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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s006834081
|
p02397
|
u780025254
| 1,000 | 131,072 |
Wrong Answer
| 60 | 7,548 | 158 |
Write a program which reads two integers x and y, and prints them in ascending order.
|
while True:
a, b = map(int, input().split())
if a == 0 and b == 0:
break
if a > b:
print(a, b)
elif a < b:
print(b, a)
|
s863966123
|
Accepted
| 60 | 5,616 | 164 |
while True:
x = list(map(int, input().split()))
if x[0] == 0 and x[1] == 0:
break
else:
x.sort()
print(" ".join(map(str, x)))
|
s429598566
|
p02397
|
u239237743
| 1,000 | 131,072 |
Wrong Answer
| 40 | 5,616 | 126 |
Write a program which reads two integers x and y, and prints them in ascending order.
|
while True:
a,b = map(int, input().split())
if a == 0 and b == 0: break
if a > b:
print(b, a)
else:
print(a, b)
|
s774273004
|
Accepted
| 40 | 6,120 | 265 |
a_list=[]
b_list=[]
while True:
a,b = map(int, input().split())
if a == 0 and b == 0:
break
a_list.append(a)
b_list.append(b)
for (alist, blist) in zip(a_list,b_list):
if alist < blist:
print(alist, blist)
else:
print(blist,alist)
|
s023123847
|
p03494
|
u050428930
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 159 |
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.
|
ans=10**9
n=int(input())
s=list(map(int,input().split()))
for i in s:
p=0
while i%2==0:
i//=2
p+=1
if p<ans:
ans=p
print(p)
|
s439219199
|
Accepted
| 18 | 3,060 | 161 |
ans=10**9
n=int(input())
s=list(map(int,input().split()))
for i in s:
p=0
while i%2==0:
i//=2
p+=1
if p<ans:
ans=p
print(ans)
|
s085136289
|
p03721
|
u113750443
| 2,000 | 262,144 |
Wrong Answer
| 344 | 27,344 | 336 |
There is an empty array. The following N operations will be performed to insert integers into the array. In the i-th operation (1≤i≤N), b_i copies of an integer a_i are inserted into the array. Find the K-th smallest integer in the array after the N operations. For example, the 4-th smallest integer in the array \\{1,2,2,3,3,3\\} is 3.
|
def nyu():
N,K = map(int,input().split())
A = [list(map(int,input().split())) for _ in range(N)]
return N,K,A
def kansu(N,K,A):
sum = 0
ans = -1
for a in range(len(A)):
sum += A[a][1]
if (K >= sum):
ans = A[a][0]
break
print(ans)
N,K,A = nyu()
kansu(N,K,A)
|
s705633755
|
Accepted
| 507 | 27,876 | 377 |
def nyu():
N,K = map(int,input().split())
A = [list(map(int,input().split())) for _ in range(N)]
A.sort()
return N,K,A
def kansu(N,K,A):
sum = 0
ans = -1
for a in range(len(A)):
sum += A[a][1]
# print(sum,A[a][1])
if (K <= sum):
ans = A[a][0]
break
print(ans)
N,K,A = nyu()
kansu(N,K,A)
|
s131388525
|
p03993
|
u527261492
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 14,008 | 211 |
There are N rabbits, numbered 1 through N. The i-th (1≤i≤N) rabbit likes rabbit a_i. Note that no rabbit can like itself, that is, a_i≠i. For a pair of rabbits i and j (i＜j), we call the pair (i，j) a _friendly pair_ if the following condition is met. * Rabbit i likes rabbit j and rabbit j likes rabbit i. Calculate the number of the friendly pairs.
|
n=int(input())
a=list(map(int,input().split()))
cnt=0
for i in range(n):
for j in range(n):
if a[i]==j+1 and a[j]==i+1:
cnt+=1
else:
cnt=cnt
if cnt>n:
print(cnt-n)
else:
print(0)
|
s800334239
|
Accepted
| 68 | 14,008 | 147 |
n=int(input())
a=list(map(int,input().split()))
cnt=0
for i in range(n):
if a[a[i]-1]==i+1:
cnt+=1
else:
cnt=cnt
print(cnt//2)
|
s120083145
|
p03759
|
u147492202
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,056 | 183 |
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a,b,c = map(int,input().split())
if b-a == c-b:
print('Yes')
else:
print('No')
|
s378695149
|
Accepted
| 27 | 9,152 | 183 |
a,b,c = map(int,input().split())
if b-a == c-b:
print('YES')
else:
print('NO')
|
s174088624
|
p03352
|
u732870425
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 2,940 | 192 |
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
x = int(input())
ans = 0
for i in range(1, x):
for j in range(2, x):
if i**j < x:
ans = i**j
continue
else:
break
break
print(ans)
|
s413805798
|
Accepted
| 19 | 3,060 | 172 |
x = int(input())
ans = 1
for i in range(1, x):
for j in range(2, x):
if i**j <= x:
ans = max(ans, i**j)
else:
break
print(ans)
|
s220151945
|
p03605
|
u270350963
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 83 |
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
a = input()
a.find("9")
if a == 0 or a == 1:
print("Yes")
else:
print("No")
|
s060989912
|
Accepted
| 18 | 3,060 | 72 |
a = input()
if a.find("9") != -1:
print("Yes")
else:
print("No")
|
s711290106
|
p03352
|
u038216098
| 2,000 | 1,048,576 |
Time Limit Exceeded
| 2,206 | 9,044 | 126 |
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
X=int(input())
res=[]
for b in range(1,1000):
for p in range(2,1000):
if b**p<=X :
res.append(b**p)
print(res[-1])
|
s787230341
|
Accepted
| 28 | 9,144 | 144 |
X=int(input())
res=0
for b in range(1,1000):
for p in range(2,1000):
if b**p<=X :
res=max(res,b**p)
else:
break
print(res)
|
s847786175
|
p02608
|
u250554058
| 2,000 | 1,048,576 |
Wrong Answer
| 505 | 9,172 | 263 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
n = int(input())
ans = [0 for _ in range(10050)]
for i in range(1, 105):
for j in range(1, 105):
for k in range(1, 105):
v = i * i + j * j + k * k + i * j + j * k + k * j
if v < 10050:
ans[v] += 1
for i in range(n):
print(ans[i + 1])
|
s662360479
|
Accepted
| 517 | 9,192 | 263 |
n = int(input())
ans = [0 for _ in range(10050)]
for i in range(1, 105):
for j in range(1, 105):
for k in range(1, 105):
v = i * i + j * j + k * k + i * j + j * k + k * i
if v < 10050:
ans[v] += 1
for i in range(n):
print(ans[i + 1])
|
s024276130
|
p04044
|
u071868443
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 174 |
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
N, L = map(int, input().split())
strs = [[]] * N
for i in range(N):
strs[i] = str(input())
strs.sort()
print(strs)
x = strs[0]
for j in range(1,N):
x += strs[j]
print(x)
|
s053993310
|
Accepted
| 17 | 3,060 | 162 |
N, L = map(int, input().split())
strs = [[]] * N
for i in range(N):
strs[i] = str(input())
strs.sort()
x = strs[0]
for j in range(1,N):
x += strs[j]
print(x)
|
s728413063
|
p02396
|
u747594996
| 1,000 | 131,072 |
Wrong Answer
| 140 | 6,724 | 154 |
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
def main():
i = 1
while True:
num = int(input())
print('Case ', i ,': ', num)
i += 1
if num == 0:
break
if __name__=="__main__":
main()
|
s831592221
|
Accepted
| 120 | 6,720 | 175 |
def main():
i = 1
while True:
num = int(input())
if num == 0:
break
else:
print('Case {0}: {1}'.format(i, num))
i += 1
if __name__=="__main__":
main()
|
s223224760
|
p03229
|
u509368316
| 2,000 | 1,048,576 |
Wrong Answer
| 256 | 11,716 | 229 |
You are given N integers; the i-th of them is A_i. Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.
|
N=int(input())
A=sorted([int(input()) for i in range(N)])
r=[0]*N
for i in range(N//2):
r[i*2+1]=A[i]
if N%2:
r[-1]=A[N//2]
for i in range(N//2):
r[i*2]=A[i+N//2+N%2]
print(sum([abs(r[i]-r[i+1]) for i in range(N-1)]))
|
s806351834
|
Accepted
| 261 | 13,264 | 323 |
N=int(input())
a=sorted([int(input()) for i in range(N)])
r=[0]*N
r[1:-1:2]=a[:(N-1)//2]
r[-1]=a[(N-1)//2]
r[:-1:2]=a[(N-1)//2+1:]
A=list(reversed(a))
R=[0]*N
R[1:-1:2]=A[:(N-1)//2]
R[-1]=A[(N-1)//2]
R[:-1:2]=A[(N-1)//2+1:]
print(max(sum([abs(r[i]-r[i+1]) for i in range(N-1)]),sum([abs(R[i]-R[i+1]) for i in range(N-1)])))
|
s202719807
|
p03399
|
u131405882
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 147 |
You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses.
