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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s579136694
|
p02407
|
u483716678
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 124 |
Write a program which reads a sequence and prints it in the reverse order.
|
n = int(input())
y =list(map(int,input().split()))
x = sorted(y,reverse=True)
for i in range(n):
print(x[i],end=' ')
|
s158307840
|
Accepted
| 20 | 5,604 | 240 |
n = int(input())
y =list(map(int,input().split()))
x = list()
for i in range(n-1,-1,-1):
arry = y[i]
x.append(arry)
for i in range(n):
if(i == n-1):
print('{}'.format(x[i]))
else:
print(x[i],end=' ')
|
s874018942
|
p03129
|
u219015402
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 98 |
Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.
|
N, K = map(int,input().split())
if N>=(K-1)*2 + 1:
ans = 'yes'
else:
ans = 'no'
print(ans)
|
s274520207
|
Accepted
| 17 | 2,940 | 98 |
N, K = map(int,input().split())
if N>=(K-1)*2 + 1:
ans = 'YES'
else:
ans = 'NO'
print(ans)
|
s479097895
|
p03644
|
u455533363
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 295 |
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n = int(input())
a = []
b=[]
for i in range(1,n+1):
a.append(i)
count = 0
print(a)
print(len(a),n)
for i in range(1,n+1):
b.append(count)
count = 0
while True:
if (a[i-1]%2) == 0:
count += 1
a[i-1] = a[i-1]/2
print(a[i-1])
else:
break
print(max(b)+1)
|
s467903707
|
Accepted
| 17 | 3,064 | 274 |
n = int(input())
a = []
b=[]
for i in range(1,n+1):
a.append(i)
count = 0
for i in range(1,n+1):
count = 0
while True:
if (a[i-1]%2) == 0:
count += 1
a[i-1] = a[i-1]/2
else:
break
b.append(count)
c =max(b)
print(b.index(c)+1)
|
s219259630
|
p03110
|
u266171694
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 161 |
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
n = int(input())
total = 0
for _ in range(n):
x, u = input().split()
if u == 'JPY':
total += int(x)
else:
total += float(x) * 38e4
|
s407716656
|
Accepted
| 17 | 2,940 | 175 |
n = int(input())
total = 0
for _ in range(n):
x, u = input().split()
if u == 'JPY':
total += int(x)
else:
total += float(x) * 38e4
print(total)
|
s899783685
|
p03610
|
u371467115
| 2,000 | 262,144 |
Wrong Answer
| 20 | 4,268 | 43 |
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s=list(input())
s=s[1::2]
print("".join(s))
|
s429091903
|
Accepted
| 19 | 4,264 | 43 |
s=list(input())
s=s[::2]
print("".join(s))
|
s874599285
|
p03699
|
u111525113
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 234 |
You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?
|
n = int(input())
ans = []
for i in range(n):
ans.append(int(input()))
nm = 0
for i in ans:
if i %10 != 0:
nm = i
print(i)
break
if sum(ans) % 10 != 0:
print(sum(ans))
elif nm !=0:
print(sum(ans)-i)
else:
print(0)
|
s918031033
|
Accepted
| 17 | 3,060 | 232 |
n = int(input())
ans = []
for i in range(n):
ans.append(int(input()))
nm = 0
ans.sort()
for i in ans:
if i %10 != 0:
nm = i
break
if sum(ans) % 10 != 0:
print(sum(ans))
elif nm !=0:
print(sum(ans)-i)
else:
print(0)
|
s950300639
|
p03836
|
u982591663
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,064 | 1,085 |
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx, sy, tx, ty = map(int, input().split())
ans = ""
current = [sx, sy]
# (0,0), (1,2)
while current[0] != tx:
current[0] += 1
ans += "R"
while current[1] != ty:
current[1] += 1
ans += "U"
while current[0] != sx:
current[0] -= 1
ans += "L"
while current[1] != sy:
current[1] -= 1
ans += "D"
current[1] -= 1
ans += "D"
while current[0] != tx+1:
current[0] += 1
ans += "R"
while current[1] != ty:
current[1] += 1
ans += "U"
current[0] -= 1
ans += "L"
current[1] += 1
ans += "U"
while current[0] != tx+1:
current[0] += 1
ans += "L"
while current[1] != ty:
current[1] -= 1
ans += "D"
current[0] += 1
ans += "R"
print(ans)
|
s140243269
|
Accepted
| 17 | 3,060 | 611 |
sx, sy, tx, ty = map(int, input().split())
ans = ""
diff_x = abs(tx-sx)
diff_y = abs(ty-sy)
ans += "U" * diff_y + "R" * diff_x
ans += "D" * diff_y + "L" * diff_x
ans += "L" * 1 + "U" * (diff_y+1)
ans += "R" * (diff_x+1) + "D" * 1
ans += "R" * 1 + "D" * (diff_y+1)
ans += "L" * (diff_x+1) + "U" * 1
print(ans)
|
s798966801
|
p03577
|
u334712262
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 32 |
Rng is going to a festival. The name of the festival is given to you as a string S, which ends with `FESTIVAL`, from input. Answer the question: "Rng is going to a festival of what?" Output the answer. Here, assume that the name of "a festival of s" is a string obtained by appending `FESTIVAL` to the end of s. For example, `CODEFESTIVAL` is a festival of `CODE`.
|
print(input()[len('FESTIVAL'):])
|
s617520467
|
Accepted
| 17 | 2,940 | 34 |
print(input()[:-len('FESTIVAL')])
|
s816826973
|
p03854
|
u808373096
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 33,096 | 305 |
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 = list(input())
while len(S) > 0:
print(S)
if ''.join(S[-5:]) == 'dream' or ''.join(S[-5:]) == 'erase':
S = S[:-5]
elif ''.join(S[-6:]) == 'eraser':
S = S[:-6]
elif ''.join(S[-7:]) == 'dreamer':
S = S[-7:]
else:
print('NO')
quit()
print('YES')
|
s120629858
|
Accepted
| 67 | 3,188 | 250 |
S = input()
while len(S) > 0:
if S[-5:] == 'dream' or S[-5:] == 'erase':
S = S[:-5]
elif S[-6:] == 'eraser':
S = S[:-6]
elif S[-7:] == 'dreamer':
S = S[:-7]
else:
print('NO')
quit()
print('YES')
|
s687757382
|
p03139
|
u010090035
| 2,000 | 1,048,576 |
Wrong Answer
| 32 | 8,976 | 77 |
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
n,a,b=map(int,input().split())
high=min(a,b)
low=max(n-a-b,0)
print(high,low)
|
s236118940
|
Accepted
| 27 | 9,108 | 77 |
n,a,b=map(int,input().split())
high=min(a,b)
low=max(0,a+b-n)
print(high,low)
|
s350231973
|
p03407
|
u021548497
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 114 |
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.
|
import sys
a, b, c = map(int, input().split())
if a == c or b == c or a+b == c:
print("Yes")
else:
print("No")
|
s572431048
|
Accepted
| 17 | 2,940 | 94 |
import sys
a, b, c = map(int, input().split())
if c <= a+b:
print("Yes")
else:
print("No")
|
s177735295
|
p03337
|
u574053975
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 192 |
You are given two integers A and B. Find the largest value among A+B, A-B and A \times B.
|
a=list(map(int,input().split()))
if a[0]+a[1]>a[0]-a[1]&a[0]+a[1]>a[0]*a[1]:
print(a[0]+a[1])
elif a[0]*a[1]>a[0]+a[1]&a[0]*a[1]>a[0]-a[1]:
print(a[0]*a[1])
else :
print(a[0]-a[1])
|
s103927458
|
Accepted
| 17 | 3,064 | 200 |
a=list(map(int,input().split()))
if (a[0]+a[1]>a[0]-a[1])&(a[0]+a[1]>a[0]*a[1]):
print(a[0]+a[1])
elif (a[0]*a[1]>a[0]+a[1])&(a[0]*a[1]>a[0]-a[1]):
print(a[0]*a[1])
else :
print(a[0]-a[1])
|
s673987175
|
p02400
|
u539789745
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,628 | 131 |
Write a program which calculates the area and circumference of a circle for given radius r.
|
import math
def main():
r = float(input())
print(2 * math.pi * r, math.pi * r * r)
if __name__ == "__main__":
main()
|
s059943655
|
Accepted
| 20 | 5,640 | 175 |
import math
def main():
r = float(input())
print("{:.6f}".format(math.pi * r * r),
"{:.6f}".format(2 * math.pi * r))
if __name__ == "__main__":
main()
|
s225232274
|
p02850
|
u102275718
| 2,000 | 1,048,576 |
Wrong Answer
| 2,106 | 61,968 | 813 |
Given is a tree G with N vertices. The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i. Consider painting the edges in G with some number of colors. We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different. Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.
