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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s823920468
|
p03853
|
u691018832
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 260
|
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
h, w = map(int, readline().split())
for i in range(h):
c = read().rstrip().decode()
print(c + '\n' + c)
|
s169632016
|
Accepted
| 17
| 3,064
| 264
|
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)
h, w = map(int, readline().split())
for i in range(h):
c = readline().rstrip().decode()
print(c + '\n' + c)
|
s934028473
|
p03636
|
u669770658
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 57
|
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.
|
s = str(input())
print(s[0] + str(len(s[1:-2])) + s[-1])
|
s824614407
|
Accepted
| 17
| 2,940
| 57
|
s = str(input())
print(s[0] + str(len(s[1:-1])) + s[-1])
|
s059514290
|
p04012
|
u614550445
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,316
| 160
|
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.
|
from collections import Counter
w = input()
c = Counter(w)
for k, v in c.items():
if v % 2 != 0:
print("NO")
exit()
print("YES")
|
s198843093
|
Accepted
| 20
| 3,316
| 152
|
from collections import Counter
w = input()
c = Counter(w)
for k, v in c.items():
if v % 2 != 0:
print("No")
exit()
print("Yes")
|
s559584711
|
p03565
|
u231079882
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 339
|
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
s = input()
t = input()
ans = "UNRESTORABLE"
for i in range(len(s)-len(t)+1):
for j in range(len(t)):
if s[i+j] != t[j] and s[i+j] != "?":
break
if j == len(t) - 1:
mazai = (s[:i] + t + s[i+len(t):]).replace("?","a")
if mazai < "UNRESTORABLE":
ans = mazai
print(ans)
|
s017736567
|
Accepted
| 22
| 3,064
| 347
|
s = input()
t = input()
ans = "UNRESTORABLE"
for i in range(len(s)-len(t)+1):
for j in range(len(t)):
if s[i+j] != t[j] and s[i+j] != "?":
break
if j == len(t) - 1:
mazai = (s[:i] + t + s[i+len(t):]).replace("?","a")
if True or mazai < "UNRESTORABLE":
ans = mazai
print(ans)
|
s689472726
|
p03591
|
u215743476
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 71
|
Ringo is giving a present to Snuke. Ringo has found out that Snuke loves _yakiniku_ (a Japanese term meaning grilled meat. _yaki_ : grilled, _niku_ : meat). He supposes that Snuke likes grilled things starting with `YAKI` in Japanese, and does not like other things. You are given a string S representing the Japanese name of Ringo's present to Snuke. Determine whether S starts with `YAKI`.
|
s = input()
if s[0:3] == 'YAKI':
print('Yes')
else:
print('No')
|
s408952492
|
Accepted
| 17
| 2,940
| 71
|
s = input()
if s[0:4] == 'YAKI':
print('Yes')
else:
print('No')
|
s676091988
|
p03486
|
u675073679
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 205
|
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
def main():
s = sorted(input())
t = sorted(input())
S = ','.join(s)
T = ','.join(t)
if S < T:
print('Yes')
else:
print('No')
if __name__ == '__main__':
main()
|
s226484279
|
Accepted
| 17
| 2,940
| 211
|
def main():
s = sorted(input())
t = sorted(input())[::-1]
S = ','.join(s)
T = ','.join(t)
if S < T:
print('Yes')
else:
print('No')
if __name__ == '__main__':
main()
|
s336743473
|
p03150
|
u411923565
| 2,000
| 1,048,576
|
Wrong Answer
| 46
| 9,068
| 325
|
A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.
|
#15 B - KEYENCE String
S = list(input())
result = 'NO'
for i in range(len(S)):
Scpy = S.copy()
for j in range(i,len(S)):
Scond = Scpy[:i]+Scpy[j:]
print(Scond)
if (''.join(Scond)) == ('keyence'):
result = 'YES'
break
else:
continue
break
print(result)
|
s986260429
|
Accepted
| 31
| 9,096
| 303
|
#15 B - KEYENCE String
S = list(input())
result = 'NO'
for i in range(len(S)):
Scpy = S.copy()
for j in range(i,len(S)):
Scond = Scpy[:i]+Scpy[j:]
if (''.join(Scond)) == ('keyence'):
result = 'YES'
break
else:
continue
break
print(result)
|
s164631380
|
p03545
|
u316386814
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 345
|
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
li = list(map(int, input()))
for ops in product('+-', repeat=3):
tmp = li[0]
for o, n in zip(ops, li[1:]):
if o == '+':
tmp += n
else:
tmp -= n
if tmp == 7:
ans = str(li[0])
for o, n in zip(ops, li[1:]):
ans += o + str(n)
print(ans)
|
s211610879
|
Accepted
| 18
| 3,060
| 379
|
from itertools import product
li = list(map(int, input()))
for ops in product('+-', repeat=3):
tmp = li[0]
for o, n in zip(ops, li[1:]):
if o == '+':
tmp += n
else:
tmp -= n
if tmp == 7:
ans = str(li[0])
for o, n in zip(ops, li[1:]):
ans += o + str(n)
ans += '=7'
break
print(ans)
|
s370123355
|
p02678
|
u350491208
| 2,000
| 1,048,576
|
Wrong Answer
| 618
| 36,400
| 534
|
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
from collections import deque
n, m = map(int, input().split())
graph = [[] for _ in range(n + 1)]
for i in range(m):
a, b = map(int, input().split())
graph[a].append(b)
graph[b].append(a)
# print(graph)
mark = [-1]*(n+1)
mark[0] = 0
mark[1] = 0
d = deque()
d.append(1)
while d:
v = d.popleft()
for i in graph[v]:
if mark[i] != -1:
continue
mark[i] = v
d.append(i)
if mark.count(-1) > 0:
print("No")
else:
print("Yes")
ans = mark[1:]
print(*ans, sep="\n")
|
s657229744
|
Accepted
| 620
| 36,404
| 534
|
from collections import deque
n, m = map(int, input().split())
graph = [[] for _ in range(n + 1)]
for i in range(m):
a, b = map(int, input().split())
graph[a].append(b)
graph[b].append(a)
# print(graph)
mark = [-1]*(n+1)
mark[0] = 0
mark[1] = 0
d = deque()
d.append(1)
while d:
v = d.popleft()
for i in graph[v]:
if mark[i] != -1:
continue
mark[i] = v
d.append(i)
if mark.count(-1) > 0:
print("No")
else:
print("Yes")
ans = mark[2:]
print(*ans, sep="\n")
|
s576560720
|
p03827
|
u252828980
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 163
|
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
n = int(input())
s = input()
cnt = 0
n = 0
for i in range(n):
if s[i] == "I":
n += 1
elif s[i] =="D":
n -= 1
cnt = max(cnt,n)
print(cnt)
|
s148711859
|
Accepted
| 18
| 3,060
| 196
|
n = int(input())
s = input()
max1 = -100
n = 0
for i in range(len(s)):
if s[i] == "I":
n += 1
elif s[i] =="D":
n -= 1
max1 = max(max1,n)
print(max(max1,0))
|
s932623124
|
p02694
|
u460229551
| 2,000
| 1,048,576
|
Wrong Answer
| 25
| 9,164
| 85
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
x=int(input())
m=100
cnt=0
while x>=m:
cnt+=1
m=(m*1.01)//1
print(str(cnt))
|
s327379467
|
Accepted
| 24
| 9,228
| 83
|
x=int(input())
m=100
cnt=0
while x>m:
cnt+=1
m=(m*1.01)//1
print(str(cnt))
|
s908353824
|
p03644
|
u882370611
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 60
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n=int(input())
ans=0
while 2**ans <n:
ans+=1
print(2**ans)
|
s231274110
|
Accepted
| 17
| 2,940
| 66
|
n=int(input())
ans=0
while 2**ans <=n:
ans+=1
print(2**(ans-1))
|
s859563967
|
p03693
|
u506287026
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 103
|
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 int(r + g + b) % 4 == 0:
print('YES')
else:
print('NO')
|
s372692577
|
Accepted
| 17
| 2,940
| 93
|
r, g, b = input().split()
if int(r + g + b) % 4 == 0:
print('YES')
else:
print('NO')
|
s876005342
|
p03377
|
u816587940
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 86
|
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()))
k = a+b>=x and a<=x
print('Yes' if k else 'No')
|
s639594637
|
Accepted
| 17
| 2,940
| 90
|
a, b, x = (map(int, input().split()))
k = (a+b>=x) and (a<=x)
print('YES' if k else 'NO')
|
s684277945
|
p03920
|
u668785999
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 109
|
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())
for k in range(N):
if(N >= 2 * k + 1):
N -= k
else:
break
print(N)
|
s240965685
|
Accepted
| 24
| 3,572
| 178
|
N = int(input())
vec = []
k = 0
sum = 0
while(k**2 + k < 2*N):
k += 1
sum += k
vec.append(k)
if(sum - N):
vec.remove(sum - N)
for i in range(len(vec)):
print(vec[i])
|
s105097977
|
p03485
|
u108377418
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,060
| 153
|
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.
|
import math
if __name__ == "__main__":
print("--2つの整数を入力する--")
a, b = map(int, input().split())
print( math.ceil( (a + b ) / 2) )
|
s302019445
|
Accepted
| 18
| 3,188
| 110
|
import math
if __name__ == "__main__":
a, b = map(int, input().split())
print( math.ceil( (a + b ) / 2) )
|
s491789617
|
p02659
|
u743420240
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 9,148
| 101
|
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
def main():
a, b = input().split()
a = int(a)
b = float(b)
print(a*b)
main()
|
s940178600
|
Accepted
| 29
| 10,068
| 150
|
from decimal import Decimal
def main():
a, b = input().split()
a = Decimal(a)
b = Decimal(b)
res = int(a*b)
print(res)
main()
|
s327485765
|
p03644
|
u448720391
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 151
|
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.
