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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s885461286
|
p03720
|
u676264453
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 324
|
There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city?
|
N,M = map(int, input().split())
a = []
for i in range(M):
tmp_l = map(int, input().split())
for i in tmp_l:
a.append(i)
itr = 0
count = 0
d = {}
print(list(range(1,N+1,1)))
print(a)
for j in list(range(1,N+1,1)):
d[j] = 0
for k in a:
d[k] = d[k] + 1
for o in list(range(1,N+1,1)):
print(d[o])
|
s544722719
|
Accepted
| 17
| 3,064
| 287
|
N,M = map(int, input().split())
a = []
for i in range(M):
tmp_l = map(int, input().split())
for i in tmp_l:
a.append(i)
itr = 0
count = 0
d = {}
for j in list(range(1,N+1,1)):
d[j] = 0
for k in a:
d[k] = d[k] + 1
for o in list(range(1,N+1,1)):
print(d[o])
|
s148932156
|
p03635
|
u952708174
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 119
|
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?
|
string=input('')
first=string[0]
last=string[len(string)-1]
mid=str(len(string[1:len(string)-1]))
print(first+mid+last)
|
s610978522
|
Accepted
| 17
| 2,940
| 59
|
n, m = (int(i) for i in input().split())
print((n-1)*(m-1))
|
s443575638
|
p03223
|
u543954314
| 2,000
| 1,048,576
|
Wrong Answer
| 231
| 7,764
| 198
|
You are given N integers; the i-th of them is A_i. Find the maximum possible sum of the absolute differences between the adjacent elements after arranging these integers in a row in any order you like.
|
n = int(input())
a = [int(input()) for _ in range(n)]
a.sort()
a1 = a[:n//2]
a2 = a[n-1:n//2-1:-1]
ans = 0
if n%2==0:
a2 += [a1[-1]]
for i in range(n//2):
ans += a2[i]+a2[i+1]-a1[i]*2
print(ans)
|
s578011827
|
Accepted
| 214
| 8,532
| 338
|
n = int(input())
l = [int(input()) for _ in range(n)]
l.sort()
if n%2:
s1 = l[:(n-1)//2]
b1 = l[(n-1)//2:]
t1 = sum(b1)*2 - sum(s1)*2 -sum(b1[:2])
s2 = l[:(n+1)//2]
b2 = l[(n+1)//2:]
t2 = sum(b2)*2 - sum(s2)*2 + sum(s2[-2:])
t = max(t1,t2)
else:
s = l[:n//2]
b = l[n//2:]
t = sum(b)*2 - sum(s)*2 -b[0] + s[-1]
print(t)
|
s484818270
|
p02927
|
u927530154
| 2,000
| 1,048,576
|
Wrong Answer
| 20
| 3,060
| 315
|
Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have?
|
def seki_day(m, d):
d10 = d // 10
d1 = d - d10 * 10
if d1 >= 2 and m == d10 * d1:
print(m, d)
return 1
else:
return 0
M, D = [int(x) for x in input().split()]
count = 0
for m in range(1, M + 1):
for d in range(22, D + 1):
count += seki_day(m, d)
print(count)
|
s371241270
|
Accepted
| 19
| 2,940
| 295
|
def seki_day(m, d):
d10 = d // 10
d1 = d - d10 * 10
if d1 >= 2 and m == d10 * d1:
return 1
else:
return 0
M, D = [int(x) for x in input().split()]
count = 0
for m in range(1, M + 1):
for d in range(22, D + 1):
count += seki_day(m, d)
print(count)
|
s616066216
|
p03625
|
u816631826
| 2,000
| 262,144
|
Wrong Answer
| 115
| 18,216
| 337
|
We have N sticks with negligible thickness. The length of the i-th stick is A_i. Snuke wants to select four different sticks from these sticks and form a rectangle (including a square), using the sticks as its sides. Find the maximum possible area of the rectangle.
|
n = int(input())
vals = list(map(int, input().split()))
d= {}
a = [0, 0]
for i in range(n):
if(vals[i] in d.keys()):
d[vals[i]] += 1
else:
d[vals[i]] = 1
print(d)
for i in d.keys():
if(d[i] >= 4):
a.append(i)
a.append(i)
elif(d[i] >= 2):
a.append(i)
a.sort()
print(a[-1]*a[-2])
|
s784062218
|
Accepted
| 111
| 14,252
| 256
|
n=int(input())
a=[int(x) for x in input().split()]
a.sort()
lar=sec=i=0
while(i<n-1):
if a[i]==a[i+1]:
i+=1
if a[i]>lar:
sec=lar
lar=a[i]
elif a[i]>sec:
sec=a[i]
i+=1
print(lar*sec)
|
s664924598
|
p03578
|
u513900925
| 2,000
| 262,144
|
Wrong Answer
| 2,105
| 35,420
| 441
|
Rng is preparing a problem set for a qualification round of CODEFESTIVAL. He has N candidates of problems. The difficulty of the i-th candidate is D_i. There must be M problems in the problem set, and the difficulty of the i-th problem must be T_i. Here, one candidate of a problem cannot be used as multiple problems. Determine whether Rng can complete the problem set without creating new candidates of problems.
