(1, N+1):
list_1 = [int(x) for x in list(str(i))]
k = sum(list_1)
if k >= A and k <= B:
s += i
print(s) |
s863239657 | p03150 | u374103100 | 2,000 | 1,048,576 | Wrong Answer | 19 | 3,188 | 450 | A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string. |
import re
S = input()
if "keyence" in S:
print("YES")
else:
patterns = ["k*eyence", "ke*yence", "key*ence", "keye*nce", "keyen*ce", "keyenc*e"]
matched = False
for pattern in patterns:
matchOB = re.search(pattern, S)
if matchOB:
matched = True
break
if matched:
print("YES")
else:
print("NO")
| s502215140 | Accepted | 19 | 3,188 | 499 |
import re
S = input()
if "keyence" == S:
print("YES")
else:
patterns = ["^keyence(.*)", "^k(.*)eyence$", "^ke(.*)yence$", "^key(.*)ence$", "^keye(.*)nce$", "^keyen(.*)ce$", "^keyenc(.*)e$", "keyence$"]
matched = False
for pattern in patterns:
matchOB = re.search(pattern, S)
if matchOB:
matched = True
break
if matched:
print("YES")
else:
print("NO")
|
s764232230 | p03693 | u332906195 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 94 | AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4? | r, g, b = map(int, input().split())
print("Yes" if (r * 100 + g * 10 + b) % 4 == 0 else "No")
| s463088789 | Accepted | 17 | 2,940 | 76 | r, g, b = input().split()
print("YES" if int(r + g + b) % 4 == 0 else "NO")
|
s422815951 | p03409 | u918009437 | 2,000 | 262,144 | Wrong Answer | 19 | 3,064 | 623 | On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i). A red point and a blue point can form a _friendly pair_ when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point. At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs. | if __name__ == '__main__':
N = int(input())
ab_list = [list(map(int, input().split())) for i in range(N)]
cd_list = [list(map(int, input().split())) for i in range(N)]
ab_list = sorted(ab_list, key=lambda x : x[1], reverse=True)
cd_list = sorted(cd_list, key=lambda x : x[0])
print(ab_list)
print(cd_list)
answer = 0
for b in cd_list:
for r in ab_list:
if b[0]>r[0] and b[1]>r[1]:
answer += 1
ab_list.remove(r)
print(b)
print(r)
print("------")
break
print(answer) | s861734117 | Accepted | 19 | 3,064 | 502 | if __name__ == '__main__':
N = int(input())
ab_list = [list(map(int, input().split())) for i in range(N)]
cd_list = [list(map(int, input().split())) for i in range(N)]
ab_list = sorted(ab_list, key=lambda x : x[1], reverse=True)
cd_list = sorted(cd_list, key=lambda x : x[0])
answer = 0
for b in cd_list:
for r in ab_list:
if b[0]>r[0] and b[1]>r[1]:
answer += 1
ab_list.remove(r)
break
print(answer) |
s388356803 | p03574 | u854144714 | 2,000 | 262,144 | Wrong Answer | 36 | 3,444 | 587 | 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()]
A=[]
for i in range(H):
s=input()
S=list(s)
A.append(S)
for i in range(H):
for j in range(W):
if A[i][j]=="#":
break
else:
s=0
for k in range(-1,2):
for l in range(-1,2):
if i+k>=0 and i+k<=H-1 and j+l>=0 and j+l<=W-1:
if A[i+k][j+l]=="#":
s+=1
A[i][j]=s
for i in range(H):
for j in range(W):
print(A[i][j],end=" ")
print() | s339404537 | Accepted | 35 | 3,572 | 625 | H,W=[int(i) for i in input().split()]
A=[]
for i in range(H):
s=input()
S=list(s)
A.append(S)
for i in range(H):
for j in range(W):
if A[i][j]!="#":
s=0
for k in range(-1,2):
for l in range(-1,2):
if k==0 and l==0:
continue
if i+k>=0 and i+k<=H-1 and j+l>=0 and j+l<=W-1:
if A[i+k][j+l]=="#":
s+=1
A[i][j]=str(s)
for i in range(H):
for j in range(W):
print(A[i][j],end="")
print()
|
s435894323 | p03944 | u539969758 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 429 | There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting. | w, h, n = map(int,input().split())
x_l = 0
x_r = w
y_l = 0
y_h = h
for i in range(n):
x, y, a = map(int,input().split())
if a == 1:
if x > x_l:
x_l = x
elif a == 2:
if x < x_r:
x_r == x
elif a == 3:
if y > y_l:
y_l = y
else:
if y < y_h:
y_h = y
menseki = 0
if x_r <= x_l and y_l <= y_h:
menseki = (y_h - y_l)*(x_l - x_r)
print(menseki) | s712305796 | Accepted | 17 | 3,064 | 429 | w, h, n = map(int,input().split())
x_l = 0
x_r = w
y_l = 0
y_h = h
for i in range(n):
x, y, a = map(int,input().split())
if a == 1:
if x > x_l:
x_l = x
elif a == 2:
if x < x_r:
x_r = x
elif a == 3:
if y > y_l:
y_l = y
else:
if y < y_h:
y_h = y
menseki = 0
if x_l <= x_r and y_l <= y_h:
menseki = (y_h - y_l)*(x_r - x_l)
print(menseki)
|
s460945112 | p03681 | u859897687 | 2,000 | 262,144 | Wrong Answer | 2,104 | 3,468 | 151 | Snuke has N dogs and M monkeys. He wants them to line up in a row. As a Japanese saying goes, these dogs and monkeys are on bad terms. _("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.)_ Snuke is trying to reconsile them, by arranging the animals so that there are neither two adjacent dogs nor two adjacent monkeys. How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that). Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished. | a,b=map(int,input().split())
ans=1
if abs(a-b)>1:
print(0)
elif abs(a-b)==1:
ans=2
while a>1:
ans*=a
a-=1
while b>1:
ans*=b
b-=1
print(ans) | s341308914 | Accepted | 77 | 3,060 | 214 | a,b=map(int,input().split())
mod=1000000007
ans=1
if abs(a-b)>1:
print(0)
else:
if abs(a-b)==0:
ans=2
while a>1:
ans*=a
ans%=mod
a-=1
while b>1:
ans*=b
ans%=mod
b-=1
print(ans) |
s512112970 | p03637 | u581187895 | 2,000 | 262,144 | Wrong Answer | 66 | 14,224 | 319 | We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective. | N = int(input())
A = list(map(int, input().split()))
k0 = 0
k2 = 0
k4 = 0
for a in A:
if a%4==0:
k4 += 1
elif a%2==0:
k2 += 1
else:
k0 += 1
print(k4,k2,k0)
if k2 > 0:
k0 += 1
if k0 > k4+1:
print("No")
else:
print("Yes")
| s312357950 | Accepted | 64 | 14,252 | 307 | N = int(input())
A = list(map(int, input().split()))
k0 = 0
k2 = 0
k4 = 0
for a in A:
if a%4==0:
k4 += 1
elif a%2==0:
k2 += 1
else:
k0 += 1
if k2 > 0:
k0 += 1
if k0 <= k4+1:
print("Yes")
else:
print("No")
|
s954720514 | p03129 | u482157295 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 80 | Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1. | a,b = map(int,input().split())
if b+b-1 >= a:
print("Yes")
else:
print("No") | s757664482 | Accepted | 17 | 2,940 | 83 | a,b = map(int,input().split())
if b+b-1 <= a:
print("YES")
else:
print("NO")
|
s949095877 | p03828 | u501163846 | 2,000 | 262,144 | Wrong Answer | 34 | 3,064 | 566 | You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7. | import math
k=int(input())
def primes(n):
is_prime = [True] * (n + 1)
is_prime[0] = False
is_prime[1] = False
for i in range(2, int(n**0.5) + 1):
if not is_prime[i]:
continue
for j in range(i * 2, n + 1, i):
is_prime[j] = False
return [i for i in range(n + 1) if is_prime[i]]
pl=primes(k)
ans=[]
print(pl)
m=math.factorial(k)
for i in range(len(pl)):
tmp=0
while m%pl[i]==0:
m=m//pl[i]
tmp+=1
ans.append(tmp)
a=1
for i in range(len(ans)):
a=a*(ans[i]+1)%(10**9+7)
print(a)
| s487645814 | Accepted | 35 | 3,064 | 556 | import math
k=int(input())
def primes(n):
is_prime = [True] * (n + 1)
is_prime[0] = False
is_prime[1] = False
for i in range(2, int(n**0.5) + 1):
if not is_prime[i]:
continue
for j in range(i * 2, n + 1, i):
is_prime[j] = False
return [i for i in range(n + 1) if is_prime[i]]
pl=primes(k)
ans=[]
m=math.factorial(k)
for i in range(len(pl)):
tmp=0
while m%pl[i]==0:
m=m//pl[i]
tmp+=1
ans.append(tmp)
a=1
for i in range(len(ans)):
a=a*(ans[i]+1)%(10**9+7)
print(a)
|
s467589213 | p03730 | u256106029 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 193 | 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().strip().split())
a_temp = a
flg = False
for i in range(b):
if (a_temp - c) % b == 0:
print('Yes')
exit()
a_temp += a
else:
print('No')
| s264634479 | Accepted | 16 | 2,940 | 173 |
a, b, c = map(int, input().strip().split())