|
a = int(input())
b = int(input())
c = int(input())
d = int(input())
sum = 0
if a > b:
sum += b
else:
sum += a
if c > d:
sum += d
else:
sum += c
|
s378480955
|
Accepted
| 17 | 2,940 | 159 |
a = int(input())
b = int(input())
c = int(input())
d = int(input())
sum = 0
if a > b:
sum += b
else:
sum += a
if c > d:
sum += d
else:
sum += c
print(sum)
|
s429158772
|
p03943
|
u391059484
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 112 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a,b,c = map(int, input().split())
if a + b != c and a+c != b and b+c != a:
print('NO')
else:
print('YES')
|
s872800143
|
Accepted
| 17 | 2,940 | 109 |
a,b,c = map(int, input().split())
if a + b != c and a+c != b and b+c != a:
print('No')
else:
print('Yes')
|
s397760064
|
p03693
|
u276115223
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 125 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
# A - RGB Cards
r, g, b = [int(s) for s in input().split()]
number = int(r + g + b)
print('YES' if number % 4 == 0 else 'NO')
|
s256752714
|
Accepted
| 17 | 2,940 | 118 |
# ABC 064: A – RGB Cards
r, g, b = [int(s) for s in input().split()]
print('YES' if (g * 10 + b) % 4 == 0 else 'NO')
|
s685345512
|
p03943
|
u452015170
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 135 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a, b, c = map(int, input().split())
list = sorted([a, b, c])
if (list[0] + list[1]) == list[2] :
print("YES")
else:
print("NO")
|
s646277737
|
Accepted
| 17 | 2,940 | 135 |
a, b, c = map(int, input().split())
list = sorted([a, b, c])
if (list[0] + list[1]) == list[2] :
print("Yes")
else:
print("No")
|
s245206950
|
p03637
|
u322568242
| 2,000 | 262,144 |
Wrong Answer
| 97 | 14,252 | 316 |
We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective.
|
n=int(input())
a=[int(i) for i in input().split()]
flag2=[0]*n
flag4=[0]*n
for i in range(len(a)):
if a[i]%4==0:flag4[i]=1
elif a[i]%2==0 :flag2[i]=1
cnt2=0
cnt4=0
for i in range(len(a)):
cnt2+=flag2[i]
cnt4+=flag4[i]
print(cnt2,cnt4)
if cnt2>1:n-=cnt2
if cnt4>=int(n/2):print("Yes")
else:print("No")
|
s613306813
|
Accepted
| 66 | 14,252 | 237 |
n=int(input())
a=[int(i)for i in input().split()]
count2=0
count4=0
for i in a:
if i%4==0:
count4+=1
continue
if i%2==0:
count2+=1
if count2!=0:n-=count2-1
if count4>=int(n/2):print("Yes")
else:print("No")
|
s193981410
|
p02613
|
u948911484
| 2,000 | 1,048,576 |
Wrong Answer
| 137 | 16,168 | 125 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n = int(input())
s = [input() for _ in range(n)]
l = ["AC","WA","TLE","RE"]
for i in range(4): print(l[i],"X",s.count(l[i]))
|
s934174712
|
Accepted
| 139 | 16,160 | 100 |
s=[input() for _ in range(int(input()))]
l=["AC","WA","TLE","RE"]
for i in l:print(i,"x",s.count(i))
|
s087624905
|
p04011
|
u771405286
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 179 |
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 = int(input())
K = int(input())
X = int(input())
Y = int(input())
retval = 0
for i in range(N+1):
if i <= K:
retval += X
else:
retval += Y
print(retval)
|
s748575438
|
Accepted
| 19 | 2,940 | 182 |
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
retval = 0
for i in range(1, N+1):
if i <= K:
retval += X
else:
retval += Y
print(retval)
|
s847011270
|
p03006
|
u145600939
| 2,000 | 1,048,576 |
Wrong Answer
| 100 | 3,188 | 316 |
There are N balls in a two-dimensional plane. The i-th ball is at coordinates (x_i, y_i). We will collect all of these balls, by choosing two integers p and q such that p \neq 0 or q \neq 0 and then repeating the following operation: * Choose a ball remaining in the plane and collect it. Let (a, b) be the coordinates of this ball. If we collected a ball at coordinates (a - p, b - q) in the previous operation, the cost of this operation is 0. Otherwise, including when this is the first time to do this operation, the cost of this operation is 1. Find the minimum total cost required to collect all the balls when we optimally choose p and q.
|
from itertools import combinations
n = int(input())
pos = []
for i in range(n):
pos.append(tuple(map(int,input().split())))
ans = 1e100
def cost(p,q):
ret = 0
for i,j in pos:
ret += 1 - ((i+p,j+q) in pos)
return ret
for a,b in combinations(pos, 2):
ans = max(ans, cost(a[0]-b[0],a[1]-b[1]))
print(ans)
|
s030865341
|
Accepted
| 98 | 3,064 | 348 |
from itertools import combinations
n = int(input())
if n == 1:
print(1)
exit(0)
pos = []
for i in range(n):
pos.append(tuple(map(int,input().split())))
ans = 1e100
def cost(p,q):
ret = 0
for i,j in pos:
ret += 1 - ((i+p,j+q) in pos)
return ret
for a,b in combinations(pos, 2):
ans = min(ans, cost(a[0]-b[0],a[1]-b[1]))
print(ans)
|
s822045297
|
p02608
|
u167647458
| 2,000 | 1,048,576 |
Wrong Answer
| 781 | 9,456 | 381 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
import math
def main():
N = int(input())
ans = [0]*(N+1)
n = int(math.sqrt(N))
for x in range(1, n+1):
for y in range(1, n+1):
for z in range(1, n+1):
s =x**2 + y**2 + z**2 + x*y + y*z + z*x
if s < N+1:
ans[s] += 1
for i in ans:
print(i)
if __name__ == '__main__':
main()
|
s810980079
|
Accepted
| 811 | 9,068 | 383 |
import math
def main():
N = int(input())
ans = [0]*(N)
n = int(math.sqrt(N))
for x in range(1, n+1):
for y in range(1, n+1):
for z in range(1, n+1):
s =x**2 + y**2 + z**2 + x*y + y*z + z*x
if s < N+1:
ans[s - 1] += 1
for i in ans:
print(i)
if __name__ == '__main__':
main()
|
s968755490
|
p03448
|
u705857261
| 2,000 | 262,144 |
Wrong Answer
| 50 | 3,060 | 285 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
# print(a, b, c, x)
ans = 0
for i_a in range(a):
for i_b in range(b):
for i_c in range(c):
sum = 500 * i_a + 100 * i_b + 50 * i_c
if sum == x:
ans += 1
print(ans)
|
s979606378
|
Accepted
| 56 | 3,060 | 291 |
a = int(input())
b = int(input())
c = int(input())
x = int(input())
# print(a, b, c, x)
ans = 0
for i_a in range(a+1):
for i_b in range(b+1):
for i_c in range(c+1):
sum = 500 * i_a + 100 * i_b + 50 * i_c
if sum == x:
ans += 1
print(ans)
|
s391347746
|
p03162
|
u239528020
| 2,000 | 1,048,576 |
Wrong Answer
| 502 | 31,348 | 310 |
Taro's summer vacation starts tomorrow, and he has decided to make plans for it now. The vacation consists of N days. For each i (1 \leq i \leq N), Taro will choose one of the following activities and do it on the i-th day: * A: Swim in the sea. Gain a_i points of happiness. * B: Catch bugs in the mountains. Gain b_i points of happiness. * C: Do homework at home. Gain c_i points of happiness. As Taro gets bored easily, he cannot do the same activities for two or more consecutive days. Find the maximum possible total points of happiness that Taro gains.
|
### C
N = int(input())
A = [list(map(int, input().split())) for i in range(N)]
dp = [[0]*3]*N
dp[0] = A[0]
for i in range(1, N):
dp[i][0]= max(A[i-1][0], A[i-1][1]) + dp[i-1][0]
dp[i][1] = max(A[i-1][0], A[i-1][2]) + dp[i-1][1]
dp[i][2] = max(A[i-1][1], A[i-1][2]) + dp[i-1][2]
print(max(dp[-1]))
|
s752017844
|
Accepted
| 440 | 30,580 | 262 |
# C
N = int(input())
abc = [list(map(int, input().split())) for i in range(N)]
da, db, dc = 0, 0, 0
for a, b, c in abc:
na = max(db, dc) + a
nb = max(da, dc) + b
nc = max(da, db) + c
da = na
db = nb
dc = nc
print(max([na, nb, nc]))
|
s100508283
|
p03494
|
u588526762
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 10 |
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
print('a')
|
s461333312
|
Accepted
| 19 | 3,060 | 234 |
ans = 10**9
N = int(input())
A = list(map(int, input().split()))
for i in A:
p = 0
while i % 2 == 0:
i /= 2
p += 1
if p < ans:
ans = p
print(ans)
|
s850229122
|
p03860
|
u097700948
| 2,000 | 262,144 |
Wrong Answer
| 27 | 9,032 | 48 |
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
S = input()
s = 'A'+S[9].upper() + 'C'
print(s)
|
s234024641
|
Accepted
| 22 | 9,080 | 47 |
S = input()
s = 'A'+S[8].upper() + 'C'
print(s)
|
s550933563
|
p03455
|
u729911693
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 89 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
if(a*b/2):
print('0dd')
else:
print('Even')
|
s071251678
|
Accepted
| 17 | 2,940 | 104 |
a, b = map(int, input().split())
if(a*b%2 == 0):
print('Even')
else:
print('Odd')
|
s863076841
|
p03110
|
u143492911
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 220 |
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
n=int(input())
a=[list(input().split())for i in range(n)]
print(a)
total=0
for i in range(n):
if a[i][1]=="JPY":
total+=int(a[i][0])
elif a[i][1]=="BTC":
total+=float(a[i][0])*380000
print(total)
|
s844952145
|
Accepted
| 17 | 3,060 | 221 |
n=int(input())
a=[list(input().split())for i in range(n)]
#print(a)
total=0
for i in range(n):
if a[i][1]=="JPY":
total+=int(a[i][0])
elif a[i][1]=="BTC":
total+=float(a[i][0])*380000
print(total)
|
s615503772
|
p03457
|
u118211443
| 2,000 | 262,144 |
Wrong Answer
| 966 | 3,828 | 568 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
n = int(input())
for i in range(n):
t, x, y = map(int, input().split())
if i == 0:
if x + y != t:
print('No')
exit()
else:
bx = x
by = y
bt = t
else:
d = abs(x - bx) + abs(y - by)
dt = t - bt
bx = x
by = y
bt = t
print(d, dt)
if d % 2 == 1 and dt % 2 == 0 and d <= dt:
print('No')
exit()
elif d % 2 == 0 and dt % 2 == 1 and d <= dt:
print('No')
exit()
print('Yes')
|
s070305778
|
Accepted
| 378 | 3,064 | 610 |
n = int(input())
for i in range(n):
t, x, y = map(int, input().split())
if i == 0:
if x + y != t:
print('No')
exit()
else:
bx = x
by = y
bt = t
else:
d = abs(x - bx) + abs(y - by)
dt = t - bt
bx = x
by = y
bt = t
# print(d, dt)
if d > dt:
print('No')
exit()
elif d % 2 == 1 and dt % 2 == 0:
print('No')
exit()
elif d % 2 == 0 and dt % 2 == 1:
print('No')
exit()
print('Yes')
|
s634618000
|
p03673
|
u021548497
| 2,000 | 262,144 |
Wrong Answer
| 227 | 25,412 | 187 |
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
from collections import deque
n = int(input())
a = [int(x) for x in input().split()]
b = deque([])
for i in range(n):
if i%2:
b.append(a[i])
else:
b.appendleft(a[i])
print(*b)
|
s985552713
|
Accepted
| 236 | 25,412 | 191 |
from collections import deque
n = int(input())
a = [int(x) for x in input().split()]
b = deque([])
for i in range(n):
if (i+n)%2:
b.appendleft(a[i])
else:
b.append(a[i])
print(*b)
|
s909630406
|
p03625
|
u882209234
| 2,000 | 262,144 |
Wrong Answer
| 106 | 14,224 | 265 |
We have N sticks with negligible thickness. The length of the i-th stick is A_i. Snuke wants to select four different sticks from these sticks and form a rectangle (including a square), using the sticks as its sides. Find the maximum possible area of the rectangle.