|
N = int(input())
a = [int(i) for i in input().split()]
M = int(input())
b = [int(i) for i in input().split()]
a.sort()
b.sort()
p=0
for i in range (len(b)):
for j in range(len(a)):
if b[i] == a[j]:
a = a[j+1:N]
break
else:
if j ==len(a)-1:
print("No")
p =1
break
if p ==1:
break
if i == len(b)-1:
print("Yes")
|
s276768900
|
Accepted
| 297
| 41,316
| 352
|
N = int(input())
D = list(map(int, input().split()))
M = int(input())
T = list(map(int, input().split()))
d_dict = {}
for d in D:
if d not in d_dict:
d_dict[d] = 0
d_dict[d] += 1
count = 0
for t in T:
if t in d_dict and d_dict[t] > 0:
d_dict[t] -= 1
count += 1
if count == M:
print("YES")
else:
print("NO")
|
s238269117
|
p03680
|
u026155812
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 7,084
| 226
|
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.
|
import sys
N = int(input())
a = []
for i in range(N):
a.append(int(input()))
ls = []
cur = a[0]
ans = 0
while not cur in ls:
ls.append(cur)
cur = a[cur-1]
ans += 1
if cur == 2:
print(ans)
print(-1)
|
s042297455
|
Accepted
| 224
| 13,220
| 323
|
# 2019/6/25
import sys
N = int(input())
a = []
for i in range(N):
a.append(int(input()))
if not 2 in a:
print(-1)
sys.exit()
s = set()
cur = 1
ans = 0
while not cur in s:
if a[cur-1] == 2:
print(ans+1)
sys.exit()
else:
s.add(cur)
cur = a[cur-1]
ans += 1
print(-1)
|
s615266456
|
p03486
|
u172569352
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 142
|
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.
|
s = ''.join(sorted(list(input())))
t = ''.join(sorted(list(input()), reverse=True))
print(s)
if s < t:
print('Yes')
else:
print('No')
|
s983383643
|
Accepted
| 17
| 2,940
| 133
|
s = ''.join(sorted(list(input())))
t = ''.join(sorted(list(input()), reverse=True))
if s < t:
print('Yes')
else:
print('No')
|
s650402454
|
p02694
|
u815931251
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 9,164
| 133
|
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())
year = 0
sum = 100
while sum <= x:
year = year + 1
sum = (int)(sum * 101 / 100)
print("{}".format(year))
|
s232144653
|
Accepted
| 22
| 9,168
| 132
|
x = (int)(input())
year = 0
sum = 100
while sum < x:
year = year + 1
sum = (int)(sum * 101 / 100)
print("{}".format(year))
|
s272007542
|
p03407
|
u057463552
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 85
|
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")
|
s292300063
|
Accepted
| 17
| 2,940
| 85
|
A,B,C = map(int,input().split())
if A +B >= C:
print("Yes")
else:
print("No")
|
s180764597
|
p03623
|
u762540523
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 60
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b=map(int,input().split())
print(min(abs(x-a),abs(x-b)))
|
s640863054
|
Accepted
| 17
| 2,940
| 61
|
x,a,b=map(int,input().split())
print("BA"[abs(x-a)<abs(x-b)])
|
s466694583
|
p04044
|
u708211626
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,168
| 125
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
a=input()
b=[]
for i in range(int(a[0])):
b.append(input())
f=sorted(b)
for i in range(int(a[0])):
print(f[i],end='')
|
s147045285
|
Accepted
| 28
| 9,164
| 106
|
a,b=map(int,input().split())
c=[]
for i in range(a):
c.append(input())
d=sorted(c)
print(''.join(d))
|
s744399179
|
p02646
|
u199356004
| 2,000
| 1,048,576
|
Wrong Answer
| 24
| 9,180
| 148
|
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
A,V = [int(i) for i in input().split() ]
B,W = [int(i) for i in input().split() ]
T = int(input())
print ( "Yes" if A+(V*T) >= B+(W*T) else "No" )
|
s413400523
|
Accepted
| 20
| 9,168
| 164
|
A,V = [int(i) for i in input().split() ]
B,W = [int(i) for i in input().split() ]
T = int(input())
print ("YES" if V-W > 0 and abs(A-B) / (V-W) <= T else "NO" )
|
s841166106
|
p03565
|
u106778233
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 194
|
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,T=input(),input()
m=len(S)
for i in range(len(T)-m+1):
if all(c in"?"+d for c,d in zip(S[-i-m:],T)):
S=S.replace("?","a")
print(S[:-i-m]+T+S[-i:])
quit()
print("UNRESTORABLE")
|
s577543088
|
Accepted
| 19
| 2,940
| 183
|
S,T=input(),input();l,m=len(S),len(T)
for i in range(l-m+1):
if all(c in"?"+d for c,d in zip(S[-i-m:],T)):S=S.replace("?","a");print(S[:-i-m]+T+S[l-i:]);quit()
print("UNRESTORABLE")
|
s158337563
|
p03605
|
u783589546
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 59
|
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
n = input()
if "9" in n:
print("YES")
else:
print("NO")
|
s944738831
|
Accepted
| 17
| 2,940
| 59
|
n = input()
if "9" in n:
print("Yes")
else:
print("No")
|
s996634324
|
p03080
|
u502389123
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,064
| 133
|
There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat.