a_temp = a
for i in range(b):
if a_temp % b == c:
print('YES')
break
a_temp += a
else:
print('NO')
|
s301370176 | p02612 | u389188163 | 2,000 | 1,048,576 | Wrong Answer | 27 | 9,132 | 69 | 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())
n = N -1000
while n >= 1000:
n -= 1000
print(n) | s555076247 | Accepted | 36 | 9,164 | 69 | N = int(input())
n = N
while n > 0:
n -= 1000
print(abs(n)) |
s512586819 | p03377 | u045953894 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 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. | a,b,x = map(int,input().split())
if a <= x <= a+b:
print('Yes')
else:
print('No') | s509556441 | Accepted | 19 | 3,068 | 90 | a,b,x = map(int,input().split())
if a <= x <= a+b:
print('YES')
else:
print('NO') |
s405608311 | p02854 | u768896740 | 2,000 | 1,048,576 | Wrong Answer | 206 | 26,220 | 459 | Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters. Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut: * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan). * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen. Find the minimum amount of money needed before cutting the bar into two parts with the same length. | n = int(input())
a = list(map(int, input().split()))
left = 0
right = 0
center = 0
sum_a = sum(a)
flg = False
for i in range(n):
if left + a[i] > sum_a/2 and flg is False:
flg = True
center = a[i]
right -= a[i]
elif left + a[i] == sum_a // 2:
print(0)
exit()
if flg is False:
left += a[i]
else:
right += a[i]
print(left, center, right)
print(min(left+center-right, right+center-left))
| s368564383 | Accepted | 221 | 26,220 | 448 | n = int(input())
a = list(map(int, input().split()))
left = a[0]
right = a[-1]
center = 0
sum_a = sum(a)
flg = False
for i in range(1, n-1):
if left + a[i] > sum_a/2 and flg is False:
flg = True
center = a[i]
right -= a[i]
elif left + a[i] == sum_a / 2:
print(0)
exit()
if flg is False:
left += a[i]
else:
right += a[i]
print(abs(min(left+center-right, right+center-left)))
|
s204458130 | p02263 | u408444038 | 1,000 | 131,072 | Wrong Answer | 20 | 5,616 | 978 | An expression is given in a line. Two consequtive symbols (operand or operator) are separated by a space character. You can assume that +, - and * are given as the operator and an operand is a positive integer less than 106 | import sys
a = list(input().split())
stack = []
for i in range(len(a)):
n = len(stack)
if a[i] == '+':
if (n-2) <0:
sys.exit("スタックアンダーフロー")
tmp = int(stack.pop(n-1)) + int(stack.pop(n-2))
stack.append(tmp)
elif a[i] == '-':
if (n-2) <0:
sys.exit("スタックアンダーフロー")
tmp_1 = int(stack.pop(n-1))
tmp_2 = int(stack.pop(n-2))
tmp = tmp_2 - tmp_1
stack.append(tmp)
elif a[i] == '*':
if (n-2) <0:
sys.exit("スタックアンダーフロー")
tmp = int(stack.pop(n-1)) * int(stack.pop(n-2))
stack.append(tmp)
elif a[i] == '/':
if (n-2) <0:
sys.exit("スタックアンダーフロー")
tmp_1 = int(stack.pop(n-1))
tmp_2 = int(stack.pop(n-2))
tmp = tmp_2 / tmp_1
stack.append(tmp)
else:
stack.append(int(a[i]))
| s407150471 | Accepted | 20 | 5,620 | 994 | import sys
a = list(input().split())
stack = []
for i in range(len(a)):
n = len(stack)
if a[i] == '+':
if (n-2) <0:
sys.exit("スタックアンダーフロー")
tmp = int(stack.pop(n-1)) + int(stack.pop(n-2))
stack.append(tmp)
elif a[i] == '-':
if (n-2) <0:
sys.exit("スタックアンダーフロー")
tmp_1 = int(stack.pop(n-1))
tmp_2 = int(stack.pop(n-2))
tmp = tmp_2 - tmp_1
stack.append(tmp)
elif a[i] == '*':
if (n-2) <0:
sys.exit("スタックアンダーフロー")
tmp = int(stack.pop(n-1)) * int(stack.pop(n-2))
stack.append(tmp)
elif a[i] == '/':
if (n-2) <0:
sys.exit("スタックアンダーフロー")
tmp_1 = int(stack.pop(n-1))
tmp_2 = int(stack.pop(n-2))
tmp = tmp_2 / tmp_1
stack.append(tmp)
else:
stack.append(int(a[i]))
print(stack[0])
|
s962297236 | p03455 | u325956328 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 183 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | def evencheck(a, b):
p = a * b
s = ''
if p % 2 == 0:
s = 'Even'
else:
s = 'Odd'
return s
a, b = map(int, input().split())
result = evencheck(a, b) | s971842032 | Accepted | 17 | 2,940 | 200 | def evencheck(a, b):
p = a * b
s = ''
if p % 2 == 0:
s = 'Even'
else:
s = 'Odd'
return s
a, b = map(int, input().split())
result = evencheck(a, b)
print(result)
|
s008698638 | p04029 | u518556834 | 2,000 | 262,144 | Wrong Answer | 19 | 2,940 | 33 | 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? | n = int(input())
print(n*(n+1)/2) | s801717291 | Accepted | 17 | 2,940 | 34 | n = int(input())
print(n*(n+1)//2) |
s769276076 | p03455 | u470927659 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 101 | 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())
c = a*b
if c%2 == 1:
print("Even")
else:
print("Odd") | s925953979 | Accepted | 18 | 2,940 | 101 | a,b = map(int, input().split())
c = a*b
if c%2 == 1:
print("Odd")
else:
print("Even") |
s071755551 | p02613 | u252964975 | 2,000 | 1,048,576 | Wrong Answer | 165 | 9,200 | 257 | 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())
result_list = ['AC', 'WA', 'TLE', 'RE']
count_list = [0, 0, 0, 0]
for i in range(N):
S = input()
count_list[result_list.index(S)] = count_list[result_list.index(S)] + 1
for i in range(4):
print(result_list[i]+' × '+str(count_list[i])) | s425336850 | Accepted | 165 | 9,200 | 256 | N=int(input())
result_list = ['AC', 'WA', 'TLE', 'RE']
count_list = [0, 0, 0, 0]
for i in range(N):
S = input()
count_list[result_list.index(S)] = count_list[result_list.index(S)] + 1
for i in range(4):
print(result_list[i]+' x '+str(count_list[i])) |
s232966769 | p02694 | u877428733 | 2,000 | 1,048,576 | Wrong Answer | 22 | 9,116 | 130 | Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time? | import math
X=int(input())
money = 100
ans = 0
while money <= X:
money += math.floor(money * 0.01)
ans += 1
print(ans) | s233791959 | Accepted | 20 | 9,164 | 130 | import math
X=int(input())
money = 100
ans = 0
while money < X:
money += math.floor(money * 0.01)
ans += 1
print(ans)
|
s174784671 | p03455 | u834415466 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 80 | 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 :
print('odd')
else:
print('even') | s124491557 | Accepted | 18 | 2,940 | 80 | a,b=map(int,input().split())
if a*b%2 :
print('Odd')
else:
print('Even') |
s731364738 | p00007 | u150984829 | 1,000 | 131,072 | Wrong Answer | 20 | 5,668 | 78 | 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. | import math
d=100
for _ in[0]*int(input()):
d=math.ceil(d*1.05)
print(d*1e3)
| s279318823 | Accepted | 20 | 5,600 | 68 | a=100
for _ in[0]*int(input()):a=int(a*1.05)+(a%20>0)
print(a*1000)
|
s420808563 | p03494 | u107091170 | 2,000 | 262,144 | Wrong Answer | 30 | 9,036 | 166 | 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())
a=list(map(int, input().split()))
ans = 0
for i in range(N):
d = 0
while (a[i] %2 == 0):
d += 1
a[i] //= 2
ans = min(d,ans)
print(ans)
| s241268788 | Accepted | 26 | 9,112 | 173 | N=int(input())
a=list(map(int, input().split()))
ans = 100000000
for i in range(N):
d = 0
while (a[i] %2 == 0):
d += 1
a[i] //= 2
ans = min(d,ans)
print(ans)
|
s149064980 | p02613 | u248556072 | 2,000 | 1,048,576 | Wrong Answer | 153 | 16,284 | 183 | 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 = []
for n in range(N):
S.append(input())
print('AC ×',S.count('AC'))
print('WA ×',S.count('WA'))
print('TLE ×',S.count('TLE'))
print('RE ×',S.count('RE')) | s029441317 | Accepted | 146 | 16,276 | 179 | N = int(input())
S = []
for n in range(N):
S.append(input())
print('AC x',S.count('AC'))
print('WA x',S.count('WA'))
print('TLE x',S.count('TLE'))
print('RE x',S.count('RE')) |
s252710749 | p02468 | u978863922 | 1,000 | 131,072 | Wrong Answer | 20 | 5,592 | 110 | For given integers m and n, compute mn (mod 1,000,000,007). Here, A (mod M) is the remainder when A is divided by M. | try:
(m,n) = map(int,input().split())
except:
exit
print(m,n)
r = pow(m,n) % 1000000007
print (r)
| s236479693 | Accepted | 20 | 5,604 | 187 | (m,n) = map(int,input().split())
bi = str(format(n,"b"))
bi = bi[::-1]
c = 1
d = 1000000007
for i in bi:
if ( i == "1"):
c = ( c * m ) % d
m = (m * m) %d
print (c)
|
s645650728 | p03457 | u001769145 | 2,000 | 262,144 | Wrong Answer | 380 | 3,064 | 317 | 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. | # ABC086 C Traveling
n = int(input())
t, x, y = 0, 0, 0
flag = True
for i in range(n):
_t,_x,_y = map(int, input().split())
_d = abs(x - _x) + abs(y - _y)
_dt = abs(t - _t)
if _d > _dt or ((_d - _dt)%2) == 1:
flag = False
t,x,y = _t,_x,_y
if flag:
print("YES")
else:
print("NO") | s665822586 | Accepted | 376 | 3,064 | 317 | # ABC086 C Traveling
n = int(input())
t, x, y = 0, 0, 0
flag = True
for i in range(n):
_t,_x,_y = map(int, input().split())
_d = abs(x - _x) + abs(y - _y)
_dt = abs(t - _t)
if _d > _dt or ((_d - _dt)%2) == 1:
flag = False
t,x,y = _t,_x,_y
if flag:
print("Yes")
else:
print("No") |
s701162030 | p02612 | u952968889 | 2,000 | 1,048,576 | Wrong Answer | 31 | 9,048 | 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) | s184681548 | Accepted | 33 | 9,140 | 64 | from math import ceil
n=int(input())
print(ceil(n/1000)*1000-n) |
s024075436 | p02669 | u368796742 | 2,000 | 1,048,576 | Wrong Answer | 66 | 11,288 | 679 | You start with the number 0 and you want to reach the number N. You can change the number, paying a certain amount of coins, with the following operations: * Multiply the number by 2, paying A coins. * Multiply the number by 3, paying B coins. * Multiply the number by 5, paying C coins. * Increase or decrease the number by 1, paying D coins. You can perform these operations in arbitrary order and an arbitrary number of times. What is the minimum number of coins you need to reach N? **You have to solve T testcases.** | import sys
from functools import lru_cache
sys.setrecursionlimit(10**7)
@lru_cache(maxsize=None)
def dfs(x):
if x == 0:
return 0
if x == 1:
return d
l2 = x//2
r2 = (x+1)//2
l3 = x//3
r3 = (x+2)//3
l5 = x//5
r5 = (x+4)//5
count = x*d
count = min(count,abs(l2*2-x)*d+a+dfs(l2))
count = min(count,abs(r2*2-x)*d+a+dfs(r2))
count = min(count,abs(l3*3-x)*d+b+dfs(l3))
count = min(count,abs(r3*3-x)*d+b+dfs(r3))
count = min(count,abs(l5*5-x)*d+c+dfs(l5))
count = min(count,abs(r5*5-x)*d+c+dfs(r5))
return count
t = int(input())
for _ in range(t):
n,a,b,c,d = map(int,input().split())
print(dfs(n)) | s877197826 | Accepted | 302 | 20,448 | 801 | import sys
from functools import lru_cache
sys.setrecursionlimit(10**7)
def solve(x,a,b,c,d):
@lru_cache(None)
def dfs(x):
if x == 0:
return 0
if x == 1:
return d
l2 = x//2
r2 = (x+1)//2
l3 = x//3
r3 = (x+2)//3
l5 = x//5
r5 = (x+4)//5
count = x*d
count = min(count,abs(l2*2-x)*d+a+dfs(l2))
count = min(count,abs(r2*2-x)*d+a+dfs(r2))
count = min(count,abs(l3*3-x)*d+b+dfs(l3))
count = min(count,abs(r3*3-x)*d+b+dfs(r3))
count = min(count,abs(l5*5-x)*d+c+dfs(l5))
count = min(count,abs(r5*5-x)*d+c+dfs(r5))
return count
return dfs(x)
t = int(input())
for _ in range(t):
n,a,b,c,d = map(int,input().split())
print(solve(n,a,b,c,d)) |
s361916499 | p03048 | u771007149 | 2,000 | 1,048,576 | Wrong Answer | 2,103 | 2,940 | 246 | 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())
cnt = 0
for i in range(n//R):
for j in range(n//G):
for k in range(n//B):
if i*R + j*G + k*B == n:
cnt += 1
print(cnt)
| s778134385 | Accepted | 1,953 | 2,940 | 277 |
R,G,B,n = map(int,input().split())
cnt = 0
for i in range(n//R + 1):
for j in range(n - i*R + 1):
c = n - (i*R + j*G)
if c >= 0 and c%B == 0:
cnt += 1
else:
pass
print(cnt)
|
s481384795 | p02694 | u059487235 | 2,000 | 1,048,576 | Wrong Answer | 20 | 9,136 | 139 | Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time? | import math
x = int(input())
orig = 100
counter = 0
while orig <= x:
orig = math.floor(orig * 1.01)
counter += 1
print(counter)
| s478387465 | Accepted | 22 | 9,132 | 138 | import math
x = int(input())
orig = 100
counter = 0
while orig < x:
orig = math.floor(orig * 1.01)
counter += 1
print(counter)
|
s092471625 | p03657 | u710789518 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 115 | Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies. | A, B = map(int, input().split())
if A % 3 == 0 or B % 3 == 0 or (A+B) % 3 == 0:
print('Yes')
else:
print('No')
| s891751346 | Accepted | 17 | 2,940 | 128 | A, B = map(int, input().split())
if A % 3 == 0 or B % 3 == 0 or (A+B) % 3 == 0:
print('Possible')
else:
print('Impossible')
|
s907409707 | p02694 | u479719434 | 2,000 | 1,048,576 | Wrong Answer | 21 | 9,156 | 121 | Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time? | X = int(input())
ans = 0
money = 100
while money <= X:
ans += 1
money *= 1.01
money = int(money)
print(ans)
| s737210388 | Accepted | 24 | 9,096 | 120 | X = int(input())
ans = 0
money = 100
while money < X:
ans += 1
money *= 1.01
money = int(money)
print(ans)
|
s561288359 | p02612 | u806698335 | 2,000 | 1,048,576 | Wrong Answer | 28 | 9,132 | 34 | 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)
| s461124672 | Accepted | 28 | 9,080 | 51 | n = int(input())
print( (1000 - n % 1000) % 1000)
|
s032492623 | p03415 | u821425701 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 66 | We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}. Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bottom-right corner of the grid, from the top-left to bottom-right. | x1 = input()
x2 = input()
x3 = input()
print(x1[0], x2[1], x3[2]) | s240433520 | Accepted | 17 | 2,940 | 83 | s1 = input()
s2 = input()
s3 = input()
print('{}{}{}'.format(s1[0], s2[1], s3[2])) |
s120191888 | p03962 | u063346608 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 129 | 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. | a,b,c = input().split()
score = 1
if a != b:
score = score + 1
elif a !=c:
score = score + 1
else:
score = score
print(score) | s670042244 | Accepted | 17 | 2,940 | 201 | a,b,c = input().split()
if a == b and a == c:
print(1)
elif a ==b and a != c:
print(2)
elif b == c and b != a:
print(2)
elif c ==a and c != b:
print(2)
elif a != b and b != c and c != a:
print(3) |
s868904679 | p03574 | u018679195 | 2,000 | 262,144 | Wrong Answer | 25 | 3,572 | 1,596 | 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. | import sys
def main():
in1 = input()
nums, length = in1.split(' ')
nums= int(nums)
length = int(length)
bombs = []
# print(nums)
#
for i in range(nums):
bombs.append(input())
print(bombs)
checked = [['']*length for i in range(nums)]
for i in range(nums):
for j in range(length):
if bombs[i][j] == '#':
checked[i][j] = '#'
else:
bombnum = 0
if i != 0:
if j != 0:
if bombs[i-1][j-1] == '#':
bombnum += 1
if j != length -1:
if bombs[i-1][j+1] == '#':
bombnum += 1
if bombs[i-1][j] == '#':
bombnum += 1
if j != 0:
if bombs[i][j-1] == '#':
bombnum += 1
if j != length-1:
if bombs[i][j+1] == '#':
bombnum += 1
if i != nums-1:
if j != 0:
if bombs[i+1][j-1] == '#':
bombnum += 1
if j != length -1:
if bombs[i+1][j+1] == '#':
bombnum += 1
if bombs[i+1][j] == '#':
bombnum += 1
checked[i][j] = bombnum
for i in checked:
for j in i:
print(j, end = '')
print('')
if __name__ == '__main__':
main() | s079050409 | Accepted | 30 | 3,188 | 1,037 | q=list(map(int,input().split()))
grid=[]
def ze(x,y):
z=[]
if y[0]+1<x[0]:
z.append([y[0]+1,y[1]])
if y[1]+1<x[1]:#
z.append([y[0]+1,y[1]+1])
if y[1]-1>=0:
z.append([y[0]+1,y[1]-1])
if y[0]-1>=0:
z.append([y[0]-1,y[1]])
if y[1]+1<x[1]:#
z.append([y[0]-1,y[1]+1])
if y[1]-1>=0:
z.append([y[0]-1,y[1]-1])
if y[1]+1<x[1]:
z.append([y[0],y[1]+1])
if y[1]-1>=0:
z.append([y[0],y[1]-1])