|
N = int(input())
A = list(map(int,input().split()))
A.sort(reverse=True)
ans = []
i = 0
while i < N-2:
if A[i] == A[i+1]:
ans.append(A[i])
i += 1
if len(ans) == 2: break
i += 1
if len(ans) in (0,1): print(0)
else: print(ans[0]*ans[1])
|
s233177901
|
Accepted
| 104 | 14,252 | 265 |
N = int(input())
A = list(map(int,input().split()))
A.sort(reverse=True)
ans = []
i = 0
while i < N-1:
if A[i] == A[i+1]:
ans.append(A[i])
i += 1
if len(ans) == 2: break
i += 1
if len(ans) in (0,1): print(0)
else: print(ans[0]*ans[1])
|
s810546653
|
p02972
|
u114920558
| 2,000 | 1,048,576 |
Wrong Answer
| 978 | 10,304 | 210 |
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
N = int(input())
A = list(map(int, input().split()))
B = [0]*N
for i in range(N):
cnt = N - i
ans = 0
while(cnt <= N):
ans += A[cnt-1]
cnt += N - i
B[N-i-1] = ((cnt%2) - A[N-i-1])%2
print(*B)
|
s991764751
|
Accepted
| 752 | 14,016 | 276 |
N = int(input())
A = list(map(int, input().split()))
afterA = [0] * N
ans = list()
for i in range(N, 0, -1):
cnt = 0
for j in range(2, (N//i) + 1):
cnt += afterA[j*i-1]
if(cnt % 2 != A[i-1]):
afterA[i-1] = 1
ans.append(i)
print(sum(afterA))
print(*ans)
|
s904422504
|
p03599
|
u613658617
| 3,000 | 262,144 |
Wrong Answer
| 46 | 3,188 | 696 |
Snuke is making sugar water in a beaker. Initially, the beaker is empty. Snuke can perform the following four types of operations any number of times. He may choose not to perform some types of operations. * Operation 1: Pour 100A grams of water into the beaker. * Operation 2: Pour 100B grams of water into the beaker. * Operation 3: Put C grams of sugar into the beaker. * Operation 4: Put D grams of sugar into the beaker. In our experimental environment, E grams of sugar can dissolve into 100 grams of water. Snuke will make sugar water with the highest possible density. The beaker can contain at most F grams of substances (water and sugar combined), and there must not be any undissolved sugar in the beaker. Find the mass of the sugar water Snuke will make, and the mass of sugar dissolved in it. If there is more than one candidate, any of them will be accepted. We remind you that the sugar water that contains a grams of water and b grams of sugar is \frac{100b}{a + b} percent. Also, in this problem, pure water that does not contain any sugar is regarded as 0 percent density sugar water.
|
A, B, C, D, E, F = map(int, input().split())
s, w = (0, 0)
water_list = [False] * (6000)
sugar_list = [False] * (6000)
water_list[0] = True
sugar_list[0] = True
for i in range(F):
water_list[i + A * 100] = water_list[i + A * 100] or water_list[i]
water_list[i + B * 100] = water_list[i + B * 100] or water_list[i]
sugar_list[i + C] = sugar_list[i + C] or sugar_list[i]
sugar_list[i + D] = sugar_list[i + D] or sugar_list[i]
for si in range(0, F + 1, 100):
for wi in range(F + 1):
if si <= (wi / 100) * E and si + wi <= F and water_list[i] and sugar_list[j]:
if si * (s + w) > s * (si + wi):
s, w = (si, wi)
print('{} {}'.format(s + w, s))
|
s438286782
|
Accepted
| 39 | 3,188 | 702 |
A, B, C, D, E, F = map(int, input().split())
s, w = (0, 100 * A)
water_list = [False] * (6000)
sugar_list = [False] * (6000)
water_list[0] = True
sugar_list[0] = True
for i in range(F):
water_list[i + A * 100] = water_list[i + A * 100] or water_list[i]
water_list[i + B * 100] = water_list[i + B * 100] or water_list[i]
sugar_list[i + C] = sugar_list[i + C] or sugar_list[i]
sugar_list[i + D] = sugar_list[i + D] or sugar_list[i]
for wi in range(0, F + 1, 100):
for si in range(F + 1):
if si * 100 <= wi * E and si + wi <= F and water_list[wi] and sugar_list[si]:
if si * (s + w) > s * (si + wi):
s, w = (si, wi)
print('{} {}'.format(s + w, s))
|
s956537794
|
p04029
|
u478266845
| 2,000 | 262,144 |
Wrong Answer
| 24 | 3,316 | 44 |
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())
ans =N*(N+1)/2
print(ans)
|
s917857937
|
Accepted
| 16 | 2,940 | 50 |
N = int(input())
ans =int(N*(N+1)/2)
print(ans)
|
s678111224
|
p02271
|
u279605379
| 5,000 | 131,072 |
Wrong Answer
| 20 | 7,672 | 378 |
Write a program which reads a sequence _A_ of _n_ elements and an integer _M_ , and outputs "yes" if you can make _M_ by adding elements in _A_ , otherwise "no". You can use an element only once. You are given the sequence _A_ and _q_ questions where each question contains _M i_.
|
def abc(a,b,c):
if a==n:
pass
else:
if b == 1:
c += A[a]
s.add(c)
abc(a+1,0,c)
abc(a+1,1,c)
n = int(input())
A = [int(a) for a in input().split()]
q = int(input())
M = [int(m) for m in input().split()]
s = set()
abc(0,0,0)
abc(0,1,0)
for m in M:
if m in s:
print("Yes")
else:
print("No")
|
s258764854
|
Accepted
| 1,160 | 7,748 | 378 |
def abc(a,b,c):
if a==n:
pass
else:
if b == 1:
c += A[a]
s.add(c)
abc(a+1,0,c)
abc(a+1,1,c)
n = int(input())
A = [int(a) for a in input().split()]
q = int(input())
M = [int(m) for m in input().split()]
s = set()
abc(0,0,0)
abc(0,1,0)
for m in M:
if m in s:
print("yes")
else:
print("no")
|
s581299743
|
p03643
|
u548545174
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 204 |
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.
|
N = int(input())
ans = 0
num = 0
for i in range(1, N + 1):
now = i
cnt = 0
while i % 2 == 0:
i /= 2
cnt += 1
if cnt >= ans:
ans = cnt
num = now
print(num)
|
s502970127
|
Accepted
| 17 | 2,940 | 22 |
print('ABC' + input())
|
s353693546
|
p02865
|
u952968889
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 69 |
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
n = int(input())
if n % 2:
print((n-1) / 2)
else:
print(n/2 + 1)
|
s042749309
|
Accepted
| 17 | 2,940 | 75 |
n = int(input())
if n%2:
print(int((n-1) /2))
else:
print(int(n/2 -1))
|
s736628175
|
p03505
|
u200239931
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 3,064 | 922 |
_ButCoder Inc._ runs a programming competition site called _ButCoder_. In this site, a user is given an integer value called rating that represents his/her skill, which changes each time he/she participates in a contest. The initial value of a new user's rating is 0, and a user whose rating reaches K or higher is called _Kaiden_ ("total transmission"). Note that a user's rating may become negative. Hikuhashi is a new user in ButCoder. It is estimated that, his rating increases by A in each of his odd-numbered contests (first, third, fifth, ...), and decreases by B in each of his even-numbered contests (second, fourth, sixth, ...). According to this estimate, after how many contests will he become Kaiden for the first time, or will he never become Kaiden?