|
import sys
input = sys.stdin.readline
n = int(input())
ab = [list(map(int, input().split())) for _ in range(n-1)]
graph = [[] for _ in range(n)]
for a, b in ab:
graph[a-1].append((b-1, -1))
graph[b-1].append((a-1, -1))
print(graph)
stack = [(0, -1)]
visited = [-1]*n
k = 0
while stack:
node, color = stack.pop()
visited[node] = 1
if k < len(graph[node]):
k = len(graph[node])
tmp = 1
for i, (child, child_color) in enumerate(graph[node]):
if visited[child] == 1:
continue
if child_color == -1:
if tmp == color:
tmp += 1
graph[node][i] = (child, tmp)
stack.append((child, tmp))
tmp +=1
print(k)
for a, b in ab:
print(list(filter(lambda x: x[0]==(b-1), graph[a-1]))[0][1])
|
s276423829
|
Accepted
| 558 | 44,104 | 741 |
import sys
input = sys.stdin.readline
from collections import deque
n = int(input())
ab = [list(map(int, input().split())) for _ in range(n-1)]
graph = [[] for _ in range(n)]
for a, b in ab:
graph[a-1].append(b-1)
graph[b-1].append(a-1)
stack = deque([0])
parent = [None]*n
colors = [0]*n
while stack:
node = stack.popleft()
color = 1
for child in graph[node]:
if child == parent[node]:
continue
parent[child] = node
if color == colors[node]:
color += 1
colors[child] = color
color += 1
stack.append(child)
k = max(colors)
print(k)
for a, b in ab:
if (a-1) == parent[b-1]:
print(colors[b-1])
else:
print(colors[a-1])
|
s440937858
|
p00031
|
u024715419
| 1,000 | 131,072 |
Wrong Answer
| 30 | 5,592 | 328 |
็ฅๆฏใๅคฉ็งคใไฝฟใฃใฆใใพใใๅคฉ็งคใฏใไบใคใฎ็ฟใฎไธกๆนใซๅใ็ฎๆนใฎใใฎใ่ผใใใจ้ฃๅใใใใใงใชใๅ ดๅใซใฏใ้ใๆนใซๅพใใพใใ10 ๅใฎๅ้
ใฎ้ใใฏใ่ปฝใ้ ใซ 1g, 2g, 4g, 8g, 16g, 32g, 64g, 128g, 256g, 512g ใงใใ ็ฅๆฏใฏใใ1kg ใใใใพใงใฐใฉใ ๅไฝใง้ใใใฎใใใใจ่จใใพใใใใใใใ่ฉฆใใซใใใใซใใใธใฅใผในใฎ้ใใ้ใฃใฆใใใจ่จใฃใฆใฟใใจใ็ฅๆฏใฏๅทฆใฎ็ฟใซใธใฅใผในใใๅณใฎ็ฟใซ 8g ใจ64g ใจ128g ใฎๅ้
ใ่ผใใฆ้ฃๅใใใฆใใใใๅ้
ใฎ็ฎๆนใฎๅ่จใฏ 200g ใ ใใใใธใฅใผในใฎ็ฎๆนใฏ 200g ใญใใฉใใๆญฃใใใงใใใ๏ผใใจ็ญใใพใใใ ๅทฆใฎ็ฟใซ่ผใใๅ็ฉใฎ้ใใไธใใใฎใงใๅคฉ็งคใงไธใใใใ้ใฟใฎๅ็ฉใจ้ฃๅใใใใจใใซใๅณใฎ็ฟใซ่ผใใๅ้
ใ่ปฝใ้ ใซๅบๅใใใใญใฐใฉใ ใไฝๆใใฆไธใใใใใ ใใ้ใในใๅ็ฉใฎ้ใใฏใใในใฆใฎๅ้
ใฎ้ใใฎๅ่จ (=1023g) ไปฅไธใจใใพใใ
|
w_a = [2,4,8,16,32,64,128,256,512]
while True:
try:
w_b = int(input())
b = list(bin(w_b))[::-1]
w_c = []
for i in range(len(b)):
if b[i] == "b":
break
elif b[i] == "1":
w_c.append(w_a[i])
print(*w_c)
except:
break
|
s771257681
|
Accepted
| 20 | 5,600 | 330 |
w_a = [1,2,4,8,16,32,64,128,256,512]
while True:
try:
w_b = int(input())
b = list(bin(w_b))[::-1]
w_c = []
for i in range(len(b)):
if b[i] == "b":
break
elif b[i] == "1":
w_c.append(w_a[i])
print(*w_c)
except:
break
|
s635860419
|
p02902
|
u994521204
| 2,000 | 1,048,576 |
Wrong Answer
| 676 | 8,436 | 1,558 |
Given is a directed graph G with N vertices and M edges. The vertices are numbered 1 to N, and the i-th edge is directed from Vertex A_i to Vertex B_i. It is guaranteed that the graph contains no self-loops or multiple edges. Determine whether there exists an induced subgraph (see Notes) of G such that the in-degree and out-degree of every vertex are both 1. If the answer is yes, show one such subgraph. Here the null graph is not considered as a subgraph.
|
from collections import deque
import sys
input = sys.stdin.buffer.readline
n, m = map(int, input().split())
graph = [[] for _ in range(n + 1)]
for _ in range(m):
a, b = map(int, input().split())
graph[a].append(b)
infi = 10 ** 10
def bfs(start):
res = (infi, -1)
dist = [-1] * (n + 1)
par = [-1] * (n + 1)
que = deque([])
que.append(start)
dist[start] = 0
while que:
v = que.popleft()
for u in graph[v]:
if dist[u] >= 0:
continue
dist[u] = dist[v] + 1
par[u] = v
que.append(u)
for i in range(1, n + 1):
if i == start:
continue
if dist[i] == -1:
continue
if start in graph[i]:
if res[0] > dist[i] + 1:
res = (dist[i] + 1, i)
if res[0] == infi:
return -1, []
cycle_num = res[0]
v = res[1]
CYCLE = deque([])
while v > 0:
CYCLE.appendleft(v)
v = par[v]
return cycle_num, CYCLE
jisho = {}
ans = (-1, [])
cycle_num = infi
kouho_v = -1
for i in range(1, n + 1):
temp_num, temp_cycle = bfs(i)
print(i, temp_num, temp_cycle)
if temp_num == -1:
continue
if cycle_num > temp_num:
cycle_num = min(cycle_num, temp_num)
kouho_v = i
jisho[i] = temp_cycle
if kouho_v == -1:
print(-1)
exit()
else:
ans_num = cycle_num
ans_list = jisho[kouho_v]
print(cycle_num)
print(*ans_list, sep="\n")
|
s253831309
|
Accepted
| 570 | 3,700 | 1,523 |
from collections import deque
import sys
input = sys.stdin.buffer.readline
n, m = map(int, input().split())
graph = [[] for _ in range(n + 1)]
for _ in range(m):
a, b = map(int, input().split())
graph[a].append(b)
infi = 10 ** 10
def bfs(start):
res = (infi, -1)
dist = [-1] * (n + 1)
par = [-1] * (n + 1)
que = deque([])
que.append(start)
dist[start] = 0
while que:
v = que.popleft()
for u in graph[v]:
if dist[u] >= 0:
continue
dist[u] = dist[v] + 1
par[u] = v
que.append(u)
for i in range(1, n + 1):
if i == start:
continue
if dist[i] == -1:
continue
if start in graph[i]:
if res[0] > dist[i] + 1:
res = (dist[i] + 1, i)
if res[0] == infi:
return -1, []
cycle_num = res[0]
v = res[1]
CYCLE = deque([])
while v > 0:
CYCLE.appendleft(v)
v = par[v]
return cycle_num, CYCLE
jisho = {}
ans = (-1, [])
cycle_num = infi
kouho_v = -1
for i in range(1, n + 1):
temp_num, temp_cycle = bfs(i)
if temp_num == -1:
continue
if cycle_num > temp_num:
cycle_num = min(cycle_num, temp_num)
kouho_v = i
jisho[i] = temp_cycle
if kouho_v == -1:
print(-1)
exit()
else:
ans_num = cycle_num
ans_list = jisho[kouho_v]
print(cycle_num)
print(*ans_list, sep="\n")
|
s301906433
|
p02607
|
u848535504
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,096 | 142 |
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())
n = list(map(int,input().split()))
ans = 0
for i in range(N):
if i % 2 == 0 and (i+1)%2 == 0:
ans += 1
print(ans)
|
s184585458
|
Accepted
| 23 | 8,996 | 146 |
N = int(input())
n = list(map(int,input().split()))
ans = 0
for i in range(N):
if n[i] % 2 == 1 and (i-1)%2 == 1:
ans += 1
print(ans)
|
s627785407
|
p03493
|
u243572357
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 47 |
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.
|
print(list(map(int, input().split())).count(1))
|
s790083395
|
Accepted
| 17 | 2,940 | 37 |
a = list(input())
print(a.count('1'))
|
s767216921
|
p03564
|
u724707209
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 236 |
Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.
|
if __name__ == "__main__":
n = int(input())
k = int(input())
result=1
while(n==0):
if (result+k > result*2):
result=result*2
else:
result=result+k
n=n-1
print(result)
|
s263188319
|
Accepted
| 17 | 2,940 | 236 |
if __name__ == "__main__":
n = int(input())
k = int(input())
result=1
while(n!=0):
if (result+k > result*2):
result=result*2
else:
result=result+k
n=n-1
print(result)
|
s247345663
|
p03487
|
u064246852
| 2,000 | 262,144 |
Wrong Answer
| 96 | 17,780 | 262 |
You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N). Your objective is to remove some of the elements in a so that a will be a **good sequence**. Here, an sequence b is a **good sequence** when the following condition holds true: * For each element x in b, the value x occurs exactly x times in b. For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not. Find the minimum number of elements that needs to be removed so that a will be a good sequence.
|
n = int(input())
a = list(map(int,input().split()))
dic = {}
for i in range(n):
if a[i] in dic:
dic[a[i]] += 1
else:
dic[a[i]] = 1
ans = 0
for i in dic:
if dic[i] < i:
ans += dic[i]
else:
ans += i-dic[i]
print(ans)
|
s348623466
|
Accepted
| 94 | 17,780 | 263 |
n = int(input())
a = list(map(int,input().split()))
dic = {}
for i in range(n):
if a[i] in dic:
dic[a[i]] += 1
else:
dic[a[i]] = 1
ans = 0
for i in dic:
if dic[i] < i:
ans += dic[i]
else:
ans += dic[i]-i
print(ans)
|
s334951504
|
p03456
|
u711295009
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 602 |
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.