|
import sys
n = int(input())
a = [2**i for i in range(0,7)]
for i in range(len(a)):
if n <= a[i]:
print(a[i])
sys.exit()
print(64)
|
s677709130
|
Accepted
| 18
| 3,060
| 210
|
import sys
n = int(input())
a = [2**i for i in range(0,7)]
for i in range(len(a)):
if n == a[i]:
print(a[i])
sys.exit()
elif n < a[i]:
print(a[i-1])
sys.exit()
print(64)
|
s099632381
|
p03377
|
u691018832
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 99
|
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
a, b, x = map(int, input().split())
if x-a > 0 and a+b >= x:
print('Yes')
else:
print('No')
|
s577089985
|
Accepted
| 17
| 2,940
| 100
|
a, b, x = map(int, input().split())
if x-a >= 0 and a+b >= x:
print('YES')
else:
print('NO')
|
s293663184
|
p03697
|
u481026841
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 92
|
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.
|
a,b = map(int,input().split())
n = a + b
if n >= 10:
print('error')
else:
print('n')
|
s948305643
|
Accepted
| 17
| 2,940
| 90
|
a,b = map(int,input().split())
n = a + b
if n >= 10:
print('error')
else:
print(n)
|
s169502867
|
p03944
|
u408958033
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,188
| 849
|
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
def cin():
return map(int,input().split())
def cino(test=False):
if not test:
return int(input())
else:
return input()
def cina():
return list(map(int,input().split()))
a,b,c = cin()
p = -1
q = 1e9
r = -1
s = 1e9
for i in range(c):
x,y,z = cin()
if z==1:
p = max(x,p)
elif z==2:
q = min(x,q)
elif z==3:
r = max(r,y)
else:
s = min(s,y)
if p==-1:
p = 0
if q==1e9:
q = a
if r==-1:
r=0
if s==1e9:
s=b
# w = b-(b-r+b-s)
# print(v,w)
l1 = list(range(0,p+1))
l2 = list(range(q,a))
l1 = set(l1)
l2 = set(l2)
l3 = l1.union(l2)
l4 = list(range(0,r))
l5 = list(range(s,b+1))
l4 = set(l4)
l5 = set(l5)
l6 = l4.union(l5)
print(l3,l6)
# print(l1,l2)
print(p,q,r,s)
if(len(l3)==a or len(l6)==b):
print(0)
else:
print((a-len(l3)+1)*(b-len(l6)+1))
|
s015178478
|
Accepted
| 18
| 3,064
| 540
|
def cin():
return map(int,input().split())
def cino(test=False):
if not test:
return int(input())
else:
return input()
def cina():
return list(map(int,input().split()))
a,b,c = cin()
p = -1
q = 1e9
r = -1
s = 1e9
for i in range(c):
x,y,z = cin()
if z==1:
p = max(x,p)
elif z==2:
q = min(x,q)
elif z==3:
r = max(r,y)
else:
s = min(s,y)
if p==-1:
p = 0
if q==1e9:
q = a
if r==-1:
r=0
if s==1e9:
s=b
ans = (q-p)*(s-r)
if p>q or r>s:
print(0)
else:
print(ans)
|
s471039742
|
p03738
|
u088974156
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 108
|
You are given two positive integers A and B. Compare the magnitudes of these numbers.
|
a=int(input())
b=int(input())
if(a>b):
print("LESS")
elif(a==b):
print("EQUAL")
else:
print("GREATER")
|
s669028615
|
Accepted
| 17
| 2,940
| 108
|
a=int(input())
b=int(input())
if(a<b):
print("LESS")
elif(a==b):
print("EQUAL")
else:
print("GREATER")
|
s579917055
|
p03997
|
u717626627
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 66
|
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)
|
s406633431
|
Accepted
| 17
| 2,940
| 69
|
a = int(input())
b = int(input())
h = int(input())
print((a+b)*h//2)
|
s952888570
|
p02608
|
u143441425
| 2,000
| 1,048,576
|
Wrong Answer
| 605
| 106,508
| 593
|
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).
|
from numba import njit
@njit('i8(i8)', cache=True)
def fun(n):
mx = int((n**.5) // 3) + 1
pairs = 0
for x in range(1, mx + 1):
for y in range(x, mx + 1):
for z in range(y, mx + 1):
v = x*x + y*y + z*z + x*y + y*z + z*x
if v == n:
if x == y == z:
pairs += 1
elif x == y or y == z:
pairs += 3
else:
pairs += 6
return pairs
n = int(input())
for i in range(n):
ps = fun(i + 1)
print(ps)
|
s820555569
|
Accepted
| 1,126
| 106,588
| 622
|
from numba import njit
@njit('i8(i8)', cache=True)
def fun(n):
#mx = n#int((n**.5) // 3) + 1
mx = int(n**.5 + 0.5)
pairs = 0
for x in range(1, mx + 1):
for y in range(x, mx + 1):
for z in range(y, mx + 1):
v = x*x + y*y + z*z + x*y + y*z + z*x
if v == n:
if x == y == z:
pairs += 1
elif x == y or y == z:
pairs += 3
else:
pairs += 6
return pairs
n = int(input())
for i in range(n):
ps = fun(i + 1)
print(ps)
|
s962592133
|
p02601
|
u409974118
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,164
| 179
|
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())
k = int(input())
an = 0
while a >= b:
b *= 2
an +=1
while b>=c:
c *=2
an +=1
if an > k:
print("NO")
else :
print("YES")
|
s136577532
|
Accepted
| 29
| 9,160
| 179
|
a,b,c = map(int,input().split())
k = int(input())
an = 0
while a >= b:
b *= 2
an +=1
while b>=c:
c *=2
an +=1
if an > k:
print("No")
else :
print("Yes")
|
s343812192
|
p02612
|
u234007117
| 2,000
| 1,048,576
|
Wrong Answer
| 32
| 9,080
| 33
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
print(n % 1000)
|
s598639376
|
Accepted
| 29
| 9,088
| 78
|
n = int(input())
if n % 1000 == 0:
print(0)
else:
print(1000 - (n % 1000))
|
s693440130
|
p00008
|
u661290476
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,604
| 322
|
Write a program which reads an integer n and identifies the number of combinations of a, b, c and d (0 ≤ a, b, c, d ≤ 9) which meet the following equality: a + b + c + d = n For example, for n = 35, we have 4 different combinations of (a, b, c, d): (8, 9, 9, 9), (9, 8, 9, 9), (9, 9, 8, 9), and (9, 9, 9, 8).
|
INF=-1
memo=[[INF]*51 for i in range(5)]
def rec(i,n):
if memo[i][n]!=INF:
return memo[i][n]
if i==4:
return 1 if n==0 else 0
for m in range(10):
memo[i][n]+=rec(i+1,n-m)
return memo[i][n]
while True:
try:
n=int(input())
print(rec(0,n))
except:
break
|
s492362451
|
Accepted
| 200
| 7,496
| 280
|
while True:
try:
n=int(input())
except:
break
cnt=0
for i in range(10):
for j in range(10):
for k in range(10):
for l in range(10):
if i+k+j+l==n:
cnt+=1
print(cnt)
|
s576928587
|
p03729
|
u127856129
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 98
|
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
a,b,c=input().split()
s=[a,b,c]
if a[-0]==b[0] and b[-0]==c[0]:
print("YES")
else:
print("NO")
|
s552266774
|
Accepted
| 17
| 2,940
| 99
|
a,b,c=input().split()
s=[a,b,c]
if a[-1]==b[0] and b[-1]==c[0]:
print("YES")
else:
print("NO")
|
s304333631
|
p03386
|
u580404776
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 177
|
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())
ans=[]
for i in range(K):
if A+i<=B:
ans.append(A+i)
if B-i>=A:
ans.append(B-i)
ans=list(set(ans))
print(*ans, sep='\n')
|
s322677312
|
Accepted
| 17
| 3,060
| 186
|
A,B,K=map(int,input().split())
ans=[]
for i in range(K):
if A+i<=B:
ans.append(A+i)
if B-i>=A:
ans.append(B-i)
ans=sorted(list(set(ans)))
print(*ans, sep='\n')
|
s742056105
|
p03565
|
u927534107
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 294
|
E869120 found a chest which is likely to contain treasure. However, the chest is locked. In order to open it, he needs to enter a string S consisting of lowercase English letters. He also found a string S', which turns out to be the string S with some of its letters (possibly all or none) replaced with `?`. One more thing he found is a sheet of paper with the following facts written on it: * Condition 1: The string S contains a string T as a contiguous substring. * Condition 2: S is the lexicographically smallest string among the ones that satisfy Condition 1. Print the string S. If such a string does not exist, print `UNRESTORABLE`.