|
def getinputdata():
array_result = []
data = input()
array_result.append(data.split(" "))
flg = 1
try:
while flg:
data = input()
array_temp = []
if(data != ""):
array_result.append(data.split(" "))
flg = 1
else:
flg = 0
finally:
return array_result
arr_data = getinputdata()
k = int(arr_data[0][0])
a = int(arr_data[0][1])
b = int(arr_data[0][2])
print(k, a, b)
rate = 0
chkflg = True
cnt = 0
while rate < k:
if chkflg:
rate += a
chkflg = False
cnt += 1
else:
if a < b:
cnt = -1
break
else:
rate -= b
chkflg = True
cnt += 1
# print(rate)
print(cnt)
|
s453640845
|
Accepted
| 17 | 3,064 | 927 |
def getinputdata():
array_result = []
data = input()
array_result.append(data.split(" "))
flg = 1
try:
while flg:
data = input()
array_temp = []
if(data != ""):
array_result.append(data.split(" "))
flg = 1
else:
flg = 0
finally:
return array_result
arr_data = getinputdata()
k = int(arr_data[0][0])
a = int(arr_data[0][1])
b = int(arr_data[0][2])
cnt = a - b
rate = a
if rate >= k:
print(1)
elif cnt==0:
print(-1)
else:
if (k-rate) % cnt == 0:
num = (k-rate) // cnt
print(1+ num*2)
else:
num = ((k-rate) // cnt) + 1
if num>=1 :
print(num*2+1)
else:
print(-1)
|
s044989904
|
p03567
|
u413254802
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 96 |
Snuke built an online judge to hold a programming contest. When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.) Determine whether the judge can return the string `AC` as the verdict to a program.
|
# coding: utf-8
data = input()
if data.find('AC') > -1:
print("OK")
else:
print("NG")
|
s076224946
|
Accepted
| 18 | 2,940 | 97 |
# coding: utf-8
data = input()
if data.find('AC') > -1:
print("Yes")
else:
print("No")
|
s195111605
|
p03448
|
u984989720
| 2,000 | 262,144 |
Wrong Answer
| 55 | 3,064 | 275 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
count = 0
for i in range(0,a+1):
for j in range(0,b+1):
for k in range(0,c+1):
total = a * 500 + b * 100 + c * 50
if total == x:
count += 1
print(count)
|
s336739976
|
Accepted
| 57 | 3,060 | 276 |
a = int(input())
b = int(input())
c = int(input())
x = int(input())
count = 0
for i in range(0,a+1):
for j in range(0,b+1):
for k in range(0,c+1):
total = i * 500 + j * 100 + k * 50
if total == x:
count += 1
print(count)
|
s268237912
|
p04011
|
u201234972
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 128 |
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 = int( input())
K = int( input())
X = int( input())
Y = int( input())
if N <= X:
print(N*X)
else:
print(X*K + Y*(N-K))
|
s865302001
|
Accepted
| 18 | 2,940 | 129 |
N = int( input())
K = int( input())
X = int( input())
Y = int( input())
if N <= K:
print(N*X)
else:
print(X*K + Y*(N-K))
|
s528550797
|
p03673
|
u220345792
| 2,000 | 262,144 |
Wrong Answer
| 112 | 31,224 | 191 |
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
n = int(input())
A = list(map(int, input().split()))
odd = A[0::2]
even = A[1::2]
if n % 2 == 0:
ans = even[::-1]+odd
else:
ans = odd[::-1]+even
ans = map(str, ans)
print("".join(ans))
|
s512724759
|
Accepted
| 113 | 31,300 | 192 |
n = int(input())
A = list(map(int, input().split()))
odd = A[0::2]
even = A[1::2]
if n % 2 == 0:
ans = even[::-1]+odd
else:
ans = odd[::-1]+even
ans = map(str, ans)
print(" ".join(ans))
|
s471621088
|
p03476
|
u722189950
| 2,000 | 262,144 |
Wrong Answer
| 442 | 21,656 | 628 |
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 ≤ x ≤ r_i.
|
Q = int(input())
ans = []
searchList = [i for i in range(2,10**5+1)]
primeNum = []
primeNumLst = [False]*(10**5+1)
mx = 10**5**0.5+1
while searchList[0] <= mx:
primeNum.append(searchList[0])
tmp = searchList[0]
searchList = [i for i in searchList if i%tmp != 0]
primeNum.extend(searchList)
lst = [0]*(10**5+1)
sumlst = [0]
for p in primeNum:
primeNumLst[p] = True
if primeNumLst[(p+1)//2]:
lst[p] = 1
for i in range(10**5+1):
sumlst.append(sumlst[-1] + i)
for _ in range(Q):
l, r = map(int, input().split())
ans.append(sumlst[r+1] - sumlst[l])
print("\n".join(map(str,ans)))
|
s327906341
|
Accepted
| 423 | 17,928 | 729 |
from itertools import accumulate
Q = int(input())
ans = []
def eratosthenes(N):
searchList = [i for i in range(2,N)]
primeNum = []
mx = int(N**0.5+1)
while searchList[0] <= mx:
primeNum.append(searchList[0])
tmp = searchList[0]
searchList = [i for i in searchList if i%tmp != 0]
primeNum.extend(searchList)
return primeNum
ok = [0]*(10**5+1)
primeNum = eratosthenes(10**5+1)
primeNumLst = [False] * (10**5+1)
for p in primeNum:
primeNumLst[p] = True
for p in primeNum:
if primeNumLst[(p+1)//2]:
ok[p] = 1
ok = list(accumulate(ok))
for _ in range(Q):
l, r = map(int, input().split())
ans.append(ok[r] - ok[l-1])
print("\n".join(map(str,ans)))
|
s855178042
|
p03828
|
u246770003
| 2,000 | 262,144 |
Wrong Answer
| 2,206 | 21,736 | 275 |
You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.
|
n = int(input())
factorial = 1
divisor = []
ans = 0
for i in range(1,n+1):
factorial = factorial * i
print(factorial)
for i in range(1,factorial+1):
if factorial % i == 0:
divisor.append(i)
ans = len(divisor) % (10 ** 9 + 7)
print(ans)
|
s406045762
|
Accepted
| 56 | 9,408 | 731 |
import math
import collections
n = int(input())
factorial = 1
divisor = []
ans = 1
def get_prime(num):
if num <= 1:
return
for i in range(2, num+1):
while num % i == 0:
divisor.append(i)
num //= i
for j in range(2, n+1):
get_prime(j)
count = collections.Counter(divisor)
for k in count.most_common():
temp = list(k)[1]
ans = ans * (temp + 1)
ans = ans % (10 ** 9 + 7)
print(ans)
|
s472896878
|
p03170
|
u803848678
| 2,000 | 1,048,576 |
Wrong Answer
| 990 | 3,828 | 300 |
There is a set A = \\{ a_1, a_2, \ldots, a_N \\} consisting of N positive integers. Taro and Jiro will play the following game against each other. Initially, we have a pile consisting of K stones. The two players perform the following operation alternately, starting from Taro: * Choose an element x in A, and remove exactly x stones from the pile. A player loses when he becomes unable to play. Assuming that both players play optimally, determine the winner.
|
n, k = map(int, input().split())
a = list(map(int, input().split()))
dp = [False]*(k+1)
for i in range(1, k):
flag = False
for ai in a:
if i-ai >= 0 and not dp[i-ai]:
flag = True
break
dp[i] = flag
if dp[-1]:
print("First")
else:
print("Second")
|
s566558763
|
Accepted
| 984 | 3,828 | 302 |
n, k = map(int, input().split())
a = list(map(int, input().split()))
dp = [False]*(k+1)
for i in range(1, k+1):
flag = False
for ai in a:
if i-ai >= 0 and not dp[i-ai]:
flag = True
break
dp[i] = flag
if dp[-1]:
print("First")
else:
print("Second")
|
s774147667
|
p03846
|
u026102659
| 2,000 | 262,144 |
Wrong Answer
| 84 | 13,880 | 345 |
There are N people, conveniently numbered 1 through N. They were standing in a row yesterday, but now they are unsure of the order in which they were standing. However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person. According to their reports, the difference above for person i is A_i. Based on these reports, find the number of the possible orders in which they were standing. Since it can be extremely large, print the answer modulo 10^9+7. Note that the reports may be incorrect and thus there may be no consistent order. In such a case, print 0.
|
n = int(input())
nums = {0:0}
for i in list(map(int, input().split())):
if i in nums:
if nums[i] <= 1:
nums[i] += 1
else:
print(0)
break
else:
nums[i] = 1
del nums[0]
if 1 in nums.values():
print(0)
else:
print(2**len(nums.values()))
print(nums)
|
s747413620
|
Accepted
| 76 | 14,436 | 515 |
n = int(input())
nums = {0:0}
for i in list(map(int, input().split())):
if i in nums:
nums[i] += 1
if nums[i] > 2:
ans = 0
break
else:
nums[i] = 1
if n % 2 == 0:
if 1 in nums.values():
ans = 0
else:
ans = 2**(n//2)
else:
if nums[0] == 1:
del nums[0]
if 1 in nums.values():
ans = 0
else:
ans = 2**((n-1)//2)
else:
ans = 0
print(ans % (10**9+7))
|
s398577086
|
p03854
|
u400596590
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,188 | 146 |
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
s = input().replace("eraser", "").replace("erase", "").replace("dreamer", "").replace("dreamer", "")
if s:
print("NO")
else:
print("YES")
|
s838642903
|
Accepted
| 18 | 3,188 | 144 |
s = input().replace("eraser", "").replace("erase", "").replace("dreamer", "").replace("dream", "")
if s:
print("NO")
else:
print("YES")
|
s957513092
|
p03597
|
u813450984
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 51 |
We have an N \times N square grid. We will paint each square in the grid either black or white. If we paint exactly A squares white, how many squares will be painted black?