|
import math
import itertools
a, b = map(str, input().split())
C = a+b
c = int(C)
I=c
s=0
R=int(I)
L=[]
while s==0:
for i in range(2,R+1):
if I%i==0:
I=I/i
if I==1:
s=1
L.append(i)
break
index =0
sameCCount =0
flag =0
while index < len(L)-1:
if L[index] == L[index+1]:
sameCCount+=1
print(sameCCount, index)
else:
if sameCCount%2 ==0 or sameCCount==0:
flag=1
break
else:
sameCCount=0
index+=1
if flag ==0:
print("Yes")
else:
print("No")
|
s311858307
|
Accepted
| 18 | 3,064 | 702 |
import math
import itertools
a, b = map(str, input().split())
C = a+b
c = int(C)
I=c
s=0
R=int(I)
L=[]
while s==0:
for i in range(2,R+1):
if I%i==0:
I=I/i
if I==1:
s=1
L.append(i)
break
beforeC =""
count=0
index=0
flag=0
while index < len(L):
if beforeC =="":
beforeC = L[index]
count+=1
else:
if L[index] == beforeC:
count+=1
else:
if count%2==0:
count =1
beforeC = L[index]
else:
flag=1
break
index+=1
if count==1:
flag=1
if flag ==1:
print("No")
else:
print("Yes")
|
s275806804
|
p03494
|
u593567568
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 152 |
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 = 10 ** 6
for a in A:
c = 0
while a % 2 == 0:
a //= 2
ans = min(c,ans)
print(ans)
|
s284657121
|
Accepted
| 18 | 2,940 | 164 |
N = int(input())
A = list(map(int,input().split()))
ans = 10 ** 6
for a in A:
c = 0
while a % 2 == 0:
a //= 2
c += 1
ans = min(c,ans)
print(ans)
|
s766007853
|
p02410
|
u801346721
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,504 | 237 |
Write a program which reads a $ n \times m$ matrix $A$ and a $m \times 1$ vector $b$, and prints their product $Ab$. A column vector with m elements is represented by the following equation. \\[ b = \left( \begin{array}{c} b_1 \\\ b_2 \\\ : \\\ b_m \\\ \end{array} \right) \\] A $n \times m$ matrix with $m$ column vectors, each of which consists of $n$ elements, is represented by the following equation. \\[ A = \left( \begin{array}{cccc} a_{11} & a_{12} & ... & a_{1m} \\\ a_{21} & a_{22} & ... & a_{2m} \\\ : & : & : & : \\\ a_{n1} & a_{n2} & ... & a_{nm} \\\ \end{array} \right) \\] $i$-th element of a $m \times 1$ column vector $b$ is represented by $b_i$ ($i = 1, 2, ..., m$), and the element in $i$-th row and $j$-th column of a matrix $A$ is represented by $a_{ij}$ ($i = 1, 2, ..., n,$ $j = 1, 2, ..., m$). The product of a $n \times m$ matrix $A$ and a $m \times 1$ column vector $b$ is a $n \times 1$ column vector $c$, and $c_i$ is obtained by the following formula: \\[ c_i = \sum_{j=1}^m a_{ij}b_j = a_{i1}b_1 + a_{i2}b_2 + ... + a_{im}b_m \\]
|
n, m = map(int, input().split())
M1 = []
M2 = []
for i in range(n):
M1.append(list(map(int, input().split())))
for i in range(m):
M2.append(int(input()))
for i in range(n):
c = 0
for s in range(m):
c += M1[i][s] + M2[s]
print(c)
|
s642294416
|
Accepted
| 40 | 7,940 | 237 |
n, m = map(int, input().split())
M1 = []
M2 = []
for i in range(n):
M1.append(list(map(int, input().split())))
for i in range(m):
M2.append(int(input()))
for i in range(n):
c = 0
for s in range(m):
c += M1[i][s] * M2[s]
print(c)
|
s681289111
|
p03564
|
u881612683
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 185 |
Square1001 has seen an electric bulletin board displaying the integer 1. He can perform the following operations A and B to change this value: * Operation A: The displayed value is doubled. * Operation B: The displayed value increases by K. Square1001 needs to perform these operations N times in total. Find the minimum possible value displayed in the board after N operations.
|
N = int(input())
K = int(input())
def ans(i,x):
print(i)
if i == 0:
return min(2*x, x+K)
else:
return min(ans(i-1, 2*x), ans(i-1, x+K))
print(ans(N-1, 1))
|
s003777489
|
Accepted
| 17 | 2,940 | 172 |
N = int(input())
K = int(input())
def ans(i,x):
if i == 0:
return min(2*x, x+K)
else:
return min(ans(i-1, 2*x), ans(i-1, x+K))
print(ans(N-1, 1))
|
s809281302
|
p03044
|
u379142263
| 2,000 | 1,048,576 |
Wrong Answer
| 625 | 46,336 | 601 |
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
import sys
def dfs(s):
color = [0]*n
visited = [False]*n
visited[s] = True
q = [s]
while q:
p = q.pop()
visited[p] = True
for d,v in edge[p]:
if visited[v]:
continue
if d%2 != 0:
color[v] = 1
q.append(v)
return color
if __name__ == "__main__":
n = int(input())
edge = [[] for i in range(n)]
for i in range(n-1):
u,v,w = map(int,input().split())
u-=1
v-=1
edge[u].append([w,v])
edge[v].append([w,u])
ans = dfs(0)
print(*ans)
|
s115893593
|
Accepted
| 752 | 50,988 | 1,706 |
import sys
INF = 10**18
import sys
from heapq import heapify,heappop,heappush,heappushpop
class PriorityQueue:
def __init__(self,heap):
self.heap = heap
heapify(self.heap)
def push(self,item):
heappush(self.heap,item)
def pop(self):
return heappop(self.heap)
def pushpop(self,item):
return heappushpop(self.heap,item)
def __call__(self):
return self.heap
def __len__(self):
return len(self.heap)
def dijkstra(s):
d = [INF] * n
used = [True] * n
d[s] = 0
used[s] = False
edgelist = []
q = PriorityQueue(edgelist)
for e in edge[s]:
q.push(e)
while len(edgelist):
minedge = q.pop()
if not used[minedge[1]]:
continue
v = minedge[1]
d[v] = minedge[0]
used[v] = False
for e in edge[v]:
if used[e[1]]:
q.push([e[0]+d[v],e[1]])
return d
if __name__ == "__main__":
n = int(input())
edge = [[] for i in range(n)]
for i in range(n-1):
u,v,w = map(int,input().split())
u-=1
v-=1
edge[u].append([w,v])
edge[v].append([w,u])
ans = dijkstra(0)
c_ans = [0]*n
for i in range(n):
if ans[i]%2 == 0:
c_ans[i] = 0
else:
c_ans[i] = 1
for i in range(n):
print(c_ans[i])
|
s727052530
|
p03610
|
u633548583
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,316 | 50 |
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
S=list(map(str,input().split()))
s=S[::2]
print(s)
|
s388478068
|
Accepted
| 17 | 3,188 | 23 |
S=input()
print(S[::2])
|
s381693619
|
p02663
|
u296984343
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 9,192 | 213 |
In this problem, we use the 24-hour clock. Takahashi gets up exactly at the time H_1 : M_1 and goes to bed exactly at the time H_2 : M_2. (See Sample Inputs below for clarity.) He has decided to study for K consecutive minutes while he is up. What is the length of the period in which he can start studying?
|
#A - Study Scheduling
def answ():
h1, m1, h2,m2,k = map(int, input().split())
h = h2 - h1
m = m2 - m1
if m < 0:
m = m + 60
h = h - 1
time = h*60 + m
return time - k
answ()
|
s619176887
|
Accepted
| 23 | 9,172 | 164 |
h1, m1, h2,m2,k = map(int, input().split())
if h1 > h2:
h2 +=24
h = h2 - h1
m = m2 - m1
if m < 0:
m = m + 60
h = h - 1
time = h*60 + m
print(time - k)
|
s247739392
|
p03378
|
u905582793
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 138 |
There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
N,M,X = map(int,input().split())
A = list(map(int,input().split()))
for i,x in enumerate(A):
if x > X:
t = i
break
print(t,M-t)
|
s449132139
|
Accepted
| 17 | 3,060 | 143 |
N,M,X = map(int,input().split())
A = list(map(int,input().split()))
for i,x in enumerate(A):
if x > X:
t = i
break
print(min(t,M-t))
|
s745631584
|
p03779
|
u729707098
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 57 |
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.
|
import math
n = int(input())
print(int(math.sqrt(2*n))+1)
|
s241052314
|
Accepted
| 25 | 2,940 | 86 |
n = int(input())
num = 0
for i in range(n+1):
num += i
if n<=num:
print(i)
break
|
s935316964
|
p02401
|
u340901659
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,564 | 334 |
Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part.
|
while True :
i = input().split()
r = int(i[0])
op = i[1]
l = int(i[2])
if op == '?':
break
else:
if op == '+':
print(str(r+l))
elif op == '-':
print(str(r-l))
elif op == '*':
print(str(r*l))
else :
print(str(r/l))
|
s562682144
|
Accepted
| 20 | 7,656 | 335 |
while True :
i = input().split()
r = int(i[0])
op = i[1]
l = int(i[2])
if op == '?':
break
else:
if op == '+':
print(str(r+l))
elif op == '-':
print(str(r-l))
elif op == '*':
print(str(r*l))
else :
print(str(r//l))
|
s906811545
|
p03110
|
u024555159
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 176 |
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
a=int(input())
res = float(0.0)
for i in range(a):
b=input().split()
if b[1] == "JPY":
res += float(b[0])
if b[1] == "BTC":
res += float(b[0])*38000
print(res)
|
s442236660
|
Accepted
| 17 | 3,060 | 177 |
a=int(input())
res = float(0.0)
for i in range(a):
b=input().split()
if b[1] == "JPY":
res += float(b[0])
if b[1] == "BTC":
res += float(b[0])*380000
print(res)
|
s991546136
|
p03079
|
u681917640
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 91 |
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
A, B, C = map(int, input().split())
if A == B == C:
print("YES")
else:
print("NO")
|
s396241785
|
Accepted
| 17 | 2,940 | 91 |
A, B, C = map(int, input().split())
if A == B == C:
print("Yes")
else:
print("No")
|
s196267434
|
p03997
|
u534450736
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 72 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a+b) * h / 2)
|
s156698757
|
Accepted
| 20 | 2,940 | 77 |
a = int(input())
b = int(input())
h = int(input())
print(int((a+b) * h / 2))
|
s713838429
|
p04030
|
u033606236
| 2,000 | 262,144 |
Wrong Answer
| 26 | 3,952 | 192 |
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
import queue
q = queue.LifoQueue()
for s in input():
if s =="B":
if q.empty():
continue
q.get()
else:q.put(s)
while not q.empty():
print(q.get(),end="")
|
s120563622
|
Accepted
| 17 | 2,940 | 126 |
a = []
for s in input():
if "B" in s:
if not a:continue
else:a.pop()
else:a.append(s)
print(*a,sep="")
|
s163253883
|
p02378
|
u798803522
| 1,000 | 131,072 |
Wrong Answer
| 30 | 8,096 | 690 |
A bipartite graph G = (V, E) is a graph in which the vertex set V can be divided into two disjoint subsets X and Y such that every edge e โ E has one end point in X and the other end point in Y. A matching M is a subset of edges such that each node in V appears in at most one edge in M. Given a bipartite graph, find the size of the matching which has the largest size.