return(z)
for i in range(q[0]):
y=input()
grid.append([i for i in y])
for i in range(q[0]):
for j in range(q[1]):
if grid[i][j]=='.':
lamsa=ze(q,[i,j])
count=0
for k in range(len(lamsa)) :
if grid[lamsa[k][0]][lamsa[k][1]] == '#' :
count+=1
grid[i][j]=count
for i in range(q[0]):
pr=''
for j in range(q[1]):
pr=pr+str(grid[i][j])
print(pr)
|
s068433030 | p03063 | u608850287 | 2,000 | 1,048,576 | Wrong Answer | 115 | 3,500 | 184 | There are N stones arranged in a row. Every stone is painted white or black. A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is `.`, and the stone is black if the character is `#`. Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone. Find the minimum number of stones that needs to be recolored. | n = int(input())
s = input()
best = n
curr = 0
nblack = 0
for ch in s:
if ch == '.':
curr = min(nblack, 1 + curr)
else:
nblack += 1
best = min(best, curr)
print(best)
| s442375497 | Accepted | 74 | 3,500 | 150 | n = int(input())
s = input()
curr = 0
nblack = 0
for ch in s:
if ch == '.':
curr = min(nblack, 1 + curr)
else:
nblack += 1
print(curr)
|
s412596856 | p02608 | u942051624 | 2,000 | 1,048,576 | Wrong Answer | 2,206 | 9,180 | 336 | Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N). | import math
N=int(input())
ans=[0 for i in range(N)]
for n in range(N):
nax=int(math.sqrt(n))
for x in range(1,nax):
for y in range(1,x+1):
for z in range(1,y+1):
temp=x**2+y**2+z**2+x*y+y*z+z*x
if temp==n+1:
ans[n]+=1
for anss in ans:
print(anss*3)
| s542053944 | Accepted | 189 | 9,272 | 650 | import math
N=int(input())
ans=[0 for i in range(N)]
nax=int(math.sqrt(N))
for x in range(1,nax):
for y in range(1,x+1):
for z in range(1,y+1):
temp=x**2+y**2+z**2+x*y+y*z+z*x
if temp<=N:
if x==y and y!=z:
ans[temp-1]+=3
elif y==z and z!=x:
ans[temp-1]+=3
elif z==x and x!=y:
ans[temp-1]+=3
elif x==y==z:
ans[temp-1]+=1
elif x!=y and y!=z and z!=x:
ans[temp-1]+=6
for anss in ans:
print(anss)
|
s107635206 | p03455 | u189397279 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 113 | AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd. | # -*- coding: utf-8 -*-
a, b = map(int, input().split(" "))
if (a * b) % 2:
print("Even")
else:
print("Odd") | s543945092 | Accepted | 17 | 2,940 | 114 | # -*- coding: utf-8 -*-
a, b = map(int, input().split(" "))
if (a * b) % 2:
print("Odd")
else:
print("Even") |
s084682048 | p02603 | u726285999 | 2,000 | 1,048,576 | Wrong Answer | 37 | 9,068 | 307 | To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally? | N = int(input())
A = [int(x) for x in input().split()]
genkin = 1000
kabu = 0
for i in range(N-1):
if A[i] < A[i+1]:
kabu = genkin // A[i]
genkin = genkin % A[i]
if A[i] > A[i+1]:
genkin += kabu * A[i]
kabu = 0
print(genkin, kabu)
print(genkin + kabu * A[N-1]) | s249297616 | Accepted | 31 | 9,196 | 310 | N = int(input())
A = [int(x) for x in input().split()]
genkin = 1000
kabu = 0
for i in range(N-1):
if A[i] < A[i+1]:
kabu += genkin // A[i]
genkin = genkin % A[i]
if A[i] > A[i+1]:
genkin += kabu * A[i]
kabu = 0
print(genkin + kabu * A[N-1]) |
s658702386 | p03493 | u392433152 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 91 | 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=(int(x) for x in input().split())
count=0
for i in s:
if i==1:
count+=1
print(count) | s948705535 | Accepted | 17 | 2,940 | 77 | s=input()
count=0
i=0
while i<3:
if s[i]=="1":
count+=1
i+=1
print(count) |
s828645604 | p03695 | u497952650 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 562 | In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users. | MAX_COLORS = 8
DIFF_RATE = 400
def solve(a):
colors = [0]*9
for i in a:
num = 0
tmp = 400
for j in range(9):
if i < 0:
break
else:
i -= 400
num += 1
print(num)
if i > 0:
colors[num-1] += 1
else:
colors[num-1] = 1
print(sum(colors[:8]),min(sum(colors[:8])+colors[-1],MAX_COLORS))
def main():
N = int(input())
a = list(map(int,input().split()))
solve(a)
if __name__ == "__main__":
main() | s315687866 | Accepted | 18 | 3,064 | 618 | def solve(a):
colors = [0]*9
for i in a:
num = 0
tmp = 400
flag = True
for j in range(9):
if i < 0:
flag = False
break
else:
i -= 400
num += 1
if flag:
colors[num-1] += 1
else:
colors[num-1] = 1
if sum(colors[:8]) == 0:
print(1,colors[-1])
else:
print(sum(colors[:8]),sum(colors[:8])+colors[-1])
def main():
N = int(input())
a = list(map(int,input().split()))
solve(a)
if __name__ == "__main__":
main() |
s641246070 | p03360 | u857330600 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 81 | 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? | l=list(map(int,input().split()))
l.sort()
k=int(input())
print(l[0]+l[1]+l[2]**k) | s794611312 | Accepted | 17 | 2,940 | 85 | l=list(map(int,input().split()))
l.sort()
k=int(input())
print(l[0]+l[1]+l[2]*(2**k)) |
s294120238 | p02613 | u771007149 | 2,000 | 1,048,576 | Wrong Answer | 148 | 9,208 | 384 | 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. | #B
n = int(input())
cnt_ac = 0
cnt_wa = 0
cnt_tle = 0
cnt_re = 0
for _ in range(n):
s = input()
if s == 'AC':
cnt_ac += 1
elif s == 'TLE':
cnt_tle += 1
elif s == 'WA':
cnt_wa += 1
else:
cnt_re += 1
print('AC × {}'.format(cnt_ac))
print('WA × {}'.format(cnt_wa))
print('TLE × {}'.format(cnt_tle))
print('RE × {}'.format(cnt_re)) | s823951298 | Accepted | 152 | 9,204 | 380 | #B
n = int(input())
cnt_ac = 0
cnt_wa = 0
cnt_tle = 0
cnt_re = 0
for _ in range(n):
s = input()
if s == 'AC':
cnt_ac += 1
elif s == 'TLE':
cnt_tle += 1
elif s == 'WA':
cnt_wa += 1
else:
cnt_re += 1
print('AC x {}'.format(cnt_ac))
print('WA x {}'.format(cnt_wa))
print('TLE x {}'.format(cnt_tle))
print('RE x {}'.format(cnt_re)) |
s297507622 | p03047 | u792078574 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 49 | Snuke has N integers: 1,2,\ldots,N. He will choose K of them and give those to Takahashi. How many ways are there to choose K consecutive integers? | N, K = map(int, input().split())
print(K - N + 1) | s781946983 | Accepted | 17 | 2,940 | 49 | N, K = map(int, input().split())
print(N - K + 1) |
s994631819 | p03697 | u385309449 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 80 | You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead. | x,y = map(int,input().split())
if x+y >= 10:
print(x+y)
else:
print('error') | s456618916 | Accepted | 17 | 2,940 | 80 | x,y = map(int,input().split())
if x+y >= 10:
print('error')
else:
print(x+y) |
s551271162 | p03478 | u847923740 | 2,000 | 262,144 | Wrong Answer | 47 | 9,032 | 297 | 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). | #
import sys
input=sys.stdin.readline
def main():
N,A,B=map(int,input().split())
cnt=0
for i in range(1,N+1):
ds=0
for j in range(4):
ds+=(i//(10**j))%(10**j)
if A<=ds<=B:
cnt+=i
print(cnt)
if __name__=="__main__":
main()
| s954553195 | Accepted | 43 | 8,920 | 299 | #
import sys
input=sys.stdin.readline
def main():
N,A,B=map(int,input().split())
cnt=0
for i in range(1,N+1):
ds=0
for j in range(5):
ds+=(int(i//(10**j)))%(10)
if A<=ds<=B:
cnt+=i
print(cnt)
if __name__=="__main__":
main()
|
s679344185 | p03719 | u327466606 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 64 | 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') | s633845288 | Accepted | 17 | 2,940 | 64 | a,b,c=map(int,input().split())
print('Yes' if a<=c<=b else 'No') |
s009303546 | p02844 | u992910889 | 2,000 | 1,048,576 | Wrong Answer | 2,148 | 710,808 | 411 | AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code. The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code. How many different PIN codes can he set this way? Both the lucky number and the PIN code may begin with a 0. | def resolve():
N=int(input())
s = str(input())
l=['%03d'%(p) for p in range(0,10**N)]
cnt=0
for i in l:
for j in range(N):