|
N = int(input())
A = int(input())
print(N * 2 - A)
|
s318863943
|
Accepted
| 20 | 2,940 | 56 |
N = int(input())
A = int(input())
print(pow(N, 2) - A)
|
s189914083
|
p02406
|
u138628845
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 119 |
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
a = input()
i = 0
for i in range(int(a)):
if ((int(i) + 1) % 3) == 0:
print(' {}'.format(int(i)+1),end='')
|
s568264557
|
Accepted
| 40 | 5,872 | 553 |
a = input()
i = 0
v = 0
for i in range(int(a)):
if ((int(i)+1) % 3) == 0:
print(' {}'.format(int(i)+1),end='')
elif((int(i)+1) % 10) == 3:
print(' {}'.format(int(i)+1),end='')
else:
x = int(i) + 1
while v == 0:
x = int(x) // 10
if (int(x) > 10) and (not(int(x) % 10) == 3):
pass
elif (int(x) % 10) == 3:
print(' {}'.format(int(i)+1),end='')
v = 1
else:
v = 1
v = 0
print('')
|
s114298599
|
p03160
|
u762420987
| 2,000 | 1,048,576 |
Wrong Answer
| 113 | 13,928 | 262 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N = int(input())
hlist = list(map(int, input().split()))
dp = [0] * N
dp[0] = 0
dp[1] = abs(hlist[0] - hlist[1])
for i in range(2, N):
cost1 = abs(hlist[i] - hlist[i - 1])
cost2 = abs(hlist[i] - hlist[i - 2])
dp[i] = min(cost1, cost2)
print(dp[-1])
|
s662819833
|
Accepted
| 128 | 13,904 | 237 |
N = int(input())
hlist = list(map(int, input().split()))
dp = [10**9]*(N)
dp[0] = 0
dp[1] = abs(hlist[1] - hlist[0])
for i in range(2, N):
dp[i] = min(dp[i-1]+abs(hlist[i]-hlist[i-1]), dp[i-2]+abs(hlist[i]-hlist[i-2]))
print(dp[-1])
|
s029856239
|
p03920
|
u856232850
| 2,000 | 262,144 |
Wrong Answer
| 22 | 3,332 | 132 |
The problem set at _CODE FESTIVAL 20XX Finals_ consists of N problems. The score allocated to the i-th (1≦i≦N) problem is i points. Takahashi, a contestant, is trying to score exactly N points. For that, he is deciding which problems to solve. As problems with higher scores are harder, he wants to minimize the highest score of a problem among the ones solved by him. Determine the set of problems that should be solved.
|
n = int(input())
a = int((2*n)**0.5)
b = int((a*(a+1))/2)-n
for i in range(1,b):
print(i)
for i in range(b+1,a+1):
print(i)
|
s081643523
|
Accepted
| 22 | 3,572 | 210 |
n = int(input())
a = 0
for i in range(n+1):
if i*(i+1)//2 >= n:
a = i
break
b=[]
c = a*(a+1)//2
for i in range(1,a+1):
b.append(i)
if c > n:
b.pop(c-n-1)
for i in b:
print(i)
|
s140896844
|
p04011
|
u328755070
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 127 |
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 = int(input())
K = int(input())
X = int(input())
Y = int(input())
lat = 0 if N < K else N - K
print((K-lat) * X + lat * Y)
|
s240780738
|
Accepted
| 17 | 2,940 | 127 |
N = int(input())
K = int(input())
X = int(input())
Y = int(input())
lat = 0 if N < K else N - K
print((N-lat) * X + lat * Y)
|
s291570498
|
p00031
|
u744114948
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,732 | 216 |
祖母が天秤を使っています。天秤は、二つの皿の両方に同じ目方のものを載せると釣合い、そうでない場合には、重い方に傾きます。10 個の分銅の重さは、軽い順に 1g, 2g, 4g, 8g, 16g, 32g, 64g, 128g, 256g, 512g です。 祖母は、「1kg くらいまでグラム単位で量れるのよ。」と言います。「じゃあ、試しに、ここにあるジュースの重さを量ってよ」と言ってみると、祖母は左の皿にジュースを、右の皿に 8g と64g と128g の分銅を載せて釣合わせてから、「分銅の目方の合計は 200g だから、ジュースの目方は 200g ね。どう、正しいでしょう？」と答えました。 左の皿に載せる品物の重さを与えるので、天秤で与えられた重みの品物と釣合わせるときに、右の皿に載せる分銅を軽い順に出力するプログラムを作成して下さい。ただし、量るべき品物の重さは、すべての分銅の重さの合計 (=1023g) 以下とします。
|
while True:
try:
n=int(input())
except:
break
l=list(format(n,'b'))
for i in range(len(l)):
if l[-i-1] == "1":
print(2**i, end="???")
print()
|
s266326993
|
Accepted
| 30 | 6,716 | 281 |
while True:
try:
n=int(input())
except:
break
l=list(format(n,'b'))
for i in range(len(l)):
if l[-i-1] == "1":
print(2**i, end="")
if i != len(l)-1:
print(" ",end="")
print()
|
s729243584
|
p02612
|
u510331904
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,136 | 36 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
num = int(input())
print(num%1000)
|
s988738210
|
Accepted
| 27 | 9,136 | 90 |
num = int(input())
if num % 1000 == 0:
print(0)
else:
print(1000 - (num % 1000))
|
s299851421
|
p04043
|
u265304824
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 207 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a, b, c = input().split(' ')
cnt5 = 0
cnt7 = 0
for num in [a, b, c]:
if num == 5:
cnt5 += 1
if num == 7:
cnt7 += 1
if cnt5 == 2 and cnt7 == 1:
print('YES')
else:
print('NO')
|
s164089420
|
Accepted
| 17 | 3,060 | 225 |
a, b, c = [int(s) for s in input().split(' ')]
cnt5 = 0
cnt7 = 0
for num in [a, b, c]:
if num == 5:
cnt5 += 1
if num == 7:
cnt7 += 1
if cnt5 == 2 and cnt7 == 1:
print('YES')
else:
print('NO')
|
s871112166
|
p03469
|
u534319350
| 2,000 | 262,144 |
Wrong Answer
| 31 | 4,436 | 146 |
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.
|
import datetime
S = input()
S = datetime.datetime.strptime(S, '%Y/%m/%d')
a = int(S.year)
if a != 2018:
S = S.replace(2018)
print(S)
|
s723685530
|
Accepted
| 30 | 4,464 | 167 |
import datetime
S = input()
S = datetime.datetime.strptime(S, '%Y/%m/%d')
a = int(S.year)
if a != 2018:
S = S.replace(2018)
print(S.strftime('%Y/%m/%d'))
|
s349730994
|
p03338
|
u352623442
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 3,064 | 271 |
You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when we cut the string at the optimal position.
|
n = int(input())
s = input()
ans = 0
for i in range(n):
a = s[:i]
b = s[i+1:]
tes = 0
c = []
for cha in a:
if cha in b:
if cha not in c:
c.append(cha)
tes += 1
ans = max(tes,ans)
print(ans)
|
s043055566
|
Accepted
| 18 | 3,060 | 272 |
n = int(input())
s = input()
ans = 0
for i in range(n):
a = s[:i]
b = s[i:]
tes = 0
c = []
for cha in a:
if cha in b:
if cha not in c:
c.append(cha)
tes += 1
ans = max(tes,ans)
print(ans)
|
s884178914
|
p03228
|
u059210959
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 303 |
In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person. Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.
|
#encoding utf-8
a,b,k = map(int, input().split())
print(a,b,k)
for i in range(k):
if i%2 == 0:
if a%2 == 1:
a -= 1
b += int(a/2)
a = int(a/2)
# turn b
else:
if b%2 == 1:
b -= 1
a += int(b/2)
b = int(b/2)
print(a,b)
|
s368705739
|
Accepted
| 42 | 10,688 | 726 |
#!/usr/bin/env python3
# encoding:utf-8
import copy
import random
import bisect
import fractions
import math
import sys
import collections
from decimal import Decimal
mod = 10**9+7
sys.setrecursionlimit(mod)
d = collections.deque()
def LI(): return list(map(int, sys.stdin.readline().split()))
A, B, K = LI()
for k in range(K):
if k % 2 == 0:
if A % 2 == 1:
A -= 1
c = A // 2
A -= c
B += c
else:
if B % 2 == 1:
B -= 1
c = B // 2
B -= c
A += c
print(A, B)
|
s975950478
|
p03997
|
u552738814
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 99 |
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())
def area():
return (a+b)*h/2
print(area())
|
s078821821
|
Accepted
| 27 | 9,168 | 114 |
list = []
for i in range(3):
list.append(int(input()))
area = (list[0]+list[1])*list[2]/2
print(int(area))
|
s337500858
|
p02842
|
u419963262
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 148 |
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
N=int(input())
import math
if math.floor(N*(100/108))+1<float((N+1)*(100/108)):
print(math.floor(math.floor(N*(100/108))+1))
else:
print(':(')
|
s648488258
|
Accepted
| 18 | 3,060 | 110 |
import math
N=int(input())
c=math.ceil(N*100/108)
f=((N+1)*100/108)
if c<f:
print(c)
else:
print(":(")
|
s081286497
|
p03997
|
u961469795
| 2,000 | 262,144 |
Wrong Answer
| 21 | 9,036 | 74 |
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)
|
s720117093
|
Accepted
| 22 | 8,932 | 79 |
a = int(input())
b = int(input())
h = int(input())
print(int((a + b) * h / 2))
|
s618854456
|
p03361
|
u527206610
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 296 |
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j). Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`. However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black. Determine if square1001 can achieve his objective.