|
from collections import defaultdict
def dfs(here):
used[here] = 1
for near in connect[here]:
if not match[near] or (not used[near] and dfs(near)):
match[here] = near
match[near] = here
return 1
return 0
a_num, b_num, e_num = (int(n) for n in input().split(" "))
connect = defaultdict(list)
all_v = a_num + b_num
match = [0 for n in range(all_v)]
used = [0 for n in range(all_v)]
for _ in range(e_num):
v1, v2 = (int(n) for n in input().split(" "))
v2 += a_num + 1
connect[v1].append(v2)
connect[v2].append(v1)
answer = 0
for i in range(all_v):
if not match[i] and dfs(i):
answer += 1
print(answer)
|
s171385119
|
Accepted
| 90 | 8,244 | 764 |
from collections import defaultdict
def dfs(here,used,match,connect):
used[here] = 1
for near in connect[here]:
m = match[near]
if m < 0 or (not used[m] and dfs(m,used,match,connect)):
match[here] = near
match[near] = here
return 1
return 0
a_num, b_num, e_num = (int(n) for n in input().split(" "))
connect = defaultdict(list)
all_v = a_num + b_num
match = [-1 for n in range(all_v)]
for _ in range(e_num):
v1, v2 = (int(n) for n in input().split(" "))
v2 += a_num
connect[v1].append(v2)
connect[v2].append(v1)
answer = 0
for i in range(all_v):
if match[i] < 0:
used = [0 for n in range(all_v)]
if dfs(i,used,match,connect):
answer += 1
print(answer)
|
s755188368
|
p03436
|
u309834353
| 2,000 | 262,144 |
Wrong Answer
| 28 | 3,192 | 1,001 |
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 `#`.
|
H,W = map(int, input().split())
field = [list(input()) for _ in range(H)]
dy = [1,-1,0,0]
dx = [0,0,1,-1]
count = 0
# for j in range(W):
# if field[i][j] == start:
# sy, sx = i, j
# elif field[i][j] == goal:
# gy, gx = i, j
sy,sx = 0,0
gy,gx = H-1,W-1
seen = [[0 for i in range(W)] for j in range(H)]
queue = [[sy,sx]]
flag = False
while queue != []:
h,w = queue.pop(0)
before = seen[h][w]
for i in range(4):
nh = h + dy[i]
nw = w + dx[i]
if (nh == gy) and (nw == gx):
count = before+1
flag = True
if (0<=nh<H) and (0<=nw<W):
if field[nh][nw] != 'X':
if seen[nh][nw] == 0:
seen[nh][nw] = before+1
queue.append([nh,nw])
if flag:
break
whites = 0
for i in range(H):
for j in range(W):
if field[i][j] == '.':
whites += 1
print(whites,count)
print(whites - count-1)
|
s594394727
|
Accepted
| 27 | 3,064 | 1,033 |
H,W = map(int, input().split())
field = [list(input()) for _ in range(H)]
dy = [1,-1,0,0]
dx = [0,0,1,-1]
count = 0
# for j in range(W):
# if field[i][j] == start:
# sy, sx = i, j
# elif field[i][j] == goal:
# gy, gx = i, j
sy,sx = 0,0
gy,gx = H-1,W-1
seen = [[0 for i in range(W)] for j in range(H)]
queue = [[sy,sx]]
flag = False
while queue != []:
h,w = queue.pop(0)
before = seen[h][w]
for i in range(4):
nh = h + dy[i]
nw = w + dx[i]
if (nh == gy) and (nw == gx):
count = before+1
flag = True
if (0<=nh<H) and (0<=nw<W):
if field[nh][nw] != '#':
if seen[nh][nw] == 0:
seen[nh][nw] = before+1
queue.append([nh,nw])
if flag:
break
whites = 0
for i in range(H):
for j in range(W):
if field[i][j] == '.':
whites += 1
if count != 0 :
print(whites - count-1)
else:
print(-1)
|
s899152725
|
p03997
|
u457554982
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 65 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a=int(input())
b=int(input())
h=int(input())
s=(a+b)*h/2
print(s)
|
s810998911
|
Accepted
| 17 | 2,940 | 74 |
a=int(input())
b=int(input())
h=int(input())
s=(a+b)*h/2
s=int(s)
print(s)
|
s199183158
|
p03476
|
u211160392
| 2,000 | 262,144 |
Wrong Answer
| 637 | 53,768 | 415 |
We say that a odd number N is _similar to 2017_ when both N and (N+1)/2 are prime. You are given Q queries. In the i-th query, given two odd numbers l_i and r_i, find the number of odd numbers x similar to 2017 such that l_i โค x โค r_i.
|
Q = int(input())
lr = list(list(map(int,input().split())for i in range(Q)))
pn = [0,0]+[1]*(10**5-2)
O = 0
for i in range(2,317):
for j in range(i*2,10**5,i):
pn[j] = 0
p = pn.copy()
for i in range(7,10**5):
if p[i]==1:
if(pn[int((i+1)/2)]==0):
p[i] = 0
tmp = 0
p[2] =0
for i in range(10**5):
p[i] = tmp+p[i]
tmp = p[i]
print(p[:30])
for l,r in lr:
print(p[r]-p[l-1])
|
s343706052
|
Accepted
| 616 | 53,772 | 401 |
Q = int(input())
lr = list(list(map(int,input().split())for i in range(Q)))
pn = [0,0]+[1]*(10**5-2)
O = 0
for i in range(2,317):
for j in range(i*2,10**5,i):
pn[j] = 0
p = pn.copy()
for i in range(7,10**5):
if p[i]==1:
if(pn[int((i+1)/2)]==0):
p[i] = 0
tmp = 0
p[2] =0
for i in range(10**5):
p[i] = tmp+p[i]
tmp = p[i]
for l,r in lr:
print(p[r]-p[l-1])
|
s926759749
|
p03474
|
u572142121
| 2,000 | 262,144 |
Wrong Answer
| 29 | 9,104 | 104 |
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
x,y=map(int, input().split())
S=input()
if len(S)==x+y and S[x]=='-':
print('Yes')
else:
print('No')
|
s427381534
|
Accepted
| 28 | 9,028 | 126 |
x,y=map(int, input().split())
S=input()
if len(S)==x+y+1 and S[x]=='-' and S.count('-')==1:
print('Yes')
else:
print('No')
|
s565108538
|
p02928
|
u477977638
| 2,000 | 1,048,576 |
Wrong Answer
| 1,080 | 3,188 | 305 |
We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.
|
n,k=map(int,input().split())
A=list(map(int,input().split()))
cnt=0
a=0
if n==1:print(0)
else:
for i in range(n-1):
for j in range(i+1,n):
if A[i]>A[j]:
cnt+=1
for i in range(n):
for j in range(n):
if A[i]>A[j]:
a+=1
print((cnt*k+a*k*(k-1)/2)%(10**9+7))
|
s225627335
|
Accepted
| 1,088 | 3,188 | 313 |
n,k=map(int,input().split())
A=list(map(int,input().split()))
cnt=0
a=0
if n==1:print(0)
else:
for i in range(n-1):
for j in range(i+1,n):
if A[i]>A[j]:
cnt+=1
for i in range(n):
for j in range(n):
if A[i]>A[j]:
a+=1
print(int(cnt*k+a*(k*(k-1)//2))%(10**9+7))
|
s450574123
|
p02259
|
u144068724
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,568 | 353 |
Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode.
|
def bubbleSort(A,N):
count = 0
flag = 1
while flag:
flag = 0
for i in range(1,N):
if A[N-i] < A[N-i-1]:
A[N-i] , A[N-i-1] = A[N-i-1] , A[N-i]
flag = 1
count += 1
return (count)
n = int(input())
data = [int(i) for i in input().split()]
print(bubbleSort(data,n))
|
s343032014
|
Accepted
| 20 | 7,740 | 390 |
def bubbleSort(A,N):
count = 0
flag = 1
while flag:
flag = 0
for i in range(1,N):
if A[N-i] < A[N-i-1]:
A[N-i] , A[N-i-1] = A[N-i-1] , A[N-i]
flag = 1
count += 1
return (count)
n = int(input())
data = [int(i) for i in input().split()]
a = bubbleSort(data,n)
print(" ".join(map(str,data)))
print(a)
|
s307036306
|
p02384
|
u395334793
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,752 | 1,426 |
Construct a dice from a given sequence of integers in the same way as
|
class Dice:
def __init__(self):
self.front = '1'
self.back = '6'
self.top = '5'
self.bottom = '2'
self. right = '3'
self.left = '4'
def turnRight(self):
temp = self.top
self.top = self.left
self.left = self.bottom
self.bottom = self.right
self.right = temp
return self.top
def turnLeft(self):
temp = self.top
self.top = self.right
self.right = self.bottom
self.bottom = self.left
self.left = temp
return self.top
def turnFront(self):
temp = self.top
self.top = self.back
self.back = self.bottom
self.bottom = self.front
self.front = temp
return self.top
def turnBack(self):
temp = self.top
self.top = self.front
self.front = self.bottom
self.bottom = self.back
self.back = temp
return self.top
mydice = Dice()
mydice.top, mydice.front, mydice.right, mydice.left, mydice.back, mydice.bottom = input().split()
n = int(input())
for q in range(n):
top, front = input().split()
for rotate in range(3):
mydice.turnFront()
if mydice.top == front:
break
mydice.turnRight()
if mydice.top == front:
break
mydice.turnBack()
while top != mydice.top:
mydice.turnRight()
print(mydice.right)
|
s843933521
|
Accepted
| 20 | 7,720 | 1,427 |
class Dice:
def __init__(self):
self.front = '1'
self.back = '6'
self.top = '5'
self.bottom = '2'
self. right = '3'
self.left = '4'
def turnRight(self):
temp = self.top
self.top = self.left
self.left = self.bottom
self.bottom = self.right
self.right = temp
return self.top
def turnLeft(self):
temp = self.top
self.top = self.right
self.right = self.bottom
self.bottom = self.left
self.left = temp
return self.top
def turnFront(self):
temp = self.top
self.top = self.back
self.back = self.bottom
self.bottom = self.front
self.front = temp
return self.top
def turnBack(self):
temp = self.top
self.top = self.front
self.front = self.bottom
self.bottom = self.back
self.back = temp
return self.top
mydice = Dice()
mydice.top, mydice.front, mydice.right, mydice.left, mydice.back, mydice.bottom = input().split()
n = int(input())
for q in range(n):
top, front = input().split()
for rotate in range(3):
mydice.turnFront()
if mydice.top == front:
break
mydice.turnRight()
if mydice.top == front:
break
mydice.turnFront()
while top != mydice.top:
mydice.turnRight()
print(mydice.right)
|
s306238097
|
p00051
|
u744114948
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,720 | 149 |
8 ๅใฎ 0 ใใ 9 ใพใงใฎๆฐๅญใๅ
ฅๅใใใจใใใใฎ 8 ๅใฎๆฐๅญใไธฆในๆฟใใฆใงใใใๆๅคงใฎๆดๆฐใจๆๅฐใฎๆดๆฐใฎๅทฎใๅบๅใใใใญใฐใฉใ ใไฝๆใใฆใใ ใใใไธฆใณๆฟใใฆใงใใๆฐใฏ 00135569 ใฎใใใซ 0 ใใๅงใพใฃใฆใใใใใฎใจใใพใใ
|
for _ in range(int(input ())):
s=list(input ())
s.sort ()
Max=int("".join (s))
s.reverse()
Min=int("".join(s))
print(Max-Min)
|
s753129303
|
Accepted
| 30 | 6,720 | 163 |
n = int(input())
for _ in range(n):
l = list(input())
l.sort()
min = int("".join(l))
l.reverse()
max = int("".join(l))
print(max-min)
|
s028838350
|
p04029
|
u095021077
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,092 | 32 |
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)
|
s735358492
|
Accepted
| 29 | 9,096 | 32 |
N=int(input())
print(N*(N+1)//2)
|
s852477899
|
p01421
|
u847467233
| 5,000 | 131,072 |
Wrong Answer
| 20 | 5,624 | 2,168 |
ICP city has an express company whose trucks run from the crossing S to the crossing T. The president of the company is feeling upset because all the roads in the city are one-way, and are severely congested. So, he planned to improve the maximum flow (edge disjoint paths) from the crossing S to the crossing T by reversing the traffic direction on some of the roads. Your task is writing a program to calculate the maximized flow from S to T by reversing some roads, and the list of the reversed roads.