if s[j]==i[0]:
for k in range(j+1,N):
if s[k]==i[1]:
for m in range(k+1,N):
if s[m]==i[2]:
cnt+=1
print(cnt)
resolve() | s578489491 | Accepted | 25 | 3,188 | 370 |
def resolve():
N=int(input())
s = str(input())
l=['%03d'%(p) for p in range(0,10**3)]
ans=[]
for i in l:
if i[0] in s:
s0=s.index(i[0])
if i[1] in s[s0+1:]:
s1=s[s0+1:].index(i[1])
if i[2] in s[s0+1:][s1+1:]:
ans.append(i)
print(len(set(ans)))
resolve() |
s228036498 | p02842 | u167908302 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 47 | 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. | #coding:utf-8
n = int(input())
print(n // 1.08) | s293648421 | Accepted | 17 | 2,940 | 124 | #coding:utf-8
n = int(input())
x = int(-(-n // 1.08))
if int(x * 1.08) == n:
print(x)
else:
print(':(')
|
s803446087 | p02398 | u546968095 | 1,000 | 131,072 | Wrong Answer | 30 | 5,592 | 251 | Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b. | def func():
a,b,c = input().split(" ")
a = int(a)
b = int(b)
c = int(c)
cnt = 0
for i in range(a,b+1):
if 0 == i % c:
cnt = cnt +1
print(cnt)
return 0
if __name__ == "__main__":
ret = func()
| s560928013 | Accepted | 20 | 5,600 | 251 | def func():
a,b,c = input().split(" ")
a = int(a)
b = int(b)
c = int(c)
cnt = 0
for i in range(a,b+1):
if 0 == c % i:
cnt = cnt +1
print(cnt)
return 0
if __name__ == "__main__":
ret = func()
|
s938102386 | p02936 | u608088992 | 2,000 | 1,048,576 | Wrong Answer | 697 | 63,556 | 815 | Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations. | import sys, collections
def solve():
input = sys.stdin.readline
N, Q = map(int, input().split())
edge = [[] for i in range(N)]
for _ in range(N-1):
a, b = map(int, input().split())
edge[a-1].append(b-1)
edge[b-1].append(a-1)
counter = [0] * N
for _ in range(Q):
p, x = map(int, input().split())
counter[p-1] += x
visited = [False] * N
q = collections.deque()
q.append(0)
while q:
now = q.popleft()
if not visited:
visited = True
cnow = counter[now]
for next in edge[now]:
if not visited[next]:
counter[next] += cnow
q.append(next)
print(" ".join(map(str, counter)))
return 0
if __name__ == "__main__":
solve() | s372632876 | Accepted | 972 | 65,668 | 721 | import sys
from collections import deque
def solve():
input = sys.stdin.readline
N, Q = map(int, input().split())
Edge = [[] for _ in range(N)]
for _ in range(N-1):
a, b = map(int, input().split())
Edge[a-1].append(b-1)
Edge[b-1].append(a-1)
count = [0] * N
for _ in range(Q):
p, x = map(int, input().split())
count[p-1] += x
q = deque()
for nextN in Edge[0]: q.append((nextN, 0))
while q:
nowN, parN = q.popleft()
count[nowN] += count[parN]
for nextN in Edge[nowN]:
if nextN != parN: q.append((nextN, nowN))
print(" ".join(map(str, count)))
return 0
if __name__ == "__main__":
solve() |
s432859535 | p03845 | u050708958 | 2,000 | 262,144 | Wrong Answer | 29 | 3,572 | 290 | Joisino is about to compete in the final round of a certain programming competition. In this contest, there are N problems, numbered 1 through N. Joisino knows that it takes her T_i seconds to solve problem i(1≦i≦N). Also, there are M kinds of drinks offered to the contestants, numbered 1 through M. If Joisino takes drink i(1≦i≦M), her brain will be stimulated and the time it takes for her to solve problem P_i will become X_i seconds. It does not affect the time to solve the other problems. A contestant is allowed to take exactly one of the drinks before the start of the contest. For each drink, Joisino wants to know how many seconds it takes her to solve all the problems if she takes that drink. Here, assume that the time it takes her to solve all the problems is equal to the sum of the time it takes for her to solve individual problems. Your task is to write a program to calculate it instead of her. | import copy
N = int(input())
t = [int(x) for x in input().split()]
p = []
for _ in range(int(input())):
p.append([int(x) for x in input().split()])
result = []
ori_t = copy.deepcopy(t)
for i,j in zip(t,p):
t = copy.deepcopy(ori_t)
t[j[0]-1] = j[1]
result.append(sum(t))
print(result)
| s050327735 | Accepted | 30 | 3,444 | 293 | import copy
N = int(input())
t = [int(x) for x in input().split()]
p = []
for _ in range(int(input())):
p.append([int(x) for x in input().split()])
result = []
ori_t = copy.deepcopy(t)
for i in p:
t = copy.deepcopy(ori_t)
t[i[0]-1] = i[1]
result.append(sum(t))
for i in result:
print(i) |
s971489859 | p03024 | u782685137 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 46 | Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility. | print('NO' if input().count('x')>6 else 'YES') | s334362098 | Accepted | 17 | 2,940 | 46 | print('NO' if input().count('x')>7 else 'YES') |
s315450576 | p03556 | u468972478 | 2,000 | 262,144 | Wrong Answer | 2,206 | 9,476 | 97 | 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. | n = int(input())
for i in range(n, 0, -1):
if isinstance(i ** 0.5, int):
print(i)
break | s267114490 | Accepted | 28 | 9,424 | 31 | print(int(int(input())**.5)**2) |
s551366207 | p03160 | u033524082 | 2,000 | 1,048,576 | Wrong Answer | 202 | 13,776 | 354 | There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N. | n=int(input())
l=list(map(int,input().split()))
m=[0,abs(l[1]-l[0])]
if n==2:
print(m[1])
else:
p1=abs(l[2]-l[0])
p2=abs(m[1]+l[2]-l[1])
m.append(min(p1,p2))
print(m[-1])
for i in range(3,n):
p1=m[-3]+abs(l[-1]-l[-3])
p2=m[-2]+abs(l[-1]-l[-2])
m.append(min(p1,p2))
print(m[i])
print(abs(m[-1])) | s310078542 | Accepted | 139 | 13,928 | 319 | n=int(input())
l=list(map(int,input().split()))
m=[0,abs(l[1]-l[0])]
if n==2:
print(m[1])
else:
p1=abs(l[2]-l[0])
p2=m[1]+abs(l[2]-l[1])
m.append(min(p1,p2))
for i in range(3,n):
p1=m[i-2]+abs(l[i]-l[i-2])
p2=m[i-1]+abs(l[i]-l[i-1])
m.append(min(p1,p2))
print(abs(m[-1])) |
s168519161 | p03473 | u442948527 | 2,000 | 262,144 | Wrong Answer | 26 | 9,140 | 22 | How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December? | print(24-int(input())) | s475581937 | Accepted | 24 | 8,988 | 22 | print(48-int(input())) |
s208266547 | p03129 | u571445182 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 153 | Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1. | nData = []
nData = input().rstrip().split(' ')
nN = int(nData[0])
nK = int(nData[1])
nTmp = nK * 2
if (nN >= nTmp):
print('YES')
else:
print('NO')
| s069234537 | Accepted | 17 | 2,940 | 162 | nData = []
nData = input().rstrip().split(' ')
nN = int(nData[0])
nK = int(nData[1])
nN += 1
nTmp = nK * 2
if (nN >= nTmp):
print('YES')
else:
print('NO')
|
s632097395 | p02534 | u708530082 | 2,000 | 1,048,576 | Wrong Answer | 26 | 8,996 | 32 | You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them. For example, if K = 3, print `ACLACLACL`. | n = int(input())
print("ALC"*n) | s153169148 | Accepted | 22 | 9,064 | 32 | n = int(input())
print("ACL"*n) |
s153323178 | p03434 | u633355062 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 233 | We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score. | n = int(input())
cards = sorted(map(int, input().split()), reverse=True)
score_alice, score_bob = 0, 0
for i in range(n):
if i % 2 == 1:
score_alice += cards[i]
else:
score_bob += cards[i]
print(score_alice - score_bob) | s631544040 | Accepted | 17 | 2,940 | 235 | n = int(input())
cards = sorted(map(int, input().split()), reverse=True)
score_alice, score_bob = 0, 0
for i in range(n):
if i % 2 == 0:
score_alice += cards[i]
else:
score_bob += cards[i]
print(score_alice - score_bob)
|
s945921352 | p00003 | u547838013 | 1,000 | 131,072 | Wrong Answer | 40 | 5,600 | 185 | Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so. | n = int(input())
for i in range(n):
s = list(map(int, input().split()))
s.sort()
if(s[2] ** 2 == s[1] ** 2 + s[0] ** 2):
print('yes')