|
s=str(input())
t=str(input())
for i in range(len(s)-len(t)+1):
l=list(s[len(s)-len(t)-i:len(s)-i])
if all([l[i]==t[i] or l[i]=="?" for i in range(len(t))]):
s_new=s[:len(s)-len(t)-i]+t+s[len(s)-i:]
break
s=s.replace("?","a")
if t in s:print(s)
else:print("UNRESTORABLE")
|
s688727034
|
Accepted
| 17
| 3,064
| 284
|
s=str(input())
t=str(input())
for i in range(len(s)-len(t)+1):
l=s[len(s)-len(t)-i:len(s)-i]
if all([l[i]==t[i] or l[i]=="?" for i in range(len(t))]):
s=s[:len(s)-len(t)-i]+t+s[len(s)-i:]
break
s=s.replace("?","a")
if t in s:print(s)
else:print("UNRESTORABLE")
|
s157684572
|
p03478
|
u694244301
| 2,000
| 262,144
|
Wrong Answer
| 549
| 9,156
| 201
|
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 i in range(1, n + 1):
t = 0
k = i
while(True):
t += k % 10
k = k / 10
if k == 0:
break
if a <= t <= b:
ans += 1
print(ans)
|
s074725989
|
Accepted
| 39
| 9,128
| 207
|
n, a, b = map(int, input().split())
ans = 0
for i in range(1, n + 1):
t = 0
k = i
while(True):
t += k % 10
k = int(k / 10)
if k == 0:
break
if a <= t <= b:
ans += i
print(ans)
|
s650460064
|
p00461
|
u546285759
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,688
| 147
|
整数 _n_ (1 ≤ _n_ ) に対し, _n_ \+ 1 個の I と _n_ 個の O を I から始めて交互に並べてできる文字列を _P n_ とする.ここで I と O はそれぞれ英大文字のアイとオーである. _P_ 1| IOI ---|--- _P_ 2| IOIOI _P_ 3| IOIOIOI | .| | .| | .| _P_ _n_| IOIOIO ... OI (O が _n_ 個) 図 1-1 本問で考える文字列 _P n_ 整数 _n_ と, I と O のみからなる文字列 _s_ が与えられた時, _s_ の中に _P n_ が何ヶ所含まれているかを出力するプログラムを作成せよ.
|
n, m, s= int(input()), int(input()), input()
ioi= "IOI" if n==3 else "IOI"+"OI"*(n-1)
n= 2*n+1
print(sum(1 for i in range(m-n+1) if s[i:i+n]==ioi))
|
s916258132
|
Accepted
| 540
| 9,452
| 198
|
while True:
n= int(input())
if n== 0: break
m, s= int(input()), input()
ioi= "IOI" if n==3 else "IOI"+"OI"*(n-1)
n= 2*n+1
print(sum(1 for i in range(m-n+1) if s[i:i+n]==ioi))
|
s709261450
|
p03455
|
u321121629
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 96
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b=[int(i) for i in input().split()]
c = a*b
if c % 2 ==0:
print('even')
else:
print('odd')
|
s363801108
|
Accepted
| 17
| 2,940
| 89
|
a,b = map(int, input().split())
if (a * b) % 2 == 0:
print('Even')
else:
print('Odd')
|
s979751236
|
p00003
|
u710016128
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,600
| 255
|
Write a program which judges wheather given length of three side form a right triangle. Print "YES" if the given sides (integers) form a right triangle, "NO" if not so.
|
n = int(input())
ans = list()
for i in range(n):
line = input().split(" ")
if (line[0] == line[1]) or (line[1] == line[2]) or (line[0] == line[2]):
ans.append("YES")
else:
ans.append("NO")
for j in range(n):
print(ans[j])
|
s288722789
|
Accepted
| 30
| 7,676
| 369
|
n = int(input())
ans = list()
for i in range(n):
line = input().split(" ")
a = float(line[0])
b = float(line[1])
c = float(line[2])
if ((a**2 + b**2) == c**2) or ((b**2 + c**2) == a**2) or ((c**2 + a**2) == b**2):
ans.append("YES")
else:
ans.append("NO")
for j in range(n):
print(ans[j])
|
s634881736
|
p03228
|
u612721349
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 210
|
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, c = map(int, input().split())
for i in range(c):
if c % 2 == 0:
if a % 2 == 1:
a -= 1
b += a // 2
a //= 2
else:
if b % 2 == 1:
b-= 1
a += b // 2
b //= 2
print(a, b)
|
s084335043
|
Accepted
| 17
| 2,940
| 211
|
a, b, c = map(int, input().split())
for i in range(c):
if i % 2 == 0:
if a % 2 == 1:
a -= 1
b += a // 2
a //= 2
else:
if b % 2 == 1:
b-= 1
a += b // 2
b //= 2
print(a, b)
|
s238811424
|
p03680
|
u565476466
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 7,084
| 164
|
Takahashi wants to gain muscle, and decides to work out at AtCoder Gym. The exercise machine at the gym has N buttons, and exactly one of the buttons is lighten up. These buttons are numbered 1 through N. When Button i is lighten up and you press it, the light is turned off, and then Button a_i will be lighten up. It is possible that i=a_i. When Button i is not lighten up, nothing will happen by pressing it. Initially, Button 1 is lighten up. Takahashi wants to quit pressing buttons when Button 2 is lighten up. Determine whether this is possible. If the answer is positive, find the minimum number of times he needs to press buttons.
|
N = int(input())
a = [int(input()) for _ in range(N)]
ans, nex = 1, a[0]
while ans < N or nex == 2:
nex = a[nex - 1]
ans += 1
print(ans if ans != N else -1)
|
s602392974
|
Accepted
| 201
| 7,084
| 165
|
N = int(input())
a = [int(input()) for _ in range(N)]
ans, nex = 1, a[0]
while ans < N and nex != 2:
nex = a[nex - 1]
ans += 1
print(ans if ans != N else -1)
|
s783671863
|
p02865
|
u148551245
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 87
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
n = int(input())
if n % 2 == 1:
print((n - 1) / 2)
else:
print((n - 1) / 2 - 1)
|
s406527640
|
Accepted
| 17
| 2,940
| 91
|
n = int(input())
if n % 2 == 1:
print(int((n - 1) / 2))
else:
print(int(n / 2 - 1))
|
s388625909
|
p00009
|
u197615397
| 1,000
| 131,072
|
Time Limit Exceeded
| 5,940
| 10,892
| 347
|
Write a program which reads an integer n and prints the number of prime numbers which are less than or equal to n. A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7.
|
from collections import deque
import bisect
from math import sqrt
import sys
primes = deque([2, 3, 5, 7, 11, 13, 17, 19])
append = primes.append
for i in range(21, 1000000, 2):
for j in range(3, int(sqrt(i))+1, 2):
if i%j == 0:
break
else:
append(i)
for l in sys.stdin:
print(bisect.bisect(primes, int(l)))
|
s455040375
|
Accepted
| 140
| 40,592
| 573
|
def get_prime_set(ub):
from itertools import chain
from math import sqrt
if ub < 4:
return ({}, {}, {2}, {2, 3})[ub]
ub, ub_sqrt = ub+1, int(sqrt(ub))+1
primes = {2, 3} | set(chain(range(5, ub, 6), range(7, ub, 6)))
du = primes.difference_update
for n in chain(range(5, ub_sqrt, 6), range(7, ub_sqrt, 6)):
if n in primes:
du(range(n*3, ub, n*2))
return primes
import sys
from bisect import bisect_right
primes = list(get_prime_set(999999))
print(*(bisect_right(primes, int(n)) for n in sys.stdin), sep="\n")
|
s213011546
|
p02614
|
u903005414
| 1,000
| 1,048,576
|
Wrong Answer
| 67
| 9,116
| 471
|
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.
|
H, W, K = map(int, input().split())
C = []
for _ in range(H):
C.append(list(input()))
# print(f'{C=}')
ans = 0
for H_bit in range(1 << H):
for W_bit in range(1 << W):
cnt = 0
for i in range(H):
for j in range(W):
if ((H_bit >> i) & 1) == 0 and ((W_bit >> i)
& 1) == 0 and C[i][j] == '#':
cnt += 1
if cnt == K:
ans += 1
print(ans)
|
s422972599
|
Accepted
| 63
| 9,216
| 764
|
H, W, K = map(int, input().split())
C = []
for _ in range(H):
C.append(list(input()))
# print(f'{C=}')
ans = 0
for H_bit in range(1 << H):
H_idx = []
for i in range(H):
if H_bit & (1 << i):
H_idx.append(i)
# print(f'{H_bit=}, {H_idx=}')
for W_bit in range(1 << W):
W_idx = []
for i in range(W):
if W_bit & (1 << i):
W_idx.append(i)
# print(f'{H_idx=}, {W_idx=}')
cnt = 0
for i in range(H):
for j in range(W):
if i not in H_idx and j not in W_idx and C[i][j] == '#':
cnt += 1
if cnt == K:
ans += 1
print(ans)
|
s749592444
|
p02613
|
u321415743
| 2,000
| 1,048,576
|
Wrong Answer
| 189
| 9,180
| 233
|
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.
|
num = int(input())
test = {'AC': 0, 'WA': 0, 'TLE': 0, 'RE': 0}
for i in range(num):
result = input()
for j in test.keys():
if result == j:
test[result] += 1
for i in test:
print(i, '×', str(test[i]))
|
s930198010
|
Accepted
| 185
| 9,184
| 232
|
num = int(input())
test = {'AC': 0, 'WA': 0, 'TLE': 0, 'RE': 0}
for i in range(num):
result = input()
for j in test.keys():
if result == j:
test[result] += 1
for i in test:
print(i, 'x', str(test[i]))
|
s037318805
|
p03635
|
u031146664
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 66
|
In _K-city_ , there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city?
|
s = input()
lis = [s[:1],str(len(s)-2),s[-1]]
print("".join(lis))
|
s452231018
|
Accepted
| 17
| 2,940
| 51
|
a, b = map(int, input().split())
print((a-1)*(b-1))
|
s151414922
|
p03478
|
u548545174
| 2,000
| 262,144
|
Wrong Answer
| 35
| 3,060
| 179
|
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 i in range(1, N + 1):
s = 0
i = str(i)
for n in i:
s += int(n)
if A <= s <= B:
ans += s
print(ans)
|
s635623539
|
Accepted
| 34
| 2,940
| 161
|
N, A, B = map(int, input().split())
ans = 0
for i in range(1, N+1):
i = str(i)
if A <= sum([int(j) for j in i]) <= B:
ans += int(i)
print(ans)
|
s172869221
|
p02853
|
u785578220
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 152
|
We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned.