|
h, w = map(int, input().split())
S = []
flag = True
for i in range(h):
r = list(input())
S.append(r)
for i in range(0,h-1):
for j in range(0,w-1):
if S[i][j]=='#' and S[i-1][j]=='.' and S[i][j-1]=='.' and S[i+1][j]=='.' and S[i][j+1]=='.':
flag = False
print(flag)
|
s523876633
|
Accepted
| 18 | 3,064 | 332 |
h, w = map(int, input().split())
S = []
flag = True
for i in range(h):
r = list(input())
S.append(r)
for i in range(0,h-1):
for j in range(0,w-1):
if S[i][j]=='#' and S[i-1][j]=='.' and S[i][j-1]=='.' and S[i+1][j]=='.' and S[i][j+1]=='.':
flag = False
if flag:
print('Yes')
else:
print('No')
|
s725596610
|
p02613
|
u799528852
| 2,000 | 1,048,576 |
Wrong Answer
| 142 | 9,208 | 319 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(n):
s = input()
if s == "AC":
ac += 1
elif s == "WA":
wa += 1
elif s == "TLE":
tle += 1
else:
re += 1
print("AC ×" + str(ac))
print("WA ×" + str(wa))
print("TLE ×" + str(tle))
print("RE ×" + str(re))
|
s752448108
|
Accepted
| 145 | 9,052 | 319 |
n = int(input())
ac = 0
wa = 0
tle = 0
re = 0
for i in range(n):
s = input()
if s == "AC":
ac += 1
elif s == "WA":
wa += 1
elif s == "TLE":
tle += 1
else:
re += 1
print("AC x " + str(ac))
print("WA x " + str(wa))
print("TLE x " + str(tle))
print("RE x " + str(re))
|
s665389728
|
p03588
|
u993622994
| 2,000 | 262,144 |
Wrong Answer
| 488 | 30,636 | 203 |
A group of people played a game. All players had distinct scores, which are positive integers. Takahashi knows N facts on the players' scores. The i-th fact is as follows: the A_i-th highest score among the players is B_i. Find the maximum possible number of players in the game.
|
n = int(input())
ab = sorted([list(map(int, input().split())) for _ in range(n)])
ans = 0
ans += ab[n-1][0]
if ab[0][0] != 1:
ans += ab[0][0] - 1
if ab[n-1][1] > 0:
ans += ab[n-1][1]
print(ans)
|
s502522517
|
Accepted
| 490 | 28,652 | 110 |
n = int(input())
ab = sorted([list(map(int, input().split())) for _ in range(n)])
print(ab[-1][0] + ab[-1][1])
|
s354689098
|
p04012
|
u875361824
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 223 |
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
def main():
w = input()
for o in range(ord('a'), ord('z')+1):
c = chr(o)
if w.count(c) % 2 == 1:
print("NO")
return
print("YES")
if __name__ == '__main__':
main()
|
s911810689
|
Accepted
| 17 | 2,940 | 223 |
def main():
w = input()
for o in range(ord('a'), ord('z')+1):
c = chr(o)
if w.count(c) % 2 == 1:
print("No")
return
print("Yes")
if __name__ == '__main__':
main()
|
s896605320
|
p03456
|
u336624604
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 235 |
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.
|
n = list(map(int,input().split()))
s = 1
for i in range(3):
s *=10
if n[1]<s:
N = n[0]*s+n[1]
for i in range(100):
if N==i*i:
print('Yes')
exit()
break
print('No')
|
s147631779
|
Accepted
| 17 | 2,940 | 99 |
a,b = input().split()
x = int(a+b)
if x == int(x**0.5)**2:
print('Yes')
else:
print('No')
|
s740827617
|
p02612
|
u473430959
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,136 | 50 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n=int(input())
x=n//1000
m=n-(1000*x)
print(m)
|
s819946719
|
Accepted
| 29 | 9,128 | 89 |
n=int(input())
x=n//1000
m=n-(1000*x)
if m==0:
print(0)
else:
y=1000-m
print(y)
|
s113494707
|
p01102
|
u896240461
| 8,000 | 262,144 |
Wrong Answer
| 20 | 5,572 | 698 |
The programming contest named _Concours de Programmation Comtemporaine Interuniversitaire_ (CPCI) has a judging system similar to that of ICPC; contestants have to submit correct outputs for two different inputs to be accepted as a correct solution. Each of the submissions should include the program that generated the output. A pair of submissions is judged to be a correct solution when, in addition to the correctness of the outputs, they include an identical program. Many contestants, however, do not stop including a different version of their programs in their second submissions, after modifying a single string literal in their programs representing the input file name, attempting to process different input. The organizers of CPCI are exploring the possibility of showing a special error message for such _close_ submissions, indicating contestants what's wrong with such submissions. Your task is to detect such close submissions.
|
while 1:
s1 = input().split('"')
l1 = len(s1)
if s1[0] == ".": break
s2 = input().split('"')
l2 = len(s2)
if l1 != l2:
print("DIFFERENT")
break
i = 0
j = 0
k = 0
flag_1 = 0
flag_2 = 0
cnt = 0
x1 = []
x2 = []
while 1:
if i > l1-2:
if cnt == 1:
print("CLOSE")
break
else:
print("IDENTICAL")
break
if s1[i] != s2[i]:
print("DIFFERENT")
break
if s1[i+1] != s2[i+1]:
if cnt == 1:
print("DIFFERENT")
break
else: cnt = 1
i += 2
|
s571739833
|
Accepted
| 20 | 5,576 | 627 |
while 1:
s1 = input().split('"')
l1 = len(s1)
if s1[0] == ".": break
s2 = input().split('"')
l2 = len(s2)
if l1 != l2:
print("DIFFERENT")
continue
i = 0
cnt = 0
while 1:
if s1[i] != s2[i]:
print("DIFFERENT")
break
if i > l1-2:
if cnt == 1:
print("CLOSE")
break
else:
print("IDENTICAL")
break
if s1[i+1] != s2[i+1]:
if cnt == 1:
print("DIFFERENT")
break
else: cnt = 1
i += 2
|
s487701168
|
p03697
|
u875361824
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 156 |
You are given two integers A and B as the input. Output the value of A + B. However, if A + B is 10 or greater, output `error` instead.
|
def main():
s = input().strip()
if len(s) == len(set(s)):
print("yes")
else:
print("no")
if __name__ == '__main__':
main()
|
s834440960
|
Accepted
| 17 | 2,940 | 190 |
def main():
a, b = list(map(int, input().split()))
answer = a + b
if answer >= 10:
print("error")
else:
print(answer)
if __name__ == '__main__':
main()
|
s241286423
|
p02414
|
u587193722
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,668 | 518 |
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 = [int(i) for i in input().split()]
A = []
B = []
C = []
D = []
for ni in range(n):
A.append([int(i) for i in input().split()])
for mi in range(m):
B.append([int(i) for i in input().split()])
print(A)
print(B)
for j in range(n):
for h in range(l):
for f in range(m):
C.append((A[j][f])*(B[f][h]))
for a in range(0,len(C),m):
D.append(C[a]+C[a+m-1])
for b in range(0,len(D)):
if (b+1) % n == 0:
print(D[b],end=' \n')
else:
print(D[b], end=' ')
|
s679849378
|
Accepted
| 530 | 8,788 | 419 |
n, m, l = [int(i) for i in input().split()]
A = []
B = []
C = []
for ni in range(n):
A.append([int(i) for i in input().split()])
for mi in range(m):
B.append([int(i) for i in input().split()])
for i in range(n):
C.append([])
for j in range(l):
C[i].append(0)
for k in range(m):
C[i][j] += A[i][k] * B[k][j]
for ni in range(n):
print(" ".join([str(s) for s in C[ni]]))
|
s428139387
|
p03228
|
u972398652
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 215 |
In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person. Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.
|
A, B, K = map(int, input().split())
for k in range(K):
if A % 2 != 0:
A -= 1
B += A/2
A -= A/2
if B % 2!= 0:
B -= 1
A += B/2
B -= B/2
print(round(A), round(B))
|
s135477898
|
Accepted
| 17 | 3,064 | 420 |
A, B, K = map(int, input().split())
k = 0
while True:
if A % 2 != 0:
A -= 1
B += round(A/2)
A -= round(A/2)
k += 1
if k == K:
break
if B % 2 != 0:
B -= 1
A += round(B/2)
B -= round(B/2)
k += 1
if k == K:
break
print(A, B)
|
s230514241
|
p02255
|
u605879293
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 363 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
def insertionSort(length, array):
for i in range(1, length):
v = array[i]
j = i - 1
while j >= 0 and array[j] > v:
array[j+1] = array[j]
j -= 1
array[j+1] = v
print(' '.join(map(str, array)))
length = int(input())
num_array = [int(x) for x in input().split()]
insertionSort(length, num_array)
|
s231509285
|
Accepted
| 20 | 5,976 | 303 |
def insertionSort(a):
n = len(a)
print(*a)
for i in range(1, n):
v = a[i]
j = i-1
while j >= 0 and a[j] > v:
a[j+1] = a[j]
j += -1
a[j+1] = v
print(*a)
n = int(input())
a = [int(x) for x in input().split()]
insertionSort(a)
|
s133663487
|
p03130
|
u767438459
| 2,000 | 1,048,576 |
Wrong Answer
| 32 | 9,116 | 187 |
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once.
|
d=[0,0,0,0]
for i in range(3):
a,b=map(int,input().split())
d[a-1]+=1
d[b-1]+=1
if d[0] <= 2 and d[1] <= 2 and d[2] <= 2 and d[3] <= 2:
print("Yes")
else:
print("No")
|
s921295895
|
Accepted
| 32 | 9,180 | 187 |
d=[0,0,0,0]
for i in range(3):
a,b=map(int,input().split())
d[a-1]+=1
d[b-1]+=1
if d[0] <= 2 and d[1] <= 2 and d[2] <= 2 and d[3] <= 2:
print("YES")
else:
print("NO")
|
s137557713
|
p02854
|
u539281377
| 2,000 | 1,048,576 |
Wrong Answer
| 466 | 26,024 | 153 |
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters. Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut: * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan). * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen. Find the minimum amount of money needed before cutting the bar into two parts with the same length.