|
# AOJ 2304 Reverse Roads
# Python3 2018.7.21 bal4u
# *******************************************
# Dinic's Max Flow Algorithm
# *******************************************
INF = 0x7fffffff
class Donic:
def __init__(self, V):
self.V = V
self.level = [0] * V
self.iter = [0] * V
self.edge = [[] for i in range(V)]
def add_edge(self, frm, to, cap):
f, t = len(self.edge[frm]), len(self.edge[to])
self.edge[frm].append([to, cap, t])
self.edge[to].append([frm, cap, f])
def bfs(self, s):
self.level = [-1] * self.V
self.level[s] = 0
Q = []
Q.append(s)
while Q:
v = Q.pop()
for to, cap, rev in self.edge[v]:
if cap > 0 and self.level[to] < 0:
self.level[to] = self.level[v] + 1
Q.append(to)
def dfs(self, v, t, f):
if v == t: return f
k = self.iter[v]
while k < len(self.edge[v]):
to, cap, rev = self.edge[v][k]
if cap > 0 and self.level[v] < self.level[to]:
d = self.dfs(to, t, f if f <= cap else cap)
if d > 0:
self.edge[v][k][1] -= d
self.edge[to][rev][1] += d
return d
self.iter[v] += 1
k += 1
return 0
def maxFlow(self, s, t):
flow = 0
while True:
self.bfs(s)
if self.level[t] < 0: break
self.iter = [0] * self.V
while True:
f = self.dfs(s, t, INF)
if f <= 0: break
flow += f
return flow
N, M = map(int, input().split())
dir = [[0 for j in range(N)] for i in range(N)]
id = [[0 for j in range(N)] for i in range(N)]
d = Donic(N)
for i in range(M):
x, y = map(int, input().split())
x, y = x-1, y-1
dir[y][x] = 1
id[y][x] = i+1
d.add_edge(x, y, 1)
S, T = map(int, input().split())
print(d.maxFlow(S-1, T-1))
ans = []
for i in range(N):
for to, cap, rev in d.edge[i]:
if cap < 1 and dir[i][to]: ans.append(id[i][to])
ans.sort()
print(len(ans))
print(*ans, sep='\n')
|
s848224057
|
Accepted
| 50 | 7,432 | 2,142 |
# AOJ 2304 Reverse Roads
# Python3 2018.7.21 bal4u
# *******************************************
# Dinic's Max Flow Algorithm
# *******************************************
class MaxFlow:
def __init__(self, V):
self.V = V
self.level = [0] * V
self.iter = [0] * V
self.edge = [[] for i in range(V)]
def add_edge(self, fr, to, cap):
f, t = len(self.edge[fr]), len(self.edge[to])
self.edge[fr].append([to, cap, t])
self.edge[to].append([fr, cap, f])
def bfs(self, s):
self.level = [-1] * self.V
self.level[s] = 0
Q = []
Q.append(s)
while Q:
v = Q.pop()
for to, cap, rev in self.edge[v]:
if cap > 0 and self.level[to] < 0:
self.level[to] = self.level[v] + 1
Q.append(to)
def dfs(self, v, t, f):
if v == t: return f
for i in range(self.iter[v], len(self.edge[v])):
to, cap, rev = self.edge[v][i]
if cap > 0 and self.level[v] < self.level[to]:
d = self.dfs(to, t, min(f, cap))
if d > 0:
self.edge[v][i][1] -= d
self.edge[to][rev][1] += d
return d
self.iter[v] = i
return 0
def maxFlow(self, s, t, INF=10**8):
flow = 0
while True:
self.bfs(s)
if self.level[t] < 0: break
self.iter = [0] * self.V
while True:
f = self.dfs(s, t, INF)
if f <= 0: break
flow += f
return flow
N, M = map(int, input().split())
dir = [[0 for j in range(N)] for i in range(N)]
id = [[0 for j in range(N)] for i in range(N)]
d = MaxFlow(N)
for i in range(M):
x, y = map(int, input().split())
x, y = x-1, y-1
dir[y][x] = 1
id[y][x] = i+1
d.add_edge(x, y, 1)
S, T = map(int, input().split())
print(d.maxFlow(S-1, T-1))
ans = []
for i in range(N):
for to, cap, rev in d.edge[i]:
if cap < 1 and dir[i][to]: ans.append(id[i][to])
ans.sort()
print(len(ans))
if len(ans) > 0: print(*ans, sep='\n')
|
s332672621
|
p03623
|
u672316981
| 2,000 | 262,144 |
Wrong Answer
| 30 | 8,984 | 100 |
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x, a, b = map(int, input().split())
if abs(a - x) < abs(a - b):
print('A')
else:
print('B')
|
s466145620
|
Accepted
| 26 | 9,160 | 100 |
x, a, b = map(int, input().split())
if abs(x - a) < abs(x - b):
print('A')
else:
print('B')
|
s897149417
|
p03779
|
u226746496
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 20,892 | 88 |
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.
|
X = int(input())
for i in range(0, X):
X -= (i+1)
if X <= 0:
print(i+1)
|
s663522078
|
Accepted
| 27 | 3,064 | 123 |
X = int(input())
result = 0
for i in range(0, X):
result += (i+1)
if X <= result:
print(i+1)
break
|
s568196720
|
p03814
|
u217303170
| 2,000 | 262,144 |
Wrong Answer
| 29 | 8,992 | 90 |
Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends with `Z`.
|
s = 'HASFJGHOGAKZZFEGA'
rs = s[::-1]
x = s.index('A')
y = rs.index('Z')
print(len(s)-y-x)
|
s927842000
|
Accepted
| 29 | 9,216 | 78 |
s = input()
rs = s[::-1]
x = s.index('A')
y = rs.index('Z')
print(len(s)-y-x)
|
s023302191
|
p04031
|
u863841238
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 130 |
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal **integers** by transforming some of them. He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (iโ j), he has to pay the cost separately for transforming each of them (See Sample 2). Find the minimum total cost to achieve his objective.
|
n = int(input())
a = list(map(int,input().split()))
avg = round(sum(a)/n)
ans = 0
for i in a:
ans += (i-avg)**2
print (avg)
|
s118861121
|
Accepted
| 18 | 3,188 | 130 |
n = int(input())
a = list(map(int,input().split()))
avg = round(sum(a)/n)
ans = 0
for i in a:
ans += (i-avg)**2
print (ans)
|
s124516138
|
p03720
|
u826263061
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 178 |
There are N cities and M roads. The i-th road (1โคiโคM) connects two cities a_i and b_i (1โคa_i,b_iโคN) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
n, m = list(map(int, input().split()))
routes = []
for i in range(m):
ri = list(map(int, input().split()))
routes.extend(ri)
for i in range(n):
print(routes.count(i))
|
s634149337
|
Accepted
| 19 | 2,940 | 180 |
n, m = list(map(int, input().split()))
routes = []
for i in range(m):
ri = list(map(int, input().split()))
routes.extend(ri)
for i in range(n):
print(routes.count(i+1))
|
s629458472
|
p03023
|
u366886346
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 68 |
Given an integer N not less than 3, find the sum of the interior angles of a regular polygon with N sides. Print the answer in degrees, but do not print units.
|
s=input()
if s.count("o")<7:
print("NO")
else:
print("YES")
|
s411627845
|
Accepted
| 17 | 2,940 | 32 |
n=int(input())
print((n-2)*180)
|
s863442749
|
p02612
|
u999287037
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,136 | 56 |
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())
while N >= 1000:
N -= 1000
print(N)
|
s809250584
|
Accepted
| 29 | 9,152 | 60 |
N = int(input())
while N > 1000:
N -= 1000
print(1000-N)
|
s127679790
|
p03359
|
u528748570
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 101 |
In AtCoder Kingdom, Gregorian calendar is used, and dates are written in the "year-month-day" order, or the "month-day" order without the year. For example, May 3, 2018 is written as 2018-5-3, or 5-3 without the year. In this country, a date is called _Takahashi_ when the month and the day are equal as numbers. For example, 5-5 is Takahashi. How many days from 2018-1-1 through 2018-a-b are Takahashi?