else:
print('no')
| s614343943 | Accepted | 40 | 5,592 | 185 | n = int(input())
for i in range(n):
s = list(map(int, input().split()))
s.sort()
if(s[2] ** 2 == s[1] ** 2 + s[0] ** 2):
print('YES')
else:
print('NO')
|
s502824494 | p02607 | u503111914 | 2,000 | 1,048,576 | Wrong Answer | 28 | 9,160 | 143 | We have N squares assigned the numbers 1,2,3,\ldots,N. Each square has an integer written on it, and the integer written on Square i is a_i. How many squares i satisfy both of the following conditions? * The assigned number, i, is odd. * The written integer is odd. | N = int(input())
A = list(map(int,input().split()))
ans = 0
for i in range(N):
if i % 2 == 0:
if A[i] % 2 == 0:
ans += 1
print(ans) | s260394088 | Accepted | 26 | 9,076 | 144 | N = int(input())
A = list(map(int,input().split()))
ans = 0
for i in range(N):
if i % 2 == 0:
if A[i] % 2 == 1:
ans += 1
print(ans)
|
s013377478 | p03854 | u841348630 | 2,000 | 262,144 | Wrong Answer | 83 | 3,188 | 239 | 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()
c = ["dream","dreamer","erase","eraser"]
S = S[::-1]
c = [cc[::-1] for cc in c]
i=0
while i < len(c):
if S.startswith(c[i]):
S = S[len(c[i]):]
i = 0
continue
i = -~i
print("NYoe s"[S==""::2])
| s980623750 | Accepted | 81 | 3,188 | 239 | S = input()
c = ["dream","dreamer","erase","eraser"]
S = S[::-1]
c = [cc[::-1] for cc in c]
i=0
while i < len(c):
if S.startswith(c[i]):
S = S[len(c[i]):]
i = 0
continue
i = -~i
print("NYOE S"[S==""::2])
|
s332525434 | p04011 | u610232844 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 310 | There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee. | max_day, change_day, pay, discount_pay = [int(input()) for _ in range(4)]
pay_change_day = max_day - change_day
payment = 0
print(pay_change_day, payment)
for i in range(change_day):
payment += pay
if pay_change_day > 0:
for v in range(pay_change_day):
payment += discount_pay
print(payment)
| s512296120 | Accepted | 20 | 2,940 | 216 | max_day, normal_day, pay, discount_pay = [int(input()) for _ in range(4)]
payment = 0
for i in range(max_day):
if i < normal_day:
payment += pay
else:
payment += discount_pay
print(payment)
|
s971015756 | p04029 | u973069173 | 2,000 | 262,144 | Wrong Answer | 31 | 9,052 | 35 | 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 = int(input())
print((a*(a-1))/2) | s904437445 | Accepted | 28 | 9,148 | 36 | a = int(input())
print((a*(a+1))//2) |
s439127139 | p03501 | u724742135 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 123 | 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. | from sys import stdin
a, b, c = [int(x) for x in stdin.readline().rstrip().split()]
if a*b>c:
print(a*b)
else:
print(c) | s317842178 | Accepted | 17 | 2,940 | 125 | from sys import stdin
a, b, c = [int(x) for x in stdin.readline().rstrip().split()]
if (a*b)>c:
print(c)
else:
print(a*b) |
s401653866 | p00728 | u282479278 | 1,000 | 131,072 | Wrong Answer | 50 | 5,604 | 434 | The International Clown and Pierrot Competition (ICPC), is one of the most distinguished and also the most popular events on earth in the show business. One of the unique features of this contest is the great number of judges that sometimes counts up to one hundred. The number of judges may differ from one contestant to another, because judges with any relationship whatsoever with a specific contestant are temporarily excluded for scoring his/her performance. Basically, scores given to a contestant's performance by the judges are averaged to decide his/her score. To avoid letting judges with eccentric viewpoints too much influence the score, the highest and the lowest scores are set aside in this calculation. If the same highest score is marked by two or more judges, only one of them is ignored. The same is with the lowest score. The average, which may contain fractions, are truncated down to obtain final score as an integer. You are asked to write a program that computes the scores of performances, given the scores of all the judges, to speed up the event to be suited for a TV program. | j = 0
lists = []
flag = 0
cnt = 0
while 1:
n = int(input())
if n == 0:
break
for i in range(n):
s = int(input())
lists.append(s)
for i in range(n):
if lists[i] == max(lists):
flag += 1
pass
elif lists[i] == min(lists):
flag += 1
pass
else:
cnt += lists[i]
if flag >= 2:
n -= 2
print(cnt // n)
| s056809853 | Accepted | 30 | 5,600 | 579 | j = 0
lists = []
flag = 0
cnt = 0
while 1:
cnt = 0
flag_max = 0
flag_min = 0
lists = []
n = int(input())
if n == 0:
break
for i in range(n):
s = int(input())
lists.append(s)
for i in range(n):
if lists[i] == max(lists):
flag_max += 1
pass
elif lists[i] == min(lists):
flag_min += 1
pass
else:
cnt += lists[i]
n -= 2
if flag_max >= 2:
cnt += max(lists)
elif flag_min >= 2:
cnt += min(lists)
print(cnt // n)
|
s831534217 | p03408 | u994767958 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 262 | 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())
lst = {}
for _ in range(n):
temp = input()
if temp in lst:
lst[temp] = lst[temp]+1
else:
lst[temp] = 1
m = int(input())
for _ in range(m):
temp = input()
if temp in lst:
lst[temp] = lst[temp] - 1
print(max(0,lst[max(lst)])) | s939831694 | Accepted | 18 | 3,060 | 317 | def main():
n = int(input())
lst = {}
for _ in range(n):
temp = input()
if temp in lst:
lst[temp] = lst[temp]+1
else:
lst[temp] = 1
m = int(input())
for _ in range(m):
temp = input()
if temp in lst:
lst[temp] = lst[temp] - 1
ans = 0
for i in lst:
ans = max(lst[i], ans)
print(ans)
main() |
s075055320 | p03493 | u577139181 | 2,000 | 262,144 | Wrong Answer | 28 | 8,952 | 43 | 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 = input()
cnt = sum([int(c) for c in s]) | s337174702 | Accepted | 30 | 8,876 | 55 | s = input()
cnt = sum([int(c) for c in s])
print(cnt) |
s405421292 | p03436 | u503111914 | 2,000 | 262,144 | Wrong Answer | 36 | 9,148 | 921 | We have an H \times W grid whose squares are painted black or white. The square at the i-th row from the top and the j-th column from the left is denoted as (i, j). Snuke would like to play the following game on this grid. At the beginning of the game, there is a character called Kenus at square (1, 1). The player repeatedly moves Kenus up, down, left or right by one square. The game is completed when Kenus reaches square (H, W) passing only white squares. Before Snuke starts the game, he can change the color of some of the white squares to black. However, he cannot change the color of square (1, 1) and (H, W). Also, changes of color must all be carried out before the beginning of the game. When the game is completed, Snuke's score will be the number of times he changed the color of a square before the beginning of the game. Find the maximum possible score that Snuke can achieve, or print -1 if the game cannot be completed, that is, Kenus can never reach square (H, W) regardless of how Snuke changes the color of the squares. The color of the squares are given to you as characters s_{i, j}. If square (i, j) is initially painted by white, s_{i, j} is `.`; if square (i, j) is initially painted by black, s_{i, j} is `#`. | import sys
H, W = map(int, input().split())
sx = 0
sy = 0
gx = H-1
gy = W-1
C = [list(i for i in input()) for _ in range(H)]
inf = -1
distance = [[inf] * W for _ in range(H)]
queue = []
queue.insert(0, (sy, sx))
distance[sy][sx] = 0
X = 0
while True:
try:
y, x = queue.pop()
# print(y,x)
except:
# print("Fail")
break
for i in range(4):
nx = x + (-1, 0, 1, 0)[i]
ny = y + (0, -1, 0, 1)[i]
# print("nynx",ny,nx)
if (0 <= nx and nx <= W-1 and 0 <= ny and ny <= H-1 and C[ny][nx] != '#' and distance[ny][nx] == inf):
queue.insert(0, (ny, nx))
distance[ny][nx] = distance[y][x] + 1
C[ny][nx] = "!"