|
l = [300000,200000,100000]
a,b = map(int,input().split())
res = 0
if a in range(1,4):
res +=l[a-1]
if b in range(1,4):
res +=l[b-1]
print(res)
|
s724329644
|
Accepted
| 17
| 3,060
| 187
|
l = [300000,200000,100000]
a,b = map(int,input().split())
res = 0
if a in range(1,4):
res +=l[a-1]
if b in range(1,4):
res +=l[b-1]
if a == b and a==1:
res+=400000
print(res)
|
s023404749
|
p03387
|
u166306121
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 439
|
You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.
|
A,B,C = map(int,input().split())
oddcnt = 0
sub = [A,B,C]
sub2 = [i - min(sub) for i in sub]
sub2 = sorted(sub2)[1:3]
print(sub2)
for i in sub2:
if i%2 != 0:
oddcnt += 1
if oddcnt == 0:
print((sub2[1]+(sub2[1]-sub2[0]))//2)
elif oddcnt == 1:
if sub2[0]%2 ==0:
print((sub2[1]+(sub2[1]-sub2[0]))//2)
else:
print((sub2[1]+(sub2[1]-sub2[0]+1))//2+1)
else:
print((sub2[1]+(sub2[1]-sub2[0]+1))//2+1)
|
s090276564
|
Accepted
| 18
| 3,064
| 441
|
A,B,C = map(int,input().split())
oddcnt = 0
sub = [A,B,C]
sub2 = [i - min(sub) for i in sub]
sub2 = sorted(sub2)[1:3]
# print(sub2)
for i in sub2:
if i%2 != 0:
oddcnt += 1
if oddcnt == 0:
print((sub2[1]+(sub2[1]-sub2[0]))//2)
elif oddcnt == 1:
if sub2[0]%2 ==0:
print((sub2[1]+(sub2[1]-sub2[0]))//2)
else:
print((sub2[1]+(sub2[1]-sub2[0]+1))//2+1)
else:
print((sub2[1]+(sub2[1]-sub2[0]+1))//2+1)
|
s460606660
|
p03777
|
u019584841
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 79
|
Two deer, AtCoDeer and TopCoDeer, are playing a game called _Honest or Dishonest_. In this game, an honest player always tells the truth, and an dishonest player always tell lies. You are given two characters a and b as the input. Each of them is either `H` or `D`, and carries the following information: If a=`H`, AtCoDeer is honest; if a=`D`, AtCoDeer is dishonest. If b=`H`, AtCoDeer is saying that TopCoDeer is honest; if b=`D`, AtCoDeer is saying that TopCoDeer is dishonest. Given this information, determine whether TopCoDeer is honest.
|
a,b=input().split()
if a==b=="H" and a==b=="D":
print("H")
else:
print("D")
|
s104666289
|
Accepted
| 17
| 2,940
| 79
|
a,b=input().split()
if a==b=="H" or a==b=="D":
print("H")
else:
print("D")
|
s757403352
|
p03644
|
u957084285
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 2,940
| 193
|
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
N = int(input())
ans = 1
max_count = 0
n = 2
while (n <= N):
k = n
count = 0
while k%2 == 0:
k /= 2
count += 1
if count > max_count:
ans = n
max_count = count
print(ans)
|
s568377447
|
Accepted
| 17
| 2,940
| 203
|
N = int(input())
ans = 1
max_count = 0
n = 2
while (n <= N):
k = n
count = 0
while k%2 == 0:
k /= 2
count += 1
if count > max_count:
ans = n
max_count = count
n += 1
print(ans)
|
s924980346
|
p03302
|
u945181840
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 101
|
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:
print('*')
else:
print('x')
|
s772876510
|
Accepted
| 17
| 2,940
| 119
|
a, b = map(int, input().split())
if a + b == 15:
print('+')
elif a * b == 15:
print('*')
else:
print('x')
|
s605197286
|
p03795
|
u026788530
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 38
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n=int(input())
print(n*800-200*n//15)
|
s782180311
|
Accepted
| 17
| 2,940
| 39
|
n=int(input())
print(n*800-200*(n//15))
|
s956772922
|
p02972
|
u620868411
| 2,000
| 1,048,576
|
Wrong Answer
| 955
| 10,708
| 382
|
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.
|
# -*- coding: utf-8 -*-
n = int(input())
al = list(map(int, input().split()))
bl = [0]*n
for i in range(n,0,-1):
if i*2>n:
if al[i-1]>0:
bl[i-1] = 1
continue
c = 2
s = 0
while i*c<=n:
s += bl[i*c-1]
c += 1
if al[i-1]==1:
bl[i-1] = 1 if s%2==0 else 0
else:
bl[i-1] = 0 if s%2==0 else 1
print(*bl)
|
s512879416
|
Accepted
| 1,016
| 14,196
| 491
|
# -*- coding: utf-8 -*-
n = int(input())
al = list(map(int, input().split()))
bl = [0]*n
for i in range(n,0,-1):
if i*2>n:
if al[i-1]>0:
bl[i-1] = 1
continue
c = 2
s = 0
while i*c<=n:
s += bl[i*c-1]
c += 1
if al[i-1]==1:
bl[i-1] = 1 if s%2==0 else 0
else:
bl[i-1] = 0 if s%2==0 else 1
res = []
for i in range(1,n+1):
if bl[i-1]>0:
res.append(i)
print(len(res))
if len(res)>0:
print(*res)
|
s714911264
|
p03437
|
u808003008
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 182
|
You are given positive integers X and Y. If there exists a positive integer not greater than 10^{18} that is a multiple of X but not a multiple of Y, choose one such integer and print it. If it does not exist, print -1.
|
x, y = list(map(int, input().split()))
if x == y:
print(-1)
else:
for i in range(1, 100):
if x * i > y and x * i % y != 0:
print(x * i)
break
|
s455831455
|
Accepted
| 18
| 2,940
| 212
|
x, y = list(map(int, input().split()))
if x == y:
print(-1)
else:
for i in range(2, 11):
if x * i % y != 0:
print(x * i)
break
elif i == 10:
print(-1)
|
s798861780
|
p03436
|
u619144316
| 2,000
| 262,144
|
Wrong Answer
| 2,114
| 185,828
| 662
|
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 = [int(i) for i in input().split(' ')]
MAP = []
SUM = 0
for i in range(H):
tmp = list(input())
SUM += tmp.count('#')
MAP.append(tmp)
print(MAP,SUM)
S = [[0,0]]
while True:
tmp = []
for s in S:
if not s[1] == W -1:
if MAP[s[0]][s[1]+1] == '.':
a = [s[0], s[1]+1]
tmp.append(a)
if not s[0] == H -1:
if MAP[s[0]+1][s[1]] == '.':
a = [s[0]+1, s[1]]
tmp.append(a)
S = tmp
if len(s) == 0:
print(-1)
exit()
if [H-1,W-1] in S:
move = H + W - 1
print(H*W - move - SUM)
exit()
|
s922615558
|
Accepted
| 40
| 9,404
| 894
|
from collections import deque
def main():
R,C= map(int,input().split())
S = [0,0]
mv = [[1,0],[-1,0],[0,-1],[0,1]]
MAP = []
flg = 0
block = 0
for _ in range(R):
t = list(input())
block += t.count('#')
MAP.append(t)
stack = deque([S])
MAP_c = [[None]*C for _ in range(R)]
MAP_c[0][0] = 1
while stack:
v = stack.popleft()
for m in mv:
u = [v[0]+m[0],v[1]+m[1]]
if u[0] >=0 and u[0] < R and u[1] >= 0 and u[1] < C and MAP[u[0]][u[1]] == '.':
MAP[u[0]][u[1]] = '#'
MAP_c[u[0]][u[1]] = MAP_c[v[0]][v[1]] + 1
stack.append(u)
if u == [R-1,C-1]:
flg = 1
if flg == 1:
break
if flg == 1:
a = MAP_c[R-1][C-1]
print(R*C - a - block)
else:
print(-1)
main()
|
s127224732
|
p03712
|
u037430802
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 201
|
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
h,w = map(int,input().split())
ans = []
ans.append("*"*(w+2))
for i in range(h):
s = input()
tmp = "*"
tmp += s
tmp += "*"
ans.append(tmp)
ans.append("*"*(w+2))
for i in ans:
print(i)
|
s712061298
|
Accepted
| 18
| 3,060
| 201
|
h,w = map(int,input().split())
ans = []
ans.append("#"*(w+2))
for i in range(h):
s = input()
tmp = "#"
tmp += s
tmp += "#"
ans.append(tmp)
ans.append("#"*(w+2))
for i in ans:
print(i)
|
s459895776
|
p03612
|
u405256066
| 2,000
| 262,144
|
Wrong Answer
| 65
| 13,880
| 193
|
You are given a permutation p_1,p_2,...,p_N consisting of 1,2,..,N. You can perform the following operation any number of times (possibly zero): Operation: Swap two **adjacent** elements in the permutation. You want to have p_i ≠ i for all 1≤i≤N. Find the minimum required number of operations to achieve this.
|
from sys import stdin
N = int(stdin.readline().rstrip())
P = [int(x) for x in stdin.readline().rstrip().split()]
cnt = 0
for i,j in enumerate(P):
if i+1 != j:
cnt += 1
print(cnt//2)
|
s747103548
|
Accepted
| 72
| 13,880
| 272
|
from sys import stdin
N = int(stdin.readline().rstrip())
P = [int(x) for x in stdin.readline().rstrip().split()]
cnt = 0
ans = 0
for i,j in enumerate(P):
if i+1 == j:
cnt += 1
else:
ans += ((cnt+1)//2)
cnt = 0
ans += ((cnt+1)//2)
print(ans)
|
s063486413
|
p00005
|
u896025703
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,716
| 215
|
Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given a and b.