|
a,b = map(int, input().split())
count=0
if a < b:
count=a
else:
count=a-1
print(count)
|
s363009766
|
Accepted
| 17 | 2,940 | 102 |
a,b = map(int, input().split())
count=0
if a <= b:
count=a
else:
count=a-1
print(count)
|
s250192683
|
p03415
|
u548514780
| 2,000 | 262,144 |
Wrong Answer
| 30 | 8,960 | 58 |
We have a 3ร3 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.
|
i1=input()
i2=input()
i3=input()
print(i1[0],i2[1], i3[2])
|
s950303097
|
Accepted
| 25 | 9,012 | 72 |
i1=input()
i2=input()
i3=input()
print(str(i1[0])+str(i2[1])+str(i3[2]))
|
s932099045
|
p03778
|
u175590965
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 91 |
AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W. If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle. Find the minimum distance it needs to be moved.
|
w,a,b = map(int,input().split())
if abs(a-b) >= w:
print(0)
else:
print(abs(a-b)-w)
|
s986941129
|
Accepted
| 17 | 2,940 | 90 |
w,a,b = map(int,input().split())
if abs(a-b) <=w:
print(0)
else:
print(abs(a-b)-w)
|
s538430691
|
p03455
|
u440904221
| 2,000 | 262,144 |
Wrong Answer
| 2,103 | 2,940 | 149 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
import sys
a,b = map(str,input().split())
s = a+b
s = int(s)
for i in range(s):
if i*i == s:
print("Yes")
sys.exit(0)
print("No")
|
s987667278
|
Accepted
| 17 | 2,940 | 89 |
a,b = map(int,input().split())
if (a*b)%2 == 1:
print("Odd")
else :
print("Even")
|
s586684467
|
p02854
|
u821251381
| 2,000 | 1,048,576 |
Wrong Answer
| 271 | 43,836 | 316 |
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 ,*A = map(int,open(0).read().split())
L = [0 for i in range(N+1)]
D = [0 for i in range(N)]
for i,a in enumerate(A,1):
L[i] = L[i-1] + a
for i in range(N):
D[i] = L[-1]-L[i]
#L.pop(0)
print(L)
print(D)
ans = 1e19
for i in range(N):
#print(L[i],D[i],L[i]-D[i])
ans = min(abs(L[i]-D[i]),ans)
print(ans)
|
s529527794
|
Accepted
| 223 | 36,292 | 318 |
N ,*A = map(int,open(0).read().split())
L = [0 for i in range(N+1)]
D = [0 for i in range(N)]
for i,a in enumerate(A,1):
L[i] = L[i-1] + a
for i in range(N):
D[i] = L[-1]-L[i]
#L.pop(0)
#print(L)
#print(D)
ans = 1e19
for i in range(N):
#print(L[i],D[i],L[i]-D[i])
ans = min(abs(L[i]-D[i]),ans)
print(ans)
|
s677884639
|
p03997
|
u822871631
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 59 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a, b, h = [int(input()) for _ in range(3)]
print((a+b)*h/2)
|
s057254602
|
Accepted
| 17 | 2,940 | 102 |
# coding: utf-8
# Your code here
a, b, h = [int(input()) for _ in range(3)]
print(int((a+b)*h/2))
|
s617535162
|
p03478
|
u153955689
| 2,000 | 262,144 |
Wrong Answer
| 43 | 9,176 | 140 |
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 = map(int, input().split())
ans = 0
for n in range(1, N+1):
if A <= sum(list(map(int,list(str(n))))) <= B:
ans =+ n
print(ans)
|
s476121580
|
Accepted
| 42 | 9,024 | 140 |
N, A, B = map(int, input().split())
ans = 0
for n in range(1, N+1):
if A <= sum(list(map(int,list(str(n))))) <= B:
ans += n
print(ans)
|
s335048862
|
p00002
|
u982618289
| 1,000 | 131,072 |
Time Limit Exceeded
| 40,000 | 8,260 | 255 |
Write a program which computes the digit number of sum of two integers a and b.
|
while True:
inNums = input().strip().split()
a = int(inNums[0])
b = int(inNums[1])
temp = a+b
x = 0
digitNum = 0
while temp // 10**x > 0:
digitNum = x
print(x)
if inNums == None:
break
break
|
s366925236
|
Accepted
| 30 | 7,468 | 81 |
import sys
for s in sys.stdin:
a,b = map(int,s.split())
print(len(str(a+b)))
|
s551947271
|
p03999
|
u578323547
| 2,000 | 262,144 |
Wrong Answer
| 27 | 3,316 | 505 |
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion. All strings that can be obtained in this way can be evaluated as formulas. Evaluate all possible formulas, and print the sum of the results.
|
s = input()
l = len(s)
patterns = []
for i in range(2**(l-1)):
pattern = [''] * (l-1)
for j in range(l-1):
if (i >> j & 1):
pattern[j] = '+'
patterns.append(pattern)
print(patterns)
ans = 0
for pattern in patterns:
a = s[0]
for i in range(l-1):
if (pattern[i] == ''):
a += s[i+1]
else:
print(pattern, a)
ans += int(a)
a = s[i+1]
else:
print(pattern, a)
ans += int(a)
print(ans)
|
s881884757
|
Accepted
| 20 | 3,064 | 432 |
s = input()
l = len(s)
patterns = []
for i in range(2**(l-1)):
pattern = [''] * (l-1)
for j in range(l-1):
if (i >> j & 1):
pattern[j] = '+'
patterns.append(pattern)
ans = 0
for pattern in patterns:
a = s[0]
for i in range(l-1):
if (pattern[i] == ''):
a += s[i+1]
else:
ans += int(a)
a = s[i+1]
else:
ans += int(a)
print(ans)
|
s475243523
|
p03862
|
u396211450
| 2,000 | 262,144 |
Wrong Answer
| 139 | 20,036 | 287 |
There are N boxes arranged in a row. Initially, the i-th box from the left contains a_i candies. Snuke can perform the following operation any number of times: * Choose a box containing at least one candy, and eat one of the candies in the chosen box. His objective is as follows: * Any two neighboring boxes contain at most x candies in total. Find the minimum number of operations required to achieve the objective.
|
n,k=map(int,input().split())
l=list(map(int,input().split()))
if k==0:
print(sum(l))
else:
c=0
for i in range(1,n):
if l[i]+l[i-1]>k:
d=abs(k-l[i-1])
if d<=0:
c=c+l[i]
l[i]=0
else:
c=c+abs(l[i]-d)
l[i]=d
print(c)
|
s078074661
|
Accepted
| 99 | 19,980 | 234 |
N,x=map(int,input().split())
a=list(map(int,input().split()))
ans=0
for i in range(1,N):
tmp=a[i-1]+a[i]
if tmp>x:
ans+=tmp-x
if a[i]>=tmp-x:
a[i]-=tmp-x
else:
a[i]=0
print(ans)
|
s575139287
|
p02381
|
u025362139
| 1,000 | 131,072 |
Wrong Answer
| 40 | 6,996 | 151 |
You have final scores of an examination for n students. Calculate standard deviation of the scores s1, s2 ... sn. The variance ฮฑ2 is defined by ฮฑ2 = (โn _i_ =1(s _i_ \- m)2)/n where m is an average of si. The standard deviation of the scores is the square root of their variance.
|
#coding: UTF-8
from statistics import mean, median,variance,stdev
N = int(input())
buf = list(map(int, input().split()))
sd = stdev(buf)
print(sd)
|
s490872661
|
Accepted
| 40 | 7,128 | 192 |
#coding: UTF-8
import statistics as st
while True:
N = int(input())
if N == 0:
break
buf = list(map(float, input().split()))
sd = st.pstdev(buf)
print("%.8f"%sd)
|
s580372722
|
p03469
|
u853418194
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 46 |
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
a = input()
print(a.replace('2018', '2017'))
|
s299147977
|
Accepted
| 17 | 2,940 | 46 |
a = input()
print(a.replace('2017', '2018'))
|
s930951858
|
p04011
|
u903005414
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 137 |
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.
|
n, k, x, y = int(input()), int(input()), int(input()), int(input())
if n <= k:
ans = n * x
else:
ans = n * x + (n - k) * y
print(ans)
|
s730238491
|
Accepted
| 17 | 2,940 | 138 |
n, k, x, y = int(input()), int(input()), int(input()), int(input())
if n <= k:
ans = n * x
else:
ans = k * x + (n - k) * y
print(ans)
|
s955027456
|
p03501
|
u779308281
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 84 |
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be AรT yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
N,A,B = map(int,input().split(" "))
planA = N * A
planB = B
print(max(planA,planB))
|
s384532968
|
Accepted
| 19 | 3,060 | 53 |
N,A,B = map(int,input().split(" "))
print(min(N*A,B))
|
s912533740
|
p04030
|
u506422818
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,020 | 109 |
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
s = ""
for i in (input()):
if i in ["0", "1"]:
s = s + i
else:
if len(s) >0:
s = s[:-1]
|
s937442902
|
Accepted
| 30 | 9,016 | 114 |
s = ""
for i in (input()):
if i in ["0", "1"]:
s = s + i
else:
if len(s) >0:
s = s[:-1]
print(s)
|
s491962301
|
p03471
|
u614875193
| 2,000 | 262,144 |
Wrong Answer
| 704 | 3,060 | 195 |
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
N,Y = map(int,input().split())
ans = [-1,-1,-1]
for i in range(N+1):
for j in range(N-i+1):
if i*10000 + j*5000 + (N-i-j)*1000 == Y:
ans = [i,j,N-i-j]
print(ans)
|
s370677874
|
Accepted
| 720 | 3,060 | 196 |
N,Y = map(int,input().split())
ans = [-1,-1,-1]
for i in range(N+1):
for j in range(N-i+1):
if i*10000 + j*5000 + (N-i-j)*1000 == Y:
ans = [i,j,N-i-j]
print(*ans)
|
s138011892
|
p03351
|
u360515075
| 2,000 | 1,048,576 |
Wrong Answer
| 20 | 2,940 | 115 |
Three people, A, B and C, are trying to communicate using transceivers. They are standing along a number line, and the coordinates of A, B and C are a, b and c (in meters), respectively. Two people can directly communicate when the distance between them is at most d meters. Determine if A and C can communicate, either directly or indirectly. Here, A and C can indirectly communicate when A and B can directly communicate and also B and C can directly communicate.