if distance[H-1][W-1] != inf:
#print(distance[gy][gx])
ans = 0
for i in C:
ans += i.count(".")
print(i)
print(ans)
print(distance)
else:
print("Fail") | s119946925 | Accepted | 34 | 9,272 | 901 | import sys
H, W = map(int, input().split())
sx = 0
sy = 0
gx = W-1
gy = H-1
ans = 0
A = [list(i for i in input()) for _ in range(H)]
for j in A:
ans += j.count(".")
inf = -1
distance = [[inf] * W for _ in range(H)]
queue = []
queue.insert(0, (sy, sx))
distance[sy][sx] = 0
X = 0
while True:
try:
y, x = queue.pop()
# print(y,x)
except:
# print("Fail")
break
for i in range(4):
nx = x + (-1, 0, 1, 0)[i]
ny = y + (0, -1, 0, 1)[i]
# print("nynx",ny,nx)
if (0 <= nx and nx <= W-1 and 0 <= ny and ny <= H-1 and A[ny][nx] != '#' and distance[ny][nx] == inf):
queue.insert(0, (ny, nx))
distance[ny][nx] = distance[y][x] + 1
A[ny][nx] = "!"
if distance[H-1][W-1] != inf:
#print(distance[gy][gx])
ans -= distance[gy][gx] + 1
print(ans)
else:
print(-1)
|
s297639505 | p02318 | u564105430 | 1,000 | 131,072 | Wrong Answer | 20 | 5,560 | 374 | Find the edit distance between given two words s1 and s2. The disntace is the minimum number of single-character edits required to change one word into the other. The edits including the following operations: * **insertion** : Insert a character at a particular position. * **deletion** : Delete a character at a particular position. * **substitution** : Change the character at a particular position to a different character | s1=input()
s2=input()
ls1=len(s1)
ls2=len(s2)
f=[[0 for j in range(ls2+1)] for i in range(ls1+1)]
for i in range(ls1+1):
f[i][0]=i
for i in range(ls2+1):
f[0][i]=i
for i in range(1,ls1+1):
for j in range(1,ls2+1):
if s1[i-1]!=s2[j-1]:
f[i][j]=min(f[i-1][j]+1,f[i][j-1]+1,f[i-1][j-1]+1)
else:
f[i][j]=f[i-1][j-1]
print(f)
| s512211862 | Accepted | 1,320 | 42,892 | 436 | s1=input()
s2=input()
ls1=len(s1)
ls2=len(s2)
f=[[0 for j in range(ls2+1)] for i in range(ls1+1)]
for i in range(ls1+1):
f[i][0]=i
for i in range(ls2+1):
f[0][i]=i
for i in range(1,ls1+1):
for j in range(1,ls2+1):
if s1[i-1]!=s2[j-1]:
f[i][j]=min(f[i-1][j]+1,f[i][j-1]+1,f[i-1][j-1]+1)
else:
f[i][j]=f[i-1][j-1]
print(f[i][j])
|
s366752007 | p03523 | u366886346 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 214 | You are given a string S. Takahashi can insert the character `A` at any position in this string any number of times. Can he change S into `AKIHABARA`? | s=input()
str=["A","K","I","H","A","B","A","R","A"]
count=0
for i in range(9):
if s[count]==str[i]:
count+=1
if count==len(s):
break
if count!=len(s):
print("No")
else:
print("Yes")
| s739971810 | Accepted | 17 | 3,064 | 367 | s=input()
str=["A","K","I","H","A","B","A","R","A"]
count=0
ans=0
for i in range(9):
if count==len(s) and i<8:
ans=1
break
elif count==len(s):
break
if s[count]==str[i]:
count+=1
elif s[count]!=str[i] and str[i]!="A":
ans=1
break
if len(s)>9:
ans=1
if ans==1:
print("NO")
else:
print("YES")
|
s210393673 | p02927 | u414699019 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 162 | Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have? | M,D=map(int,input().split())
result=0
for i in (2,D//10+1):
if i<D//10:
result+=M//i-1
else:
if D%10>=2:
result+=min(M//i,D%10)-1
print(result) | s928446904 | Accepted | 17 | 3,060 | 250 | M,D=map(int,input().split())
result=0
for i in range(2,D//10+1):
if i<D//10:
result+=max(min(M//i,9)-1,0)
#print(i,min(M//i,9)-1)
else:
if D%10>=2:
result+=max(min(M//i,D%10)-1,0)
#print(i,min(M//i,D%10)-1)
print(result) |
s404384449 | p03472 | u941753895 | 2,000 | 262,144 | Wrong Answer | 491 | 21,960 | 346 | You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order: * Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times. * Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana. The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total? | N,H=map(int,input().split())
if N==1 and H==10:
exit()
l=[]
for i in range(N):
a,b=map(int,input().split())
l.append([a,b])
l=sorted(l,key=lambda x:x[0])
ax=l[-1][0]
bx=l[-1][1]
c=0
for x in l[:-1]:
if H<=bx:
c+=1
print(c)
exit()
if x[1]>ax:
H-=x[1]
c+=1
H-=bx
c+=1
if H<=0:
print(c)
exit()
c+=-(-H//ax)
print(c) | s662276519 | Accepted | 442 | 37,052 | 852 | import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,queue,copy
sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()
def main():
n,h=LI()
l=[]
for _ in range(n):
a,b=LI()
l.append([a,-1])
l.append([b,1])
l=sorted(l,key=lambda x:x[0],reverse=True)
# print(l)
ans=0
for i,x in enumerate(l):
if x[1]==1:
ans+=1
h-=x[0]
else:
ans+=-(-h//x[0])
return ans
if h<=0:
return ans
return ans
# main()
print(main())
|
s787457914 | p02414 | u567380442 | 1,000 | 131,072 | Wrong Answer | 30 | 6,720 | 253 | Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula: \\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\] where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively. | n, m, L = map(int, input().split())
a = [list(map(int, input().split())) for _ in range(m)]
b = [list(map(int, input().split())) for _ in range(L)]
c = [[sum(ak * bk for ak, bk in zip(ai,bj)) for bj in zip(*b)] for ai in a]
for ci in c:
print(*ci) | s229321596 | Accepted | 200 | 7,768 | 253 | n, m, L = map(int, input().split())
a = [list(map(int, input().split())) for _ in range(n)]
b = [list(map(int, input().split())) for _ in range(m)]
c = [[sum(ak * bk for ak, bk in zip(ai,bj)) for bj in zip(*b)] for ai in a]
for ci in c:
print(*ci) |
s757812534 | p03377 | u945181840 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 95 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | A, B, X = map(int, input().split())
if A + B >= X >= A:
print('Yes')
else:
print('No') | s284851997 | Accepted | 17 | 2,940 | 95 | A, B, X = map(int, input().split())
if A + B >= X >= A:
print('YES')
else:
print('NO') |
s765109096 | p03606 | u598229387 | 2,000 | 262,144 | Wrong Answer | 21 | 3,188 | 144 | Joisino is working as a receptionist at a theater. The theater has 100000 seats, numbered from 1 to 100000. According to her memo, N groups of audiences have come so far, and the i-th group occupies the consecutive seats from Seat l_i to Seat r_i (inclusive). How many people are sitting at the theater now? | n = int(input())
lr = [[int(i) for i in input().split()] for j in range(n)]
ans = 0
for i in range(n):
ans += lr[i][1] - lr[i][0]
print(ans) | s440967473 | Accepted | 20 | 3,188 | 146 | n = int(input())
lr = [[int(i) for i in input().split()] for j in range(n)]
ans = 0
for i in range(n):
ans += lr[i][1] - lr[i][0]+1
print(ans) |
s977550644 | p02612 | u986942034 | 2,000 | 1,048,576 | Wrong Answer | 29 | 9,140 | 24 | 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. | print(int(input())%1000) | s401262940 | Accepted | 30 | 9,148 | 81 | N = int(input())
change = 1000 - (N % 1000) if N % 1000 != 0 else 0
print(change) |
s704199915 | p03730 | u662613022 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 163 | 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`. | N,M,R = map(int,input().split())
ans = ''
for i in range(M+1):
if (i * N) % M == R:
ans = 'Yes'
break;
if ans == '':
ans = 'No'
print(ans) | s617321448 | Accepted | 17 | 3,060 | 163 | N,M,R = map(int,input().split())
ans = ''
for i in range(M+1):
if (i * N) % M == R:
ans = 'YES'
break;
if ans == '':
ans = 'NO'
print(ans) |
s381844857 | p02612 | u801701525 | 2,000 | 1,048,576 | Wrong Answer | 26 | 9,140 | 45 | 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(1000-(N-round(N,-3))) | s512086278 | Accepted | 29 | 9,148 | 138 | N = int(input())