|
def gcd(a, b):
if b == 0: return a
else: return gcd(b, a % b)
def lcm(a, b):
return a * b / gcd(a, b)
while True:
try:
a, b = map(int, input().split())
print(gcd(a, b), lcm(a, b))
except EOFError:
break
|
s277840971
|
Accepted
| 30
| 7,612
| 225
|
def gcd(a, b):
if b == 0: return a
else: return gcd(b, a % b)
def lcm(a, b):
return a * b / gcd(a, b)
while True:
try:
a, b = map(int, input().split())
print(int(gcd(a, b)), int(lcm(a, b)))
except EOFError:
break
|
s777900045
|
p03730
|
u509214520
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,156
| 136
|
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
A, B, C = map(int, input().split())
amari = []
for i in range(1, B+100):
amari.append(A*i%B)
print('Yes' if C in amari else 'No')
|
s681649619
|
Accepted
| 26
| 9,008
| 134
|
A, B, C = map(int, input().split())
amari = []
for i in range(1, B+1):
amari.append(A*i%B)
print('YES' if C in amari else 'NO')
|
s418579128
|
p03644
|
u208120643
| 2,000
| 262,144
|
Wrong Answer
| 117
| 27,068
| 409
|
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.
|
import numpy as np
N = 100
# N=int(input())
Division_2 = []
number=[]
for i in range(1, N+1):
j = i
count = 0
while True:
count += 1
j = j/2
if j/2 != int(j/2):
break
Division_2.append(count)
number.append(i)
Division_2=np.array(Division_2)
print(number[np.argmax(Division_2)])
|
s070392965
|
Accepted
| 116
| 27,228
| 427
|
import numpy as np
#N = 100
N=int(input())
Division_2 = []
number=[]
for i in range(1, N+1):
j = i
count = 0
while True:
if j/2 != int(j/2):
break
j = j/2
count += 1
Division_2.append(count)
number.append(i)
Division_2=np.array(Division_2)
print(number[np.argmax(Division_2)])
#print(Division_2)
|
s855010924
|
p03456
|
u955547613
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 160
|
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.
|
# -*- coding: utf-8 -*-
import math
a, b = map(str, input().split())
c = int(a + b)
root = math.sqrt(c)
if root**2 == c:
print("YES")
else:
print("NO")
|
s727624261
|
Accepted
| 17
| 3,060
| 265
|
# -*- coding: utf-8 -*-
import math
a, b = map(str, input().split())
c = int(a + b)
flag = False
for i in range(1, 1000):
if (i**2 == c):
flag = True
print("Yes")
break
elif (i**2 > c):
break
if not flag:
print("No")
|
s316162565
|
p03555
|
u863044225
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 49
|
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.
|
s=input()
t=input()
print('YNeos'[s!=t[::-1]::2])
|
s948893185
|
Accepted
| 17
| 2,940
| 50
|
s=input()
t=input()
print('YNEOS'[s!=t[::-1]::2])
|
s152868514
|
p03455
|
u002459665
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 94
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a, b = map(int, input().split())
if a * b % 2 == 0:
print('Odd')
else:
print('Even')
|
s856969688
|
Accepted
| 17
| 2,940
| 94
|
a, b = map(int, input().split())
if a * b % 2 == 0:
print('Even')
else:
print('Odd')
|
s531059617
|
p02613
|
u753971348
| 2,000
| 1,048,576
|
Wrong Answer
| 151
| 9,164
| 316
|
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, re, wa, tle = 0,0,0,0
for i in range(N):
line = input()
if line == "AC":
ac += 1
elif line == "TLE":
tle += 1
elif line == "WA":
wa += 1
elif line == "RE":
re += 1
print("AC × " + str(ac))
print("WA × " + str(wa))
print("TLE × " + str(tle))
print("RE × " + str(re))
|
s633855983
|
Accepted
| 149
| 9,140
| 312
|
N = int(input())
ac, re, wa, tle = 0,0,0,0
for i in range(N):
line = input()
if line == "AC":
ac += 1
elif line == "TLE":
tle += 1
elif line == "WA":
wa += 1
elif line == "RE":
re += 1
print("AC x " + str(ac))
print("WA x " + str(wa))
print("TLE x " + str(tle))
print("RE x " + str(re))
|
s860408228
|
p03556
|
u558836062
| 2,000
| 262,144
|
Wrong Answer
| 25
| 3,060
| 132
|
Find the largest square number not exceeding N. Here, a _square number_ is an integer that can be represented as the square of an integer.
|
A,B,N = 0,1,int(input())
while (True):
A += 1
if A*A>N:
break
b = A*A
print('{0} = {1}**2' .format(b,int(b**(1/2))))
|
s004562579
|
Accepted
| 24
| 2,940
| 94
|
A,B,N = 0,1,int(input())
while (True):
A += 1
if A*A>N:
break
b = A*A
print(b)
|
s204758160
|
p02613
|
u544865362
| 2,000
| 1,048,576
|
Wrong Answer
| 158
| 9,180
| 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.
|
n = int(input())
hashmap = {
'AC' : 0,
'WA' : 0,
'TLE': 0,
'RE' : 0
}
while n > 0:
t = input()
hashmap[t] = hashmap.get(t, 0) + 1
n -= 1
for key, val in hashmap.items():
print(key + "x" + str(val))
|
s421992840
|
Accepted
| 163
| 9,476
| 292
|
n = int(input())
import collections
# hashmap = collections.OrderedDict()
hashmap = {
'AC' : 0,
'WA' : 0,
'TLE': 0,
'RE' : 0
}
while n > 0:
t = input()
hashmap[t] = hashmap.get(t, 0) + 1
n -= 1
for key, val in hashmap.items():
print(key + " x " + str(val))
|
s174597591
|
p02612
|
u531219227
| 2,000
| 1,048,576
|
Wrong Answer
| 34
| 9,080
| 31
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print(N%1000)
|
s712961196
|
Accepted
| 28
| 9,156
| 70
|
N = int(input())
if N%1000 != 0:
print(1000-N%1000)
else:
print(0)
|
s841923495
|
p03351
|
u740284863
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 138
|
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())
if abs(a-c) <= d:
print("Yes")
elif abs(a-b)+abs(b-c) <= d:
print("Yes")
else:
print("No")
|
s680215695
|
Accepted
| 17
| 2,940
| 147
|
a,b,c,d = map(int,input().split())
if abs(a-c) <= d:
print("Yes")
elif abs(a-b) <= d and abs(b-c) <= d:
print("Yes")
else:
print("No")
|
s845541187
|
p02663
|
u686230543
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 9,104
| 100
|
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?
|
h1, m1, h2, m2, k = map(int, input().split())
t1 = 60 * h1 + m1
t2 = 60 * h2 * m2
print(t2 - t1 - k)
|
s885391321
|
Accepted
| 20
| 9,164
| 100
|
h1, m1, h2, m2, k = map(int, input().split())
t1 = 60 * h1 + m1
t2 = 60 * h2 + m2
print(t2 - t1 - k)
|
s617248135
|
p03695
|
u210827208
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,188
| 556
|
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
n=int(input())
A=list(map(int,input().split()))
ov=0
res=0
C=[0]*8
for i in A:
if 1<=i and i<400:
C[0]+=1
elif 400<=i and i<800:
C[1]+=1
elif 800<=i and i<1200:
C[2]+=1
elif 1200<=i and i<1600:
C[3]+=1
elif 1600<=i and i<2000:
C[4]+=1
elif 2000<=i and i<2400:
C[5]+=1
elif 2400<=i and i<2800:
C[6]+=1
elif 2800<=i and i<3200:
C[7]+=1
else:
ov+=1
for j in C:
if j!=0:
res+=1
if res+ov>8:
print(8)
else:
print(res+ov)
|
s948783072
|
Accepted
| 17
| 3,060
| 186
|
n=int(input())
A=list(map(int,input().split()))
C=[0]*9
for i in A:
x=i//400
if x<8:
C[x]=1
else:
C[8]+=1
b=sum(C[:8])
print(str(max(b,1))+' '+str(b+C[8]))
|
s793130491
|
p02821
|
u295294832
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 14,140
| 1,244
|
Takahashi has come to a party as a special guest. There are N ordinary guests at the party. The i-th ordinary guest has a _power_ of A_i. Takahashi has decided to perform M _handshakes_ to increase the _happiness_ of the party (let the current happiness be 0). A handshake will be performed as follows: * Takahashi chooses one (ordinary) guest x for his left hand and another guest y for his right hand (x and y can be the same). * Then, he shakes the left hand of Guest x and the right hand of Guest y simultaneously to increase the happiness by A_x+A_y. However, Takahashi should not perform the same handshake more than once. Formally, the following condition must hold: * Assume that, in the k-th handshake, Takahashi shakes the left hand of Guest x_k and the right hand of Guest y_k. Then, there is no pair p, q (1 \leq p < q \leq M) such that (x_p,y_p)=(x_q,y_q). What is the maximum possible happiness after M handshakes?