|
a, b, c, d = map(int, input().split())
print ("YES" if abs(a-c) <= d or abs(a-b) <= d and abs(c-b) <= d else "NO")
|
s053806121
|
Accepted
| 17 | 2,940 | 115 |
a, b, c, d = map(int, input().split())
print ("Yes" if abs(a-c) <= d or abs(a-b) <= d and abs(c-b) <= d else "No")
|
s259778354
|
p02659
|
u937642029
| 2,000 | 1,048,576 |
Wrong Answer
| 34 | 10,844 | 992 |
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
import sys, bisect, math, itertools, string, queue, copy
# import numpy as np
from collections import Counter,defaultdict,deque
from itertools import permutations, combinations, combinations_with_replacement
from heapq import heappop, heappush
from fractions import gcd
# input = sys.stdin.readline
sys.setrecursionlimit(10**8)
mod = 10**9+7
def inp(): return int(input())
def inpm(): return map(int,input().split())
def inpl(): return list(map(int, input().split()))
def inpls(): return list(input().split())
def inplm(n): return list(int(input()) for _ in range(n))
def inplL(n): return [list(input()) for _ in range(n)]
def inplT(n): return [tuple(input()) for _ in range(n)]
def inpll(n): return [list(map(int, input().split())) for _ in range(n)]
def inplls(n): return sorted([list(map(int, input().split())) for _ in range(n)])
def main():
s = list(input().split())
a = int(s[0])
b = float(s[1])
print(int(a*b))
if __name__ == "__main__":
main()
|
s319521210
|
Accepted
| 22 | 9,212 | 179 |
import math
def main():
s = list(input().split())
a = int(s[0])
b,c = list(map(int,s[1].split('.')))
print(a*b+a*c//100)
if __name__ == "__main__":
main()
|
s178898795
|
p03826
|
u390762426
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 45 |
There are two rectangles. The lengths of the vertical sides of the first rectangle are A, and the lengths of the horizontal sides of the first rectangle are B. The lengths of the vertical sides of the second rectangle are C, and the lengths of the horizontal sides of the second rectangle are D. Print the area of the rectangle with the larger area. If the two rectangles have equal areas, print that area.
|
a,b,c,d=map(int,input().split())
max(a*b,c*d)
|
s903161268
|
Accepted
| 17 | 2,940 | 52 |
a,b,c,d=map(int,input().split())
print(max(a*b,c*d))
|
s038040530
|
p03693
|
u955251526
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 106 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r, g, b = map(int, input().split())
if (r * 100 + g * 10 + b % 4) == 0:
print('Yes')
else: print('No')
|
s450878045
|
Accepted
| 17 | 2,940 | 106 |
r, g, b = map(int, input().split())
if (r * 100 + g * 10 + b) % 4 == 0:
print('YES')
else: print('NO')
|
s625764229
|
p03448
|
u554637745
| 2,000 | 262,144 |
Wrong Answer
| 111 | 3,824 | 456 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
def read_plus_1():
return(int(input())+1)
a = read_plus_1()
b = read_plus_1()
c = read_plus_1()
x = int(input())
count = 0
for i in range(0, a):
if i * 500 > x:
continue
for j in range(0, b):
if i * 500 + j * 100 > x:
continue
for k in range(0, c):
print(i * 500 + j * 100 + k * 50)
if i * 500 + j * 100 + k * 50 == x:
count += 1
break
print(count)
|
s890208819
|
Accepted
| 37 | 3,060 | 410 |
def read_plus_1():
return(int(input())+1)
a = read_plus_1()
b = read_plus_1()
c = read_plus_1()
x = int(input())
count = 0
for i in range(0, a):
if i * 500 > x:
continue
for j in range(0, b):
if i * 500 + j * 100 > x:
continue
for k in range(0, c):
if i * 500 + j * 100 + k * 50 == x:
count += 1
break
print(count)
|
s750674049
|
p03455
|
u699699071
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 144 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
#
a,b = map(int,input().split())
if ((a+b)% 2) == 0:
print("Even")
else:
print("Odd")
|
s402069085
|
Accepted
| 17 | 2,940 | 144 |
#
a,b = map(int,input().split())
if ((a*b)% 2) == 0:
print("Even")
else:
print("Odd")
|
s893987634
|
p03302
|
u077019541
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 138 |
You are given two integers a and b. Determine if a+b=15 or a\times b=15 or neither holds. Note that a+b=15 and a\times b=15 do not hold at the same time.
|
a,b = map(int,input().split())
if a+b==15 or a*b==15:
if a*b==15:
print("+")
else:
print("*")
else:
print("x")
|
s800283923
|
Accepted
| 17 | 2,940 | 138 |
a,b = map(int,input().split())
if a+b==15 or a*b==15:
if a*b==15:
print("*")
else:
print("+")
else:
print("x")
|
s211747092
|
p02742
|
u556487440
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 279 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
h,w = map(int,input().split())
if h %2 == 0:
print(h*w /2)
else:
tmp = (h+1)/2
tmp1 = (h-1)/2
if w %2 == 0:
print((tmp+tmp1)*w/2)
else:
print(int((tmp+tmp1)*(w-1)/2 + tmp))
|
s536560267
|
Accepted
| 18 | 3,064 | 372 |
h,w = map(int,input().split())
if w == 1:
print(int(1))
elif h == 1:
print(int(1))
else:
if h %2 == 0:
print(int(h*w /2))
else:
tmp = (h+1)/2
tmp1 = (h-1)/2
if w %2 == 0:
print(int(h*w/2))
else:
print(int(h*(w-1)/2 +tmp))
|
s256879012
|
p03854
|
u474423089
| 2,000 | 262,144 |
Wrong Answer
| 19 | 4,084 | 1,344 |
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`.
|
def main():
s = list(input())
tmp = ''
n_tmp=""
max_s = len(s)
cnt = 0
flag = True
while cnt < max_s:
for i in range(cnt,cnt+5):
tmp += s[i]
if cnt +10 > max_s:
for i in s[cnt+5:]:
tmp += i
if tmp == "dreamer" or tmp == "eraser":
break
else:
flag = False
break
else:
for j in range(cnt+5,cnt+10):
n_tmp += s[j]
if (tmp == "dream" or tmp == "erase") and (n_tmp == "dream" or n_tmp =="erase"):
tmp = ""
n_tmp = ""
cnt += 5
elif tmp == "dream" and (n_tmp=="erdre" or n_tmp == "erera"):
tmp=""
n_tmp=""
cnt += 7
elif tmp =="erase" and (n_tmp == "erdre" or n_tmp == "erera"):
tmp=""
n_tmp=""
cnt += 6
if len(s) - cnt <= 7:
tmp = ""
for i in s[cnt:]:
tmp += i
if tmp == "dream" or tmp == "dreamer" or tmp == "erase" or tmp == "eraser":
break
else:
flag = False
break
flag = False
break
if flag:
print('Yes')
else:
print('No')
if __name__ == '__main__':
main()
|
s918334374
|
Accepted
| 23 | 6,516 | 203 |
import re
def main():
s = input()
pattern = "^(dream|dreamer|erase|eraser)+$"
if re.match(pattern,s):
print('YES')
else:
print('NO')
if __name__ == '__main__':
main()
|
s394440302
|
p03861
|
u152638361
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 50 |
You are given nonnegative integers a and b (a โค b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
a, b, x = map(int,input().split())
print((b-a)//x)
|
s846200964
|
Accepted
| 17 | 2,940 | 55 |
a, b, x = map(int,input().split())
print(b//x-(a-1)//x)
|
s024688411
|
p02414
|
u328199937
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 469 |
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 i in range(n)]
B = [list(map(int, input().split())) for i in range(m)]
Sum = []
key = 0
for i in range(n):
Sum_s = []
for j in range(l):
key = 0
for k in range(m):
key += A[i][k] * B[k][j]
Sum_s.append(key)
Sum.append(Sum_s)
print(Sum)
for i in range(n):
for j in range(l - 1):
print(Sum[i][j], end = ' ')
print(Sum[i][l - 1])
|
s914595925
|
Accepted
| 350 | 7,060 | 458 |
n, m, l = map(int, input().split())
A = [list(map(int, input().split())) for i in range(n)]
B = [list(map(int, input().split())) for i in range(m)]
Sum = []
key = 0
for i in range(n):
Sum_s = []
for j in range(l):
key = 0
for k in range(m):
key += A[i][k] * B[k][j]
Sum_s.append(key)
Sum.append(Sum_s)
for i in range(n):
for j in range(l - 1):
print(Sum[i][j], end = ' ')
print(Sum[i][l - 1])
|
s371087881
|
p03360
|
u759590494
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,060 | 223 |
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?
|
a,b,c = map(int,input().split())
k = int(input())
m = a
count = 0
if b>m:
m=b
count=1
if c>m:
m=c
count=2
if count==0:
print(a**k+b+c)
if count==1:
print(a+b**k+c)
if count==2:
print(a+b+c**k)
|
s626081989
|
Accepted
| 17 | 3,060 | 229 |
a,b,c = map(int,input().split())
k = int(input())
m = a
count = 0
if b>m:
m=b
count=1
if c>m:
m=c
count=2
if count==0:
print(a*2**k+b+c)
if count==1:
print(a+b*2**k+c)
if count==2:
print(a+b+c*2**k)
|
s301140634
|
p03140
|
u344959959
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,072 | 252 |
You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters. Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation: * Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter. What is the minimum number of operations required to achieve the objective?
|
a = int(input())
b = (input())
c = (input())
d = (input())
count = 0
for i in range(a):
if b[i] == c[i] == d[i]:
count += 0
elif b[i] == c[i] or b[i] == d[i] == c[i] == d[i]:
count += 1
else:
count += 2
print(count)
|
s141594612
|
Accepted
| 29 | 9,060 | 254 |
a = int(input())
b = (input())
c = (input())
d = (input())
count = 0
for i in range(a):
if b[i] == c[i] == d[i]:
count += 0
elif b[i] == c[i] or b[i] == d[i] or c[i] == d[i]:
count += 1
else:
count += 2
print(count)
|
s071403175
|
p03778
|
u878138257
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 151 |
AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W. If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle. Find the minimum distance it needs to be moved.