while True:
if N >= 1000:
N -= 1000
else:
break
if N == 0:
print(0)
else:
print(1000-N) |
s204251459 | p03852 | u138486156 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 720 | 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`. | s = input()
s_list = list(s)
while True:
l5 = s_list[-5:]
l6 = s_list[-6:]
l7 = s_list[-7:]
s5 = s6 = s7 = ''
for ch in l5:
s5 += ch
# print("s5 :", s5)
for ch in l6:
s6 += ch
# print("s6 :", s6)
for ch in l7:
s7 += ch
# print("s7 :", s7)
if (s5 == 'erase') | (s5 == 'dream'):
s_list = s_list[:-5]
# print("s_list:", s_list)
elif s6 == 'eraser':
s_list = s_list[:-6]
# print("s_list:", s_list)
elif s7 == 'dreamer':
s_list = s_list[:-7]
# print("s_list:", s_list)
else:
print('NO')
# print("s_list:", s_list)
exit()
if s_list == []:
print('YES')
exit()
| s484364682 | Accepted | 17 | 2,940 | 132 | ch = input()
if (ch == 'a') | (ch == 'i') | (ch == 'u') | (ch == 'e') | (ch == 'o'):
print('vowel')
else:
print('consonant') |
s009798890 | p03457 | u159863320 | 2,000 | 262,144 | Wrong Answer | 320 | 25,768 | 476 | AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan. | n = int(input())
t = []
p = [0, 0, 0]
while(n):
a, b, c = input().split(' ')
t.append([int(a), int(b), int(c)])
n-=1
for i in t:
while(p[0] < i[0]):
p[0]+=1
if i[1] > p[1]:
p[1]+=1
elif i[2] > p[2]:
p[2]+=1
elif i[1] < p[1]:
p[1]-=1
elif i[2] < p[2]:
p[2]-=1
else:
p[1]-=1
print(p)
if p != i:
print('No')
exit()
print('Yes') | s677448435 | Accepted | 265 | 25,384 | 459 | n = int(input())
t = []
p = [0, 0, 0]
while(n):
a, b, c = input().split(' ')
t.append([int(a), int(b), int(c)])
n-=1
for i in t:
while(p[0] < i[0]):
p[0]+=1
if i[1] > p[1]:
p[1]+=1
elif i[2] > p[2]:
p[2]+=1
elif i[1] < p[1]:
p[1]-=1
elif i[2] < p[2]:
p[2]-=1
else:
p[1]-=1
if p != i:
print('No')
exit()
print('Yes') |
s747074537 | p03643 | u532966492 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 13 | This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer. | "ABC"+input() | s423618846 | Accepted | 17 | 2,940 | 20 | print("ABC"+input()) |
s175641302 | p03068 | u092460072 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 126 | You are given a string S of length N consisting of lowercase English letters, and an integer K. Print the string obtained by replacing every character in S that differs from the K-th character of S, with `*`. | n=int(input())
s=input()
k=int(input())
j=s[k-1]
new_s=""
for i in range(n):
if s[i]!=j:
new_s+="*"
else:
new_s+=j | s739371613 | Accepted | 17 | 3,064 | 139 | n=int(input())
s=input()
k=int(input())
j=s[k-1]
new_s=""
for i in range(n):
if s[i]!=j:
new_s+="*"
else:
new_s+=j
print(new_s) |
s466441737 | p03673 | u290326033 | 2,000 | 262,144 | Wrong Answer | 2,105 | 25,156 | 123 | You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations. | n = int(input())
A = list(map(int, input().split()))
B = []
for i in range(n):
B.append(A[i])
B.reverse()
print(B) | s819262256 | Accepted | 215 | 25,920 | 307 | from collections import deque
n = int(input())
A = list(map(int, input().split()))
B = deque()
for i in range(n):
if(i%2 == 0):
B.append(A[i])
if(i%2 == 1):
B.appendleft(A[i])
if(n%2 == 1):
B.reverse()
text = ""
for b in B:
text += str(b) + " "
text = text[:-1]
print(text) |
s466006623 | p03339 | u197968862 | 2,000 | 1,048,576 | Wrong Answer | 352 | 17,656 | 604 | There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader. Here, we do not care which direction the leader is facing. The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized. Find the minimum number of people who have to change their directions. | n = int(input())
s = input()
count_e = [0] * n
count_w = [0] * n
for i in range(n):
if i == 0:
if s[i] == 'E':
count_e[i] = 1
else:
count_w[i] = 1
continue
if s[i] == 'E':
count_e[i] = count_e[i-1] + 1
count_w[i] = count_w[i-1]
else:
count_w[i] = count_e[i-1]
count_e[i] = count_e[i-1] + 1
ans = 10 ** 9
for i in range(n):
if s[i] == 'E':
e = count_e[-1] - count_e[i]
w = count_w[i]
else:
e = count_e[-1] - count_e[i]
w = count_w[i] - 1
ans = min(ans,e+w)
print(ans) | s964991100 | Accepted | 350 | 17,652 | 604 | n = int(input())
s = input()
count_e = [0] * n
count_w = [0] * n
for i in range(n):
if i == 0:
if s[i] == 'E':
count_e[i] = 1
else:
count_w[i] = 1
continue
if s[i] == 'E':
count_e[i] = count_e[i-1] + 1
count_w[i] = count_w[i-1]
else:
count_e[i] = count_e[i-1]
count_w[i] = count_w[i-1] + 1
ans = 10 ** 9
for i in range(n):
if s[i] == 'E':
e = count_e[-1] - count_e[i]
w = count_w[i]
else:
e = count_e[-1] - count_e[i]
w = count_w[i] - 1
ans = min(ans,e+w)
print(ans) |
s723642405 | p02277 | u782850499 | 1,000 | 131,072 | Wrong Answer | 20 | 7,784 | 990 | Let's arrange a deck of cards. Your task is to sort totally n cards. A card consists of a part of a suit (S, H, C or D) and an number. Write a program which sorts such cards based on the following pseudocode: Partition(A, p, r) 1 x = A[r] 2 i = p-1 3 for j = p to r-1 4 do if A[j] <= x 5 then i = i+1 6 exchange A[i] and A[j] 7 exchange A[i+1] and A[r] 8 return i+1 Quicksort(A, p, r) 1 if p < r 2 then q = Partition(A, p, r) 3 run Quicksort(A, p, q-1) 4 run Quicksort(A, q+1, r) Here, A is an array which represents a deck of cards and comparison operations are performed based on the numbers. Your program should also report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance). | def partition(A, p, r):
x = int(A[r-1][2])
i = p-1
for j in range(p,r-1):
if int(A[j][2]) <= x:
i = i + 1
A[i],A[j] = A[j],A[i]
A[i+1],A[r-1] = A[r-1],A[i+1]
return i + 1
def qsort(A,p,r):
if p < r:
q = partition(A,p,r)
qsort(A,p,q-1)
qsort(A,q+1,r)
return A
def isStable(a, b):
length = int(len(a))
for i in range(length):
for j in range(i+1,length-1) :
for x in range(length):
for y in range(x+1,length-1) :
if a[i][2]==a[j][2] and a[i] == b[y] and a[j] == b[x]:
return False
break
return True
if __name__ == '__main__':
cnt = int(input())
A = []
for i in range(cnt):
A.append(input())
A2 = A[:]
p = 0
r = len(A)
B = qsort(A,p,r)
if isStable(A2,B):
print("Stable")
else:
print("Not Stable")
for i in B:
print (i) | s449984603 | Accepted | 1,640 | 34,644 | 919 | def partition(A, p, r):
x = int(A[r-1][1])
i = p-1
for j in range(p,r-1):
if int(A[j][1]) <= x:
i = i + 1
A[i],A[j] = A[j],A[i]
A[i+1],A[r-1] = A[r-1],A[i+1]
return i + 1
def qsort(A,p,r):
if p < r:
q = partition(A,p,r)
qsort(A,p,q)
qsort(A,q+1,r)
return A
def isStable(a):
length = int(len(a))
for i in range(length):
for j in range(i+1,length-1) :
if a[i][1] == a[j][1] and a[i][2] > a[j][2]:
return False
break
return True
if __name__ == '__main__':
cnt = int(input())
A = []
for i in range(cnt):
N = input().split()
N.append(i)
A.append(N)
p = 0
r = len(A)
B = qsort(A,p,r)
if isStable(B):
print("Stable")
else:
print("Not stable")
for i in B:
print ("{0} {1}".format(i[0],i[1])) |
s713050687 | p03493 | u925364229 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 56 | 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 = input()
ans = 0
for i in S:
if i == 1:
ans += 1
| s433614348 | Accepted | 17 | 2,940 | 69 | S = input()
ans = 0
for i in S:
if i == "1":
ans += 1
print(ans) |
s077971576 | p03795 | u652569315 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 37 | 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-(n//3)*40) | s173949917 | Accepted | 17 | 2,940 | 39 | n=int(input())
print(800*n-(n//15)*200) |
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