|
N,M= [int(i) for i in input().split()]
n = sorted([int(x) for x in input().split()],reverse = True)
print(n)
MM=n[0]*2
mm=n[-1]*2
l=[0]*2
r=MM//2
while True:
c=0
for i in range(N):
for j in range(i,N):
if n[i]+n[j] > r:
if i==j:
c+=1
else:
c+=2
l[0] = i
l[1] = j
print(c,r)
if c==M: break
if abs(MM-mm) < 2:
r=mm
break
elif c>M:
mm=r
r=(MM+mm)//2
elif c<M:
MM=r
r=(mm+MM)//2
#print("fin",c,r,l[0],l[1])
v=0
q=0
for i in range(l[0]+1):
for j in range(i,N):
t=n[i]+n[j]
if t > r:
if i==j:
v+=t
q+=1
else:
v+=t*2
q+=2
if q>M:
v-=t
break
if q==M: break
print(v)
|
s742946905
|
Accepted
| 1,435
| 14,388
| 596
|
from bisect import bisect_left
N,M= [int(i) for i in input().split()]
n = sorted([int(x) for x in input().split()])
s =[0]*(N+1)
s[0]=n[0]
for i in range(1,N+1):
s[i] = s[i-1]+n[i-1]
MM=n[-1]*2
mm=n[0]*2
while True:
c=0
r=(mm+MM)//2
for i in range(N):
c += N-bisect_left(n,r-n[i])
if (MM-mm) < 2:
break
elif c>=M:
mm=r
elif c<M:
MM=r
v=0
for i in range(N):
j=bisect_left(n,r-n[i])
v+= (s[N] - s[j]) + (N-j)*n[i]
v = v - (c-M)*(r)
print(v)
|
s740754503
|
p03455
|
u709806735
| 2,000
| 262,144
|
Wrong Answer
| 17
| 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 ==0:
print("even")
else:
print("odd")
|
s628669236
|
Accepted
| 17
| 2,940
| 89
|
a ,b =map(int,input().split())
if (a*b)%2 ==0:
print("Even")
else:
print("Odd")
|
s601537488
|
p02412
|
u352203480
| 1,000
| 131,072
|
Wrong Answer
| 30
| 5,596
| 408
|
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
while True:
n, x = map(int, input().split())
if n == 0 and x == 0:
break
else:
sums = 0
count = 0
for i in range(1, n-1):
for j in range(i+1, n):
for k in range(j+1, n+1):
sums = i + j + k
print(sums)
if sums == x:
count += 1
|
s269816139
|
Accepted
| 610
| 5,596
| 396
|
while True:
n, x = map(int, input().split())
if n == 0 and x == 0:
break
else:
sums = 0
count = 0
for i in range(1, n-1):
for j in range(i+1, n):
for k in range(j+1, n+1):
sums = i + j + k
if sums == x:
count += 1
print(count)
|
s672354329
|
p04035
|
u414809621
| 2,000
| 262,144
|
Wrong Answer
| 307
| 14,724
| 685
|
We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i. At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly: * Choose a (connected) rope with a total length of at least L, then untie one of its knots. Is it possible to untie all of the N-1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots.
|
# -*- coding: utf-8 -*-
"""
Created on Fri Aug 5 22:01:07 2016
@author: dotha
"""
def say(l,r):
for i in range(l,r):
print(i+1)
if l>r:
for i in range(l,r,-1):
print(i-1)
def solver(N,L,a):
for i in range(N-2):
if a[i+1]+a[i] >= L:
break
else:
print('Impossible')
return
print('Possible')
left = 0
for i in range(N-1):
if a[i+1]+a[i] > L:
say(left,i)
left = i
if left == N-2:
print(left+1)
break
else:
say(N-1,left)
N,L=map(int,input().split(' '))
a=list(map(int,input().split(' ')))
solver(N,L,a)
|
s448829027
|
Accepted
| 221
| 14,148
| 430
|
def main():
n, l = (int(s) for s in input().strip().split(' '))
an = [int (s) for s in input().strip().split(' ')]
for i in range(1, n):
if an[i-1] + an[i] >= l:
print('Possible')
index = i
break
else:
print('Impossible')
return
for i in range(1, index):
print(i)
for i in range(n-1, index, -1):
print(i)
print(index)
main()
|
s543349955
|
p03738
|
u080364835
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 107
|
You are given two positive integers A and B. Compare the magnitudes of these numbers.
|
a = int(input())
b = int(input())
if a > b: print('GRATER')
elif a < b: print('LESS')
else: print('EQUAL')
|
s343194715
|
Accepted
| 17
| 3,060
| 108
|
a = int(input())
b = int(input())
if a > b: print('GREATER')
elif a < b: print('LESS')
else: print('EQUAL')
|
s180430427
|
p02844
|
u951401193
| 2,000
| 1,048,576
|
Wrong Answer
| 33
| 3,060
| 112
|
AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code. The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code. How many different PIN codes can he set this way? Both the lucky number and the PIN code may begin with a 0.
|
n = int(input())
s = input()
h = [0]*1000
for i in range(n-2):
tmp = int(s[i:i+3])
h[tmp]=1
print(sum(h))
|
s973263624
|
Accepted
| 584
| 3,188
| 139
|
input()
a, b, c = set(), set(), set()
for x in input():
c.update(y + x for y in b)
b.update(y + x for y in a)
a.add(x)
print(len(c))
|
s148921126
|
p03369
|
u538956308
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 53
|
In "Takahashi-ya", a ramen restaurant, a bowl of ramen costs 700 yen (the currency of Japan), plus 100 yen for each kind of topping (boiled egg, sliced pork, green onions). A customer ordered a bowl of ramen and told which toppings to put on his ramen to a clerk. The clerk took a memo of the order as a string S. S is three characters long, and if the first character in S is `o`, it means the ramen should be topped with boiled egg; if that character is `x`, it means the ramen should not be topped with boiled egg. Similarly, the second and third characters in S mean the presence or absence of sliced pork and green onions on top of the ramen. Write a program that, when S is given, prints the price of the corresponding bowl of ramen.
|
S = input()
topping = S.count("o")
print(700+topping)
|
s378227888
|
Accepted
| 17
| 2,940
| 69
|
S = input()
topping = S.count("o")
ans = 700 + topping*100
print(ans)
|
s774197795
|
p03457
|
u626331732
| 2,000
| 262,144
|
Wrong Answer
| 432
| 21,108
| 382
|
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())
txy = [[0,0,0]]
for i in range(N):
txy.append([int(x) for x in input().split()])
flg = 0
for j in range(1, N + 1):
dt = txy[j][0] - txy[j - 1][0]
dx = txy[j][1] - txy[j - 1][1]
dy = txy[j][2] - txy[j - 1][2]
d = dt - abs(dx) - abs(dy)
if d < 0 or d % 2 != 0:
print("NO")
flg = 1
break
if flg == 0:
print("YES")
|
s596371081
|
Accepted
| 415
| 21,108
| 382
|
N = int(input())
txy = [[0,0,0]]
for i in range(N):
txy.append([int(x) for x in input().split()])
flg = 0
for j in range(1, N + 1):
dt = txy[j][0] - txy[j - 1][0]
dx = txy[j][1] - txy[j - 1][1]
dy = txy[j][2] - txy[j - 1][2]
d = dt - abs(dx) - abs(dy)
if d < 0 or d % 2 != 0:
print("No")
flg = 1
break
if flg == 0:
print("Yes")
|
s467040003
|
p02614
|
u945405878
| 1,000
| 1,048,576
|
Wrong Answer
| 132
| 27,084
| 1,053
|
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.
|
import numpy as np
import itertools
n_tate, n_yoko, n_kuro = map(int, input().split())
matrix = []
for _ in range(n_tate):
yoko = [_ for _ in str(input())]
yoko2 = []
for char in yoko:
if char == "#":
yoko2.append(1)
else:
yoko2.append(0)
matrix.append(yoko2)
matrix2 = np.array(matrix)
tmp = [0, 1]
tate_use_list = list(itertools.combinations_with_replacement(tmp, n_tate))
yoko_use_list = list(itertools.combinations_with_replacement(tmp, n_yoko))
# print("original")
# print(np.array(matrix))
# print()
i = 0
counter = 0
for tate_use in tate_use_list:
for yoko_use in yoko_use_list:
# print("tate use:", tate_use)
# print("yoko_use:", yoko_use)
matrix3 = matrix2.copy()
A = np.array(tate_use).reshape(-1, 1)
B = np.array(yoko_use)
matrix4 = matrix3 * A * B
# print(matrix4)
total = np.sum(matrix4)
i += 1
if total == n_kuro:
counter += 1
print(counter)
|
s841929843
|
Accepted
| 162
| 27,156
| 1,098
|
import numpy as np
import itertools
n_tate, n_yoko, n_kuro = map(int, input().split())
matrix = []
for _ in range(n_tate):
yoko = [_ for _ in str(input())]
yoko2 = []
for char in yoko:
if char == "#":
yoko2.append(1)
else:
yoko2.append(0)
matrix.append(yoko2)
matrix2 = np.array(matrix)
tmp = [0, 1]
tate_use_list = list(itertools.product(tmp, repeat=n_tate))
yoko_use_list = list(itertools.product(tmp, repeat=n_yoko))
# print("original")
# print(np.array(matrix))
# print()
i = 0
counter = 0
for tate_use in tate_use_list:
for yoko_use in yoko_use_list:
# print(i)
# print("tate use:", tate_use)
# print("yoko_use:", yoko_use)
matrix3 = matrix2.copy()
A = np.array(tate_use).reshape(-1, 1)
B = np.array(yoko_use)
matrix4 = matrix3 * A * B
total = np.sum(matrix4)
i += 1
# print(matrix4)
if total == n_kuro:
# print(" -> OK!")
counter += 1
# print()
print(counter)
|
s013528962
|
p03827
|
u483645888
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 157
|
You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x during the operations (including before the first operation, and after the last operation).