|
w,a,b = map(int, input().split())
if a==b:
print(0)
elif min(a,b)+w >= max(a,b):
print(0)
elif min(a,b)+w < max(a,b):
print(max(a,b)-min(a,b)+w)
|
s162302657
|
Accepted
| 17 | 3,060 | 152 |
w,a,b = map(int, input().split())
if a==b:
print(0)
elif min(a,b)+w >= max(a,b):
print(0)
elif min(a,b)+w < max(a,b):
print(max(a,b)-min(a,b)-w)
|
s284910996
|
p03731
|
u030726788
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 28 |
In a public bath, there is a shower which emits water for T seconds when the switch is pushed. If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds. N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it. How long will the shower emit water in total?
|
n,t=map(int,input().split())
|
s211650722
|
Accepted
| 98 | 26,836 | 273 |
n,T=map(int,input().split())
t=list(map(int,input().split()))
ts=[]
tf=[]
ts.append(t[0])
tb=t[0]
for i in t:
if(i>tb+T):
tf.append(tb+T)
ts.append(i)
tb=i
tf.append(t[-1]+T)
time=0
#print(ts)
#print(tf)
for i in range(len(ts)):
time+=tf[i]-ts[i]
print(time)
|
s435096167
|
p04043
|
u401183062
| 2,000 | 262,144 |
Wrong Answer
| 20 | 9,088 | 138 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
def iroha():
A, B, C = map(int, input().split())
S = A + B + C
if S == 17:
print("YES")
else:
print("NO")
|
s378847138
|
Accepted
| 22 | 9,144 | 178 |
def iroha():
A, B, C = map(int, input().split())
S = A + B + C
if S == 17:
print("YES")
else:
print("NO")
if __name__ == "__main__":
iroha()
|
s798711235
|
p03068
|
u410118019
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 90 |
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())
d=s[k-1]
for a in s:
if a!=d:
a='*'
print(s)
|
s978029738
|
Accepted
| 17 | 3,060 | 145 |
n=int(input())
s=input()
k=int(input())
d=s[k-1]
s_list=list(s)
for a in range(n):
if s[a] != d:
s_list[a]="*"
s2="".join(s_list)
print(s2)
|
s665760805
|
p03377
|
u580236524
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 79 |
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:
print('No')
else:
print('Yes')
|
s812712937
|
Accepted
| 17 | 2,940 | 90 |
a,b,x = map(int,input().split())
if a > x or a + b < x:
print("NO")
else:
print("YES")
|
s747939172
|
p02613
|
u048947465
| 2,000 | 1,048,576 |
Wrong Answer
| 163 | 16,212 | 234 |
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.
|
a = int(input())
list1 = []
dict1 = {}
for x in range(a):
b = input()
list1.append(b)
for x in list1:
if x not in dict1:
dict1[x] = 1
else:
dict1[x] += 1
for x in dict1:
print(f'{x} x {dict1[x]}')
|
s473169427
|
Accepted
| 152 | 16,324 | 212 |
a = int(input())
list1 = []
dict1 = {'AC': 0, 'WA': 0, 'TLE': 0, 'RE': 0,}
for x in range(a):
b = input()
list1.append(b)
for x in list1:
dict1[x] += 1
for x in dict1:
print(f'{x} x {dict1[x]}')
|
s492083740
|
p03251
|
u102242691
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,060 | 232 |
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
n,m,x,y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
x.sort()
y.sort()
#print(x)
#print(y)
distance = y[0] - x[-1]
if distance > 1:
print("No War")
else:
print("War")
|
s313224432
|
Accepted
| 17 | 3,060 | 255 |
n,m,x,y = map(int,input().split())
X = list(map(int,input().split()))
Y = list(map(int,input().split()))
X.append(x)
Y.append(y)
X.sort()
Y.sort()
for i in range(-100,100):
if X[-1] < i <= Y[0]:
print("No War")
exit()
print("War")
|
s877516210
|
p03455
|
u675024268
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 105 |
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 == 0:
print("Odd")
elif c % 2 == 1:
print("Even")
|
s826175034
|
Accepted
| 17 | 2,940 | 94 |
a, b = map(int, input().split())
c = a * b
if c % 2 == 0:
print("Even")
else:
print("Odd")
|
s614765658
|
p03999
|
u254871849
| 2,000 | 262,144 |
Wrong Answer
| 24 | 3,360 | 433 |
You are given a string S consisting of digits between `1` and `9`, inclusive. You can insert the letter `+` into some of the positions (possibly none) between two letters in this string. Here, `+` must not occur consecutively after insertion. All strings that can be obtained in this way can be evaluated as formulas. Evaluate all possible formulas, and print the sum of the results.
|
import sys
import itertools
s = sys.stdin.readline().rstrip()
n = len(s)
ans = 0
for i in range(n):
combs = list(map(list, itertools.combinations(range(1, n), i)))
res2 = 0
for c in combs:
c.insert(0, 0); c.append(n)
print(c)
res1 = 0
for j in range(1, i+2):
res1 += int(s[c[j-1]:c[j]])
print(int(s[c[j-1]:c[j]]))
res2 += res1
ans += res2
print(ans)
|
s537068169
|
Accepted
| 17 | 3,060 | 290 |
import sys
s = sys.stdin.readline().rstrip()
n = len(s)
def c(n):
return pow(2, max(0, n - 1))
def main():
res = 0
for l in range(n):
for r in range(l, n):
res += int(s[l:r+1]) * c(l) * c(n - 1 - r)
print(res)
if __name__ == '__main__':
main()
|
s372164357
|
p03386
|
u393512980
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 104 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a,b,k=map(int,input().split())
for i in range(a,a+k+1):
print(i)
for i in range(b-k+1,b+1):
print(i)
|
s359568018
|
Accepted
| 18 | 3,060 | 166 |
a,b,k=map(int,input().split())
l=[]
for i in range(a,min(b+1,a+k)):
l.append(i)
for i in range(max(a,b-k+1),b+1):
l.append(i)
for x in sorted(set(l)):
print(x)
|
s169352790
|
p03555
|
u674569298
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 108 |
You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise.
|
a = list(input())
b = list(input())
if a[0] == a[2] and b[0] == b[2]:
print('YES')
else:
print('NO')
|
s561560331
|
Accepted
| 17 | 2,940 | 124 |
a = list(input())
b = list(input())
if a[0] == b[2] and b[0] == a[2] and a[1]==b[1]:
print('YES')
else:
print('NO')
|
s171497117
|
p02694
|
u516579758
| 2,000 | 1,048,576 |
Wrong Answer
| 21 | 9,112 | 179 |
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())
n=0
start=100
while True:
n+=1
start+=start/100
start=math.floor(start)
if start>=x:
print(start)
print(n)
break
|
s959044552
|
Accepted
| 23 | 9,164 | 158 |
import math
x=int(input())
n=0
start=100
while True:
n+=1
start+=start/100
start=math.floor(start)
if start>=x:
print(n)
break
|
s619820495
|
p03494
|
u345778634
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 254 |
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.
|
def main():
N = int(input())
A = list(map(int, input().split()))
c = 0
while True:
for i in range(N):
if A[i] % 2 == 1:
return c
else:
A[i] = A[i] // 2
c += 1
main()
|
s400317360
|
Accepted
| 18 | 2,940 | 277 |
def main():
N = int(input())
A = list(map(int, input().split()))
c = 0
while True:
for i in range(N):
if A[i] % 2 == 1:
print(c)
return
else:
A[i] = A[i] // 2
c += 1
main()
|
s807612395
|
p03494
|
u633203155
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 144 |
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.
|
input()
A = list(map(int, input().split()))
A.sort()
minval = A[0]
count = 0
while minval % 2 == 0:
minval /= 2
count += 1
print(count)
|
s490404659
|
Accepted
| 18 | 3,060 | 142 |
input()
A = list(map(int, input().split()))
count = 0
while all(a % 2 == 0 for a in A):
A = [a / 2 for a in A]
count += 1
print(count)
|
s682189337
|
p03545
|
u872887731
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 260 |
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
from itertools import product
S = input()
for string in product(["+","-"],repeat = 3):
string = list(string)
string.append("")
sum_ = [a + b for a,b in zip(S,string)]
if eval("".join(sum_)) == 7:
print("".join(sum_),"=7")
break
|
s621336879
|
Accepted
| 18 | 3,060 | 260 |
from itertools import product
S = input()
for string in product(["+","-"],repeat = 3):
string = list(string)
string.append("")
sum_ = [a + b for a,b in zip(S,string)]
if eval("".join(sum_)) == 7:
print("".join(sum_)+"=7")
break
|
s192281446
|
p03671
|
u687044304
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 71 |
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.
|
abc = list(map(int, input().split()))
abc.sort()
print(abc[-1]+abc[-2])
|
s260060104
|
Accepted
| 17 | 2,940 | 69 |
abc = list(map(int, input().split()))
abc.sort()
print(abc[0]+abc[1])
|
s484601369
|
p04043
|
u111525113
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 104 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
numbers=sorted(list(map(int,input().split())))
if numbers == [5,7,7]:
print("YES")
else:
print("NO")
|
s939458184
|
Accepted
| 17 | 2,940 | 104 |
numbers=sorted(list(map(int,input().split())))
if numbers == [5,5,7]:
print("YES")
else:
print("NO")
|
s459089130
|
p02613
|
u107915058
| 2,000 | 1,048,576 |
Wrong Answer
| 140 | 16,244 | 202 |
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 = list(input() for _ in range(n))
print("AC ร " + str(s.count("AC")))
print("WA ร " + str(s.count("WA")))
print("TLE ร " + str(s.count("TLE")))
print("RE ร " + str(s.count("RE")))
|
s235082543
|
Accepted
| 140 | 16,204 | 198 |
n = int(input())
s = list(input() for _ in range(n))
print("AC x " + str(s.count("AC")))
print("WA x " + str(s.count("WA")))
print("TLE x " + str(s.count("TLE")))
print("RE x " + str(s.count("RE")))
|
s041952805
|
p02694
|
u584520370
| 2,000 | 1,048,576 |
Wrong Answer
| 21 | 9,144 | 70 |
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 as M
a = int(input())
print(M.ceil(M.log(a / 100, 1.1)))
|
s106832799
|
Accepted
| 25 | 9,156 | 102 |
import math as M
a = int(input())
n = 0
while a > 100 :
a = M.ceil(a / 1.01)
n += 1
print(n)
|
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