|
N=int(input())
A=list(map(int,input().split()))
s=ans=sum(A)
k=0
for i in range(N-1):
k+=A[i]
ans=min(ans,abs(s-2*k))
print(k,ans)
print(ans)
|
s343148812
|
Accepted
| 183 | 26,220 | 136 |
N=int(input())
A=list(map(int,input().split()))
s=ans=sum(A)
k=0
for i in range(N-1):
k+=A[i]
ans=min(ans,abs(s-2*k))
print(ans)
|
s261966094
|
p03679
|
u924406834
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 136 |
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
import sys
x,a,b = map(int,input().split())
if b <= a:
print("delicious")
elif a + x <= b:
print("safe")
else:
print("dangerous")
|
s194576174
|
Accepted
| 17 | 2,940 | 137 |
import sys
x,a,b = map(int,input().split())
if b <= a:
print("delicious")
elif a + x >= b:
print("safe")
else:
print("dangerous")
|
s726375624
|
p03636
|
u753682919
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 43 |
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.
|
s=input();n=len(s);print(s[0]+str(n)+s[-1])
|
s108706654
|
Accepted
| 18 | 2,940 | 45 |
s=input();n=len(s)-2;print(s[0]+str(n)+s[-1])
|
s996290691
|
p04030
|
u736729525
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 101 |
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?
|
a = []
for c in input():
if c in '01':
a.append(c)
else:
a = a[:-1]
print(a)
|
s429235378
|
Accepted
| 17 | 3,064 | 110 |
a = []
for c in input():
if c in '01':
a.append(c)
else:
a = a[:-1]
print("".join(a))
|
s077885956
|
p03386
|
u798316285
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 169 |
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())
if (b-a)>2*k:
ans=[i for i in range(a,a+k+1)]+[i for i in range(b-k,b+1)]
else:
ans=[i for i in range(a,b+1)]
for i in ans:
print(i)
|
s749381164
|
Accepted
| 17 | 3,060 | 188 |
a,b,k=map(int,input().split())
if b-a+1>2*k:
ans=sorted(list(set([i for i in range(a,a+k)]+[i for i in range(b-k+1,b+1)])))
else:
ans=[i for i in range(a,b+1)]
for i in ans:
print(i)
|
s594300028
|
p03485
|
u432333240
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 94 |
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a, b = map(int, input().split())
s = a+b
if s%2!=0:
print(int(s/2)+1)
else:
print(s/2)
|
s115001149
|
Accepted
| 17 | 2,940 | 99 |
a, b = map(int, input().split())
s = a+b
if s%2!=0:
print(int(s/2)+1)
else:
print(int(s/2))
|
s953415843
|
p03456
|
u323680411
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 480 |
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.
|
from sys import stdin
def main() -> None:
a = next_int()
b = next_int()
ab = int(str(a) + str(b))
sqrt = int(ab ** 0.5)
print("YES" if sqrt ** 2 == ab else "No")
def next_int() -> int:
return int(next_str())
def next_str() -> str:
result = ""
while True:
tmp = stdin.read(1)
if tmp.strip() != "":
result += tmp
elif tmp != '\r':
break
return result
if __name__ == '__main__':
main()
|
s947228986
|
Accepted
| 17 | 3,064 | 481 |
from sys import stdin
def main() -> None:
a = next_int()
b = next_int()
ab = int(str(a) + str(b))
sqrt = int(ab ** 0.5)
print(["No", "Yes"][int(sqrt ** 2 == ab)])
def next_int() -> int:
return int(next_str())
def next_str() -> str:
result = ""
while True:
tmp = stdin.read(1)
if tmp.strip() != "":
result += tmp
elif tmp != '\r':
break
return result
if __name__ == '__main__':
main()
|
s776199794
|
p02601
|
u477837488
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,188 | 293 |
M-kun has the following three cards: * A red card with the integer A. * A green card with the integer B. * A blue card with the integer C. He is a genius magician who can do the following operation at most K times: * Choose one of the three cards and multiply the written integer by 2. His magic is successful if both of the following conditions are satisfied after the operations: * The integer on the green card is **strictly** greater than the integer on the red card. * The integer on the blue card is **strictly** greater than the integer on the green card. Determine whether the magic can be successful.
|
A, B, C = map(int, input().split())
q = int(input())
k = 0
for i in range(q+1):
for j in range(0, q-i+1):
print(i, j)
B_ = B * (2 ** i)
C_ = C * (2 ** j)
if A < B_ and B_ < C_ and A < C_:
k = 1
if k != 0:
print('Yes')
else:
print('No')
|
s314053744
|
Accepted
| 32 | 9,124 | 273 |
A, B, C = map(int, input().split())
q = int(input())
k = 0
for i in range(q+1):
for j in range(0, q-i+1):
B_ = B * (2 ** i)
C_ = C * (2 ** j)
if A < B_ and B_ < C_ and A < C_:
k = 1
if k != 0:
print('Yes')
else:
print('No')
|
s208270545
|
p03696
|
u667024514
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 331 |
You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.
|
n = int(input())
lis = []
li = []
cou = 0
s = str(input())
for i in range(n):
if s[i] == "(":
cou += 1
lis.append(s[i])
else:
cou -= 1
while cou < 0:
lis.append("(")
cou += 1
lis.append(s[i])
while cou > 0:
lis.append(")")
cou -= 1
print("".join(lis))
|
s560101061
|
Accepted
| 17 | 3,064 | 277 |
n = int(input())
s = input()
cnt = [0, 0]
add = [0, 0]
ans = ''
for i in range(n):
if s[i] == ')':
cnt[0] += 1
if cnt[0] > cnt[1]:
add[1] += 1
cnt[1] += 1
else:
cnt[1] += 1
print('(' * add[1] + s + ')'*(cnt[1] -cnt[0]))
|
s103312809
|
p03854
|
u729939940
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,188 | 151 |
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
S = input()
s = S.replace("eraser","").replace("erase","").replace("dreamer","").replace("dream","")
if s == "":
print("Yes")
else:
print("No")
|
s113623145
|
Accepted
| 18 | 3,188 | 151 |
S = input()
s = S.replace("eraser", "").replace("erase", "").replace("dreamer", "").replace("dream", "")
if s == "":
print("YES")
else:
print("NO")
|
s298735561
|
p02608
|
u067227603
| 2,000 | 1,048,576 |
Wrong Answer
| 2,205 | 9,196 | 387 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
import math
N = int(input())
def counter(N):
global count
max_num = int(math.sqrt(N))
for x in range(1, max_num + 1):
for y in range(1, max_num + 1):
for z in range(1, max_num + 1):
if x * x + y * y + z * z + x * y + y * z + z * x == N:
count += 1
print(count)
for i in range(N + 1):
count = 0
counter(i)
|
s623480918
|
Accepted
| 310 | 30,372 | 455 |
import math
import collections
N = int(input())
pre = []
for x in range(1, 101):
max_y = int(math.sqrt(10000 - x * x))
for y in range(1, max_y + 1):
max_z = int(math.sqrt(10000 - x * x - y * y))
for z in range(1, max_z + 1):
pre.append(x * x + y * y + z * z + x * y + y * z + z * x)
count = collections.Counter(pre)
for i in range(1, N + 1):
if i not in count:
print(0)
else:
print(count[i])
|
s875890234
|
p03141
|
u984276646
| 2,000 | 1,048,576 |
Wrong Answer
| 786 | 37,800 | 542 |
There are N dishes of cuisine placed in front of Takahashi and Aoki. For convenience, we call these dishes Dish 1, Dish 2, ..., Dish N. When Takahashi eats Dish i, he earns A_i points of _happiness_ ; when Aoki eats Dish i, she earns B_i points of happiness. Starting from Takahashi, they alternately choose one dish and eat it, until there is no more dish to eat. Here, both of them choose dishes so that the following value is maximized: "the sum of the happiness he/she will earn in the end" minus "the sum of the happiness the other person will earn in the end". Find the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end".
|
N = int(input())
dishes = [list(map(int, input().split())) + [0] for _ in range(N)]
dishes.sort()
cnt = 0
dT = [dishes[i][0] * N + i for i in range(N)]
dA = [dishes[i][1] * N + i for i in range(N)]
dT.sort(reverse=True)
dA.sort(reverse=True)
print(dishes)
T, A = 0, 0
t, a = 0, 0
for i in range(N):
if i % 2 == 0:
while dishes[dT[t]%N][2] == 1:
t += 1
T += dishes[dT[t]%N][0]
dishes[dT[t]%N][2] = 1
else:
while dishes[dA[a]%N][2] == 1:
a += 1
A += dishes[dA[a]%N][1]
dishes[dA[a]%N][2] = 1
print(T - A)
|
s330585539
|
Accepted
| 244 | 13,364 | 227 |
N = int(input())
AB = [0 for _ in range(N)]
SB = 0
for i in range(N):
A, B = map(int, input().split())
SB += B
AB[i] = A + B
AB.sort(reverse=True)
SAB = 0
for i in range(N):
if i % 2 == 0:
SAB += AB[i]
print(SAB-SB)
|
s092649380
|
p02393
|
u227984374
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,592 | 171 |
Write a program which reads three integers, and prints them in ascending order.