|
n = int(input())
cnt = 0
ch = 0
*w, = map(str,input().split())
for i in w:
if i == 'I':
cnt+=1
else:
cnt-=1
if cnt > ch:
ch = cnt
print(ch)
|
s582578329
|
Accepted
| 17
| 2,940
| 98
|
input()
x,mx=0,0
for w in input():
#print(w)
x+=1 if w=='I' else -1
mx = max(x,mx)
print(mx)
|
s560928623
|
p03475
|
u428132025
| 3,000
| 262,144
|
Wrong Answer
| 299
| 5,188
| 453
|
A railroad running from west to east in Atcoder Kingdom is now complete. There are N stations on the railroad, numbered 1 through N from west to east. Tomorrow, the opening ceremony of the railroad will take place. On this railroad, for each integer i such that 1≤i≤N-1, there will be trains that run from Station i to Station i+1 in C_i seconds. No other trains will be operated. The first train from Station i to Station i+1 will depart Station i S_i seconds after the ceremony begins. Thereafter, there will be a train that departs Station i every F_i seconds. Here, it is guaranteed that F_i divides S_i. That is, for each Time t satisfying S_i≤t and t％F_i=0, there will be a train that departs Station i t seconds after the ceremony begins and arrives at Station i+1 t+C_i seconds after the ceremony begins, where A％B denotes A modulo B, and there will be no other trains. For each i, find the earliest possible time we can reach Station N if we are at Station i when the ceremony begins, ignoring the time needed to change trains.
|
import sys
input = sys.stdin.readline
n = int(input())
c, s, f = [0]*(n-1), [0]*(n-1), [0]*(n-1)
for i in range(n-1):
c[i], s[i], f[i] = map(int, input().split())
ans = [0]*n
for i in range(n-1):
time = s[i]
for j in range(i, n-1):
if time < s[j]:
time = s[j]
if time % f[j] != 0:
time += f[j] - time % f[j]
time += c[j]
print(i, j, time)
ans[i] = time
for i in ans:
print(i)
|
s412184969
|
Accepted
| 106
| 3,188
| 427
|
import sys
input = sys.stdin.readline
n = int(input())
c, s, f = [0]*(n-1), [0]*(n-1), [0]*(n-1)
for i in range(n-1):
c[i], s[i], f[i] = map(int, input().split())
ans = [0]*n
for i in range(n-1):
time = s[i]
for j in range(i, n-1):
if time < s[j]:
time = s[j]
if time % f[j] != 0:
time += f[j] - time % f[j]
time += c[j]
ans[i] = time
for i in ans:
print(i)
|
s814805313
|
p02415
|
u720674978
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,532
| 24
|
Write a program which converts uppercase/lowercase letters to lowercase/uppercase for a given string.
|
print(input().swapcase)
|
s543043180
|
Accepted
| 20
| 5,540
| 26
|
print(input().swapcase())
|
s034421826
|
p03971
|
u714931250
| 2,000
| 262,144
|
Wrong Answer
| 82
| 9,308
| 396
|
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
n,a,b = map(int, input().split())
s = str(input())
a_count,b_count = 0,0
for i in range(n):
if s[i] == 'c':
print('No')
elif s[i] == 'a':
if a_count+b_count < a+b:
print('Yes')
a_count += 1
else:
print('No')
elif s[i] == 'b':
if a_count+b_count < a+b and b_count < b:
print('Yes')
b_count += 1
else:
print('No')
print(a_count,b_count)
|
s613183597
|
Accepted
| 80
| 9,312
| 373
|
n,a,b = map(int, input().split())
s = str(input())
a_count,b_count = 0,0
for i in range(n):
if s[i] == 'c':
print('No')
elif s[i] == 'a':
if a_count+b_count < a+b:
print('Yes')
a_count += 1
else:
print('No')
elif s[i] == 'b':
if a_count+b_count < a+b and b_count < b:
print('Yes')
b_count += 1
else:
print('No')
|
s361471785
|
p03471
|
u591808161
| 2,000
| 262,144
|
Wrong Answer
| 808
| 3,064
| 516
|
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.
|
import sys
input = sys.stdin.readline
def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))
n, y = MI()
result = "-1 -1 -1"
for i in range(n, -1, -1):
a = i * 10000
if a > y:
break
for j in range(n-i, -1, -1):
b = a + j * 5000
if b > y:
break
if b + 1000 * (n-i-j) == y:
result = str(a)+" "+str(b)+" "+str(n-i-j)
print(result)
sys.exit()
print(result)
|
s836735413
|
Accepted
| 514
| 3,064
| 600
|
import sys
input = sys.stdin.readline
def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))
def main():
n, y = MI()
result = "-1 -1 -1"
for i in range(n, -1, -1):
a = i * 10000
if a <= y:
for j in range(n-i, -1, -1):
b = a + j * 5000
if b <= y:
if b + 1000 * (n-i-j) == y:
result = str(i)+" "+str(j)+" "+str(n-i-j)
print(result)
sys.exit()
print(result)
main()
|
s675603505
|
p03407
|
u887207211
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 83
|
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
A, B, C = map(int,input().split())
if(A+B <= C):
print("Yes")
else:
print("No")
|
s907672658
|
Accepted
| 18
| 2,940
| 104
|
A, B, C = map(int,input().split())
ans = "No"
if(A+B >= C or A >= C or B >= C):
ans = "Yes"
print(ans)
|
s486485806
|
p03155
|
u474925961
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 68
|
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board. The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns. How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
|
n=int(input())
h=int(input())
w=int(input())
print((h-n+1)*(w-n+1))
|
s867027737
|
Accepted
| 17
| 2,940
| 68
|
n=int(input())
h=int(input())
w=int(input())
print((n-h+1)*(n-w+1))
|
s443875309
|
p03149
|
u871303155
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 115
|
You are given four digits N_1, N_2, N_3 and N_4. Determine if these can be arranged into the sequence of digits "1974".
|
l = map(int, input().split())
list = [1, 4, 7, 9]
for num in list:
if(num not in l):
print("NO")
print("YES")
|
s463020087
|
Accepted
| 19
| 2,940
| 181
|
l = list(map(int, input().split()))
list = [1, 4, 7, 9]
flg = True
for num in list:
if(num in l):
pass
else:
flg = False
if(flg):
print("YES")
else:
print("NO")
|
s248017569
|
p03672
|
u697696097
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,188
| 122
|
We will call a string that can be obtained by concatenating two equal strings an _even_ string. For example, `xyzxyz` and `aaaaaa` are even, while `ababab` and `xyzxy` are not. You are given an even string S consisting of lowercase English letters. Find the length of the longest even string that can be obtained by deleting one or more characters from the end of S. It is guaranteed that such a non-empty string exists for a given input.
|
s=input().strip()
for i in range(1000):
r=-2*i
s2 = s[:r]
half=len(s2)//2
if s2[half:]==s2[:half*-1]:
print(r)
|
s936474064
|
Accepted
| 17
| 3,060
| 157
|
import sys
s=input().strip()
for i in range(1,1000):
r=-2*i
s2 = s[:r]
half=len(s2)//2
if s2[half:]==s2[:half*-1]:
print(len(s)+r)
sys.exit()
|
s004738291
|
p03434
|
u487594898
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 199
|
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
N = int(input())
A = list(map(int,input().split()))
SA = sorted(A,reverse=True)
ALI =0
BOB=0
for i in range(1,N,2):
ALI += SA[i]
for n in range(2,N,2):
BOB += SA[n]
ans = ALI - BOB
print(ans)
|
s265639680
|
Accepted
| 18
| 3,060
| 210
|
N = int(input())
A = list(map(int,input().split()))
SA = sorted(A,reverse=True)
ALI =0
BOB=0
for i in range(0,N,2):
ALI += int(SA[i])
for n in range(1,N,2):
BOB += int(SA[n])
ans = ALI - BOB
print(ans)
|
s598336140
|
p03624
|
u193927973
| 2,000
| 262,144
|
Wrong Answer
| 27
| 3,956
| 173
|
You are given a string S consisting of lowercase English letters. Find the lexicographically (alphabetically) smallest lowercase English letter that does not occur in S. If every lowercase English letter occurs in S, print `None` instead.
|
s=list(input())
s=list(set(s))
s.sort()
print(s)
import string
l=string.ascii_lowercase
ans="None"
for i in range(len(s)):
if s[i]!=l[i]:
ans=l[i]
break
print(ans)
|
s290577518
|
Accepted
| 26
| 3,956
| 134
|
s=list(input())
s=set(s)
import string
l=string.ascii_lowercase
ans="None"
for a in l:
if a not in s:
ans=a
break
print(ans)
|
s173401026
|
p03564
|
u825528847
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,316
| 336
|
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.
|
import re
S = input()
T = input()
lenT = len(T)
can = []
for it in re.finditer(T[0], S):
a = it.span()[0]
tmp = S[a+1: a+lenT]
if len(tmp) == len(T) - 1 and all("?" == h for h in list(tmp)):
tmp = S[:a] + T + S[a+lenT:]
can.append(tmp.replace("?", "a"))
print("UNRESTORABLE" if len(can) == 0 else can[-1])
|
s863516817
|
Accepted
| 18
| 2,940
| 90
|
N = int(input())
K = int(input())
x = 1
for _ in range(N):
x = min(2*x, x+K)
print(x)
|
s777888927
|
p02743
|
u514118270
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 130
|
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
a,b,c = list(map(int,input().split()))
d = 1 / 2
A = a ** d
B = b ** d
C = c ** d
if A + B < c:
print('Yes')
else:
print('No')
|
s127345298
|
Accepted
| 43
| 5,332
| 172
|
from decimal import Decimal
a,b,c = list(map(Decimal,input().split()))
d = Decimal('0.5')
A = a ** d
B = b ** d
C = c ** d
if A + B < C:
print('Yes')
else:
print('No')
|
s915903449
|
p02388
|
u249954942
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,364
| 90
|
Write a program which calculates the cube of a given integer x.