|
a, b, c = map(int, input().split())
while b < c :
b, c = c, b
break
while a < b :
a, b = b, a
break
while b < c :
b, c = c, b
break
print(a, b, c)
|
s940716421
|
Accepted
| 20 | 5,596 | 171 |
a, b, c = map(int, input().split())
while a > b :
a, b = b, a
break
while b > c :
b, c = c, b
break
while a > b :
a, b = b, a
break
print(a, b, c)
|
s954662102
|
p03712
|
u796563423
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,112 | 187 |
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
a,b=map(int,input().split())
s=[]
for i in range(a):
s.append(input())
print("*"*(b+2))
for i in range(a):
print("*",end="")
print(s[i],end="")
print("*")
print("*"*(b+2))
|
s312032080
|
Accepted
| 29 | 9,136 | 187 |
a,b=map(int,input().split())
s=[]
for i in range(a):
s.append(input())
print("#"*(b+2))
for i in range(a):
print("#",end="")
print(s[i],end="")
print("#")
print("#"*(b+2))
|
s549561498
|
p02975
|
u833071789
| 2,000 | 1,048,576 |
Wrong Answer
| 76 | 14,116 | 516 |
Snuke has N hats. The i-th hat has an integer a_i written on it. There are N camels standing in a circle. Snuke will put one of his hats on each of these camels. If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print `Yes`; otherwise, print `No`. * The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself. What is XOR? The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non- negative integers x_1, x_2, \ldots, x_n is defined as follows: - When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even. For example, 3 \oplus 5 = 6.
|
def solve():
num = int(input())
camels = [int(i) for i in input().split()]
if len(camels) % 3 != 0:
print("no")
return(0)
camels.sort()
length = len(camels)
for i in range(3):
for j in range(int(i*length/3), int((i+1)*length/3)):
if camels[int(i*length/3)] != camels[j]:
print("no")
return(0)
if camels[0] ^ camels[int(num/3)] != camels[int(num/3)*2]:
print("no")
return(0)
print("yes")
solve()
|
s931551786
|
Accepted
| 311 | 23,160 | 676 |
# -*- coding: utf-8 -*-
import numpy
import sys
import os
def solve():
num = int(input())
camels = [int(i) for i in input().split()]
if len(camels) % 3 != 0:
for c in camels:
if c !=0:
print("No")
return(0)
print("Yes")
return
camels.sort()
length = len(camels)
for i in range(3):
for j in range(int(i*length/3), int((i+1)*length/3)):
if camels[int(i*length/3)] != camels[j]:
print("No")
return(0)
if camels[0] ^ camels[int(num/3)] != camels[int(num/3)*2]:
print("No")
return(0)
print("Yes")
solve()
|
s999216725
|
p03474
|
u916908463
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 205 |
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
A,B = map(int, input().split())
S = str(input())
for i,s in enumerate(S):
if not i == A and S[i] == '-':
print("NO")
exit()
print('YES' if S[A] == '-' and len(S) == A+B+1 else "NO")
|
s185441915
|
Accepted
| 18 | 3,060 | 205 |
A,B = map(int, input().split())
S = str(input())
for i,s in enumerate(S):
if not i == A and S[i] == '-':
print("No")
exit()
print('Yes' if S[A] == '-' and len(S) == A+B+1 else "No")
|
s754048295
|
p02607
|
u307622233
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,028 | 230 |
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.
|
def main():
N = int(input())
A = [int(i) for i in input().split()]
ans = 0
for v in A[0::2]:
if v % 2 == 1:
ans += 1
print(v)
print(ans)
if __name__ == '__main__':
main()
|
s840546273
|
Accepted
| 27 | 9,168 | 250 |
def main():
N = int(input())
# L, R, d = map(int, input().split())
A = [int(i) for i in input().split()]
ans = 0
for v in A[0::2]:
if v % 2 == 1:
ans += 1
print(ans)
if __name__ == '__main__':
main()
|
s440748667
|
p04030
|
u754553095
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 143 |
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?
|
a=list(input())
print(a)
b=[]
for i in a:
if (i=="0") or (i=="1"):
b.append(i)
elif b!=[] :
b.pop(-1)
print("".join(b))
|
s517904036
|
Accepted
| 17 | 2,940 | 134 |
a=list(input())
b=[]
for i in a:
if (i=="0") or (i=="1"):
b.append(i)
elif b!=[] :
b.pop(-1)
print("".join(b))
|
s883436108
|
p03555
|
u694370915
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 201 |
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.
|
def main():
a = input()
b = input()
if a[0] == b[2]:
if a[1] == b[1]:
if a[2] == b[0]:
print('Yes')
return
print('No')
return
if __name__ == '__main__':
main()
|
s545596196
|
Accepted
| 18 | 2,940 | 201 |
def main():
a = input()
b = input()
if a[0] == b[2]:
if a[1] == b[1]:
if a[2] == b[0]:
print('YES')
return
print('NO')
return
if __name__ == '__main__':
main()
|
s505925422
|
p03494
|
u483722302
| 2,000 | 262,144 |
Wrong Answer
| 20 | 3,060 | 255 |
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.
|
# -*- coding: utf-8 -*-
n = input()
x = list(map(int, input().split()))
cnt = 0
while True:
print(x)
s = sum(list(map(lambda elem: elem % 2, x)))
if s != 0:
break
cnt += 1
x = list(map(lambda elem: elem // 2, x))
print(cnt)
|
s731464536
|
Accepted
| 19 | 2,940 | 242 |
# -*- coding: utf-8 -*-
n = input()
x = list(map(int, input().split()))
cnt = 0
while True:
s = sum(list(map(lambda elem: elem % 2, x)))
if s != 0:
break
cnt += 1
x = list(map(lambda elem: elem // 2, x))
print(cnt)
|
s874618920
|
p02664
|
u370721525
| 2,000 | 1,048,576 |
Wrong Answer
| 89 | 11,104 | 262 |
For a string S consisting of the uppercase English letters `P` and `D`, let the _doctoral and postdoctoral quotient_ of S be the total number of occurrences of `D` and `PD` in S as contiguous substrings. For example, if S = `PPDDP`, it contains two occurrences of `D` and one occurrence of `PD` as contiguous substrings, so the doctoral and postdoctoral quotient of S is 3. We have a string T consisting of `P`, `D`, and `?`. Among the strings that can be obtained by replacing each `?` in T with `P` or `D`, find one with the maximum possible doctoral and postdoctoral quotient.
|
T = input()
l = [t for t in T]
s = len(T)
for i in range(s):
if l[i] == '?':
if i == s-1:
l[i] = 'D'
elif (i==0) or ((l[i-1]=='D') and (l[i+1]=='D')):
l[i] = 'P'
else:
l[i] = 'D'
str_changed = "".join(l)
print(str_changed)
|
s087596198
|
Accepted
| 98 | 11,160 | 349 |
T = input()
l = [t for t in T]
s = len(T)
for i in range(s):
if l[i] == '?':
if i == s-1:
l[i] = 'D'
elif i == 0 and l[i+1]!='P':
l[i] = 'P'
elif i == 0:
l[i] = 'D'
elif l[i-1]=='D' and ((l[i+1]=='D')or(l[i+1]=='?')):
l[i] = 'P'
else:
l[i] = 'D'
str_changed = "".join(l)
print(str_changed)
|
s727671288
|
p03997
|
u756420279
| 2,000 | 262,144 |
Wrong Answer
| 16 | 2,940 | 73 |
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.
|
#!/usr/bin/env python3
print((int(input())+int(input()))*int(input())/2)
|
s169891644
|
Accepted
| 17 | 2,940 | 74 |
#!/usr/bin/env python3
print((int(input())+int(input()))*int(input())//2)
|
s101744674
|
p02402
|
u343748576
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,412 | 125 |
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
n = int(input())
a = input().split()
for i in range(n):
a[i] = int(a[i])
print("{} {} {}".format(max(a), min(a), sum(a)))
|
s534187031
|
Accepted
| 20 | 8,368 | 125 |
n = int(input())
a = input().split()
for i in range(n):
a[i] = int(a[i])
print("{} {} {}".format(min(a), max(a), sum(a)))
|
s822735909
|
p03962
|
u602715823
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,064 | 30 |
AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him.
|
s =set(input().split())
len(s)
|
s003185380
|
Accepted
| 22 | 3,064 | 37 |
s =set(input().split())
print(len(s))
|
s424203227
|
p03456
|
u581403769
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 132 |
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.
|
a, b = input().split()
x = int(a + b)
import math
sqx = math.sqrt(x)
if type(sqx) == 'int':
print('Yes')
else:
print('No')
|
s124058678
|
Accepted
| 18 | 2,940 | 124 |
a, b = input().split()
x = int(a + b)
yn = 'No'
for i in range(1, 1000):
if i ** 2 == x:
yn = 'Yes'
print(yn)
|
s343310125
|
p02646
|
u127025777
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,184 | 180 |
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a = list(map(int,input().split()))
b = list(map(int,input().split()))
T = int(input())
lon = abs(a[0]-b[0])
if (b[1]*T+lon) - a[1]*T <= 0 :
print("Yes")
else :
print("No")
|
s116197403
|
Accepted
| 23 | 9,180 | 176 |
a = list(map(int,input().split()))
b = list(map(int,input().split()))
T = int(input())
lon = abs(a[0]-b[0])
if a[1]*T - b[1]*T >= lon :
print("YES")
else :
print("NO")
|
s569366300
|
p02396
|
u117053676
| 1,000 | 131,072 |
Wrong Answer
| 130 | 7,536 | 126 |
In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are given for this problem.
|
i=1
while True:
a = int(input())
if a==0:
break
else:
print("case {}: {}".format(i,a))
i += 1;
|
s720604806
|
Accepted
| 130 | 7,536 | 126 |
i=1
while True:
a = int(input())
if a==0:
break
else:
print("Case {}: {}".format(i,a))
i += 1;
|
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