|
## coding: UTF-8
def main():
print(pow(int(input()), 3))
if __name__ == "main":
main()
|
s513910199
|
Accepted
| 30
| 7,640
| 45
|
x = int(input())
y = 3
p = pow(x, y)
print(p)
|
s409850251
|
p03251
|
u397953026
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,136
| 242
|
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()
for i in range(X+1,Y+1,1):
if i <= y[0] and i > x[-1]:
print("No war")
break
else:
print("War")
|
s333229721
|
Accepted
| 31
| 9,148
| 245
|
n,m,X,Y = map(int,input().split())
x = list(map(int,input().split()))
y = list(map(int,input().split()))
x.sort()
y.sort()
for i in range(X+1,Y+1,1):
if i <= min(y) and i > max(x):
print("No War")
break
else:
print("War")
|
s827389008
|
p03377
|
u944209426
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 136
|
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 = list(map(int, input().split()))
ans=0
if a+b>=x:
if a<=x:
ans+=1
if ans==0:
print('No')
else:
print('Yes')
|
s671541043
|
Accepted
| 20
| 3,316
| 136
|
a, b, x = list(map(int, input().split()))
ans=0
if a+b>=x:
if a<=x:
ans+=1
if ans==0:
print('NO')
else:
print('YES')
|
s902270604
|
p03759
|
u133936772
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 60
|
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())
print('YNEOS'[a-b!=c-b::2])
|
s664662636
|
Accepted
| 19
| 2,940
| 61
|
a,b,c = map(int,input().split())
print('YNEOS'[a-b!=b-c::2])
|
s536657510
|
p03958
|
u422104747
| 1,000
| 262,144
|
Wrong Answer
| 25
| 3,064
| 90
|
There are K pieces of cakes. Mr. Takahashi would like to eat one cake per day, taking K days to eat them all. There are T types of cake, and the number of the cakes of type i (1 ≤ i ≤ T) is a_i. Eating the same type of cake two days in a row would be no fun, so Mr. Takahashi would like to decide the order for eating cakes that minimizes the number of days on which he has to eat the same type of cake as the day before. Compute the minimum number of days on which the same type of cake as the previous day will be eaten.
|
l=list(map(int, input().split()))
r=list(map(int, input().split()))
print(2*max(r)-l[0]-1)
|
s900825091
|
Accepted
| 22
| 3,064
| 97
|
l=list(map(int, input().split()))
r=list(map(int, input().split()))
print(max(0,2*max(r)-l[0]-1))
|
s875388731
|
p03731
|
u140251125
| 2,000
| 262,144
|
Wrong Answer
| 146
| 25,196
| 235
|
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?
|
# input
N, T = map(int, input().split())
T_lst = list(map(int, input().split()))
ans = 0
for i in range(N):
if T_lst[i] <= T_lst[i - 1] + T:
ans += T_lst[i] - T_lst[i - 1]
else:
ans += T
ans += T
print(ans)
|
s169471961
|
Accepted
| 151
| 26,832
| 238
|
# input
N, T = map(int, input().split())
T_lst = list(map(int, input().split()))
ans = 0
for i in range(1, N):
if T_lst[i] <= T_lst[i - 1] + T:
ans += T_lst[i] - T_lst[i - 1]
else:
ans += T
ans += T
print(ans)
|
s124892977
|
p04029
|
u540912766
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,096
| 40
|
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)
|
s598264672
|
Accepted
| 23
| 9,164
| 45
|
N = int(input())
print(int(N * (N + 1) / 2))
|
s326323257
|
p03673
|
u589969467
| 2,000
| 262,144
|
Wrong Answer
| 2,206
| 30,860
| 196
|
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()))
ans = []
for i in range(n):
if i%2==1:
ans.insert(0,a[i])
else:
ans.append(a[i])
if n%2==1:
ans.reverse()
print(ans)
|
s250831252
|
Accepted
| 823
| 31,504
| 445
|
import collections
def f(myList):
ans = ''
if len(myList)%2==0:
for i in range(len(myList)):
ans += str(myList[i]) + ' '
else:
for i in range(len(myList)):
ans += str(myList[-(i+1)]) + ' '
return ans[:-1]
n = int(input())
a = list(map(int,input().split()))
b = collections.deque()
for i in range(n):
if i%2==0:
b.append(a[i])
else:
b.appendleft(a[i])
print(f(b))
|
s206697307
|
p02678
|
u578464015
| 2,000
| 1,048,576
|
Wrong Answer
| 1,465
| 46,364
| 458
|
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
n,m=map(int,input().split())
d={}
for i in range(m):
x,y=map(int,input().split())
if x in d:
d[x].append(y)
else:
d[x]=[y]
if y in d:
d[y].append(x)
else:
d[y]=[x]
parent=[0]*(n+1)
bfs=[1]
visited=set([1])
while len(bfs)!=0:
x=bfs.pop(0)
visited.add(x)
for i in d[x]:
if i not in visited:
bfs.append(i)
parent[i]=x
visited.add(i)
if len(visited)==n:
print('YES')
for i in range(2,n+1):
print(parent[i])
else:
print('NO')
|
s468470816
|
Accepted
| 1,586
| 46,304
| 458
|
n,m=map(int,input().split())
d={}
for i in range(m):
x,y=map(int,input().split())
if x in d:
d[x].append(y)
else:
d[x]=[y]
if y in d:
d[y].append(x)
else:
d[y]=[x]
parent=[0]*(n+1)
bfs=[1]
visited=set([1])
while len(bfs)!=0:
x=bfs.pop(0)
visited.add(x)
for i in d[x]:
if i not in visited:
bfs.append(i)
parent[i]=x
visited.add(i)
if len(visited)==n:
print('Yes')
for i in range(2,n+1):
print(parent[i])
else:
print('No')
|
s041489461
|
p03760
|
u551109821
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 166
|
Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the characters at the even-numbered positions. You are given two strings O and E. O contains the characters at the odd- numbered positions retaining their relative order, and E contains the characters at the even-numbered positions retaining their relative order. Restore the original password.
|
O = list(input())
E = list(input())
ans = []
for i in range(len(E)):
ans.append(O[i])
ans.append(E[i])
if len(O) > len(E):
ans.append(O[-1])
print(ans)
|
s567426362
|
Accepted
| 17
| 3,060
| 175
|
O = list(input())
E = list(input())
ans = []
for i in range(len(E)):
ans.append(O[i])
ans.append(E[i])
if len(O) > len(E):
ans.append(O[-1])
print(''.join(ans))
|
s859929380
|
p02388
|
u676498528
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,716
| 4
|
Write a program which calculates the cube of a given integer x.
|
2**3
|
s272630543
|
Accepted
| 30
| 6,724
| 42
|
num = input()
num = int(num)
print(num**3)
|
s801153595
|
p02616
|
u597553490
| 2,000
| 1,048,576
|
Wrong Answer
| 235
| 33,252
| 2,230
|
Given are N integers A_1,\ldots,A_N. We will choose exactly K of these elements. Find the maximum possible product of the chosen elements. Then, print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).
|
import sys, bisect
if(__name__ == "__main__"):
N, K = map( int, input().strip().split() )
divisor = 10**9+7
list_a = list( map(int, input().strip().split()) )
multiplies = 1
if(N==K or N==1):
for i in range(N):
multiplies = multiplies * list_a[i] % divisor
print(multiplies)
else:
list_a.sort()
N_negative_numbers = bisect.bisect_left(list_a, 0)
N_select_negatives = K + N_negative_numbers -N
flag_overzero = True
if(N_select_negatives > 0 and N_select_negatives % 2 != 0):
flag_overzero = False
if(flag_overzero == True):
left = 0
right = N-1
multiply_right = list_a[right] * list_a[right-1]
flag_left = True
while(K>0):
if(K>1):
K -= 2
if(flag_left==True):
multiply_left = list_a[left] * list_a[left+1]
else:
multiply_right = list_a[right] * list_a[right-1]
if(multiply_left > multiply_right):
multiply_left %= divisor
multiplies = multiplies * multiply_left % divisor
left += 2
flag_left=True
print(multiply_left, multiplies)
else:
multiply_right %= divisor
multiplies = multiplies * multiply_right % divisor
right -= 2
flag_left=False
print(multiply_right, multiplies)
else:
multiplies = multiplies * list_a[right] % divisor
K =0
print(multiplies)
else:
list_a = [ abs(list_a[i]) for i in range(N)]
list_a.sort()
for i in range(K-1):
multiplies = multiplies * list_a[i] % divisor
multiplies = (-1) * multiplies * list_a[K-1] % divisor
print(multiplies)
|
s430855502
|
Accepted
| 147
| 30,852
| 3,294
|
import sys
if(__name__ == "__main__"):
N, K = map( int, input().strip().split() )
divisor = 10**9+7
list_a = list( map( int, input().strip().split() ) )
multiplies = 1
if(N==K):
for i in range(N):
multiplies = multiplies * list_a[i] % divisor
else:
list_a.sort()
if(list_a[-1] < 0 and K % 2 != 0):
for i in range(K):
multiplies = multiplies * list_a[N-1-i] % divisor
elif(K==1):
multiplies = list_a[N-1]
else:
left = 0
right = N-1
if(K%2!=0):
multiplies = multiplies * list_a[right] % divisor
right -= 1
multiply_left = list_a[left] * list_a[left+1]
multiply_right = list_a[right] * list_a[right-1]
flag_left = True
for s in range(K//2):
if(flag_left==True):
multiply_left = list_a[left] * list_a[left+1]
else:
multiply_right = list_a[right] * list_a[right-1]
if(multiply_left < multiply_right):
multiply_right %= divisor
multiplies = multiplies * multiply_right % divisor
right -= 2
flag_left=False
else:
multiply_left%=divisor
multiplies = multiplies * multiply_left % divisor
left += 2
flag_left=True
print(multiplies)
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.