|
N = int(input())
S = input()
cnt = 0
for s in S:
if "R" == cnt:
cnt += 1
if cnt > N - cnt:
print("Yes")
else:
print("No")
|
s440532905
|
Accepted
| 17
| 2,940
| 131
|
N = int(input())
S = input()
cnt = 0
for s in S:
if "R" == s:
cnt += 1
if cnt > N - cnt:
print("Yes")
else:
print("No")
|
s364400357
|
p02694
|
u085530099
| 2,000
| 1,048,576
|
Wrong Answer
| 23
| 9,164
| 116
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
import math
x = int(input())
t = 100
count = 0
while t <= x:
count += 1
t = t+(math.floor(t*1/100))
print(count)
|
s412245265
|
Accepted
| 21
| 9,192
| 115
|
import math
x = int(input())
t = 100
count = 0
while t < x:
count += 1
t = t+(math.floor(t*1/100))
print(count)
|
s903313028
|
p03943
|
u182047166
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 107
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a, b, c = map(int, input().split())
if a+b == c or b+c==a or c+a==b:
print("YES")
else:
print("NO")
|
s744016840
|
Accepted
| 17
| 2,940
| 117
|
a, b, c = map(int, input().split())
if a + b == c or b + c == a or c + a == b:
print("Yes")
else:
print("No")
|
s686459407
|
p03067
|
u353099601
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 119
|
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
A, B, C = map(int, input().split())
if A < C < B:
print('YES')
elif B < C < A:
print('YES')
else:
print('NO')
|
s013384574
|
Accepted
| 19
| 2,940
| 116
|
A, B, C = map(int, input().split())
if A < C < B:
print('Yes')
elif B < C < A:
print('Yes')
else:
print('No')
|
s966822256
|
p03719
|
u670180528
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 55
|
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
a,b,c=map(int,input().split());print("NYoe s"[a<=c<=b])
|
s450785024
|
Accepted
| 17
| 2,940
| 59
|
a,b,c=map(int,input().split());print("NYoe s"[a<=c<=b::2])
|
s928592328
|
p02865
|
u207576418
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 88
|
How many ways are there to choose two distinct positive integers totaling N, disregarding the order?
|
n = int(input())
a = 0
if n % 2 == 0:
a = n / 2 - 1
else:
a = (n-1)/2
print(a)
|
s679209536
|
Accepted
| 17
| 2,940
| 93
|
n = int(input())
a = 0
if n % 2 == 0:
a = n / 2 - 1
else:
a = (n-1)/2
print(int(a))
|
s623313291
|
p03844
|
u434500430
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 109
|
Joisino wants to evaluate the formula "A op B". Here, A and B are integers, and the binary operator op is either `+` or `-`. Your task is to evaluate the formula instead of her.
|
A, Op, B = input().split()
A = int(A)
B = int(B)
if A == '+':
print(A+B)
elif Op == '-':
print(A-B)
|
s332566941
|
Accepted
| 17
| 2,940
| 110
|
A, Op, B = input().split()
A = int(A)
B = int(B)
if Op == '+':
print(A+B)
elif Op == '-':
print(A-B)
|
s210193135
|
p03455
|
u058518059
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 90
|
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")
|
s198463291
|
Accepted
| 17
| 2,940
| 84
|
a, b = map(int, input().split())
if a*b%2 == 0:
print("Even")
else:
print("Odd")
|
s976430854
|
p02608
|
u532141811
| 2,000
| 1,048,576
|
Wrong Answer
| 923
| 9,116
| 283
|
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
N = int(input())
ans =[0]*(N+1)
for x in range(1, 101):
for y in range(1, 101):
for z in range(1, 101):
s = x**2 + y**2 + z**2
s += x*y + y*z + z*x
if s > N:
continue
ans[s] += 1
for i in ans:
print(i)
|
s733519384
|
Accepted
| 824
| 9,112
| 263
|
N = int(input())
ans =[0]*(N+1)
for x in range(1, 101):
for y in range(1, 101):
for z in range(1, 101):
s = x**2 + y**2 + z**2 + x*y + y*z + z*x
if s <= N:
ans[s] += 1
for i in range(1, N+1):
print(ans[i])
|
s958724203
|
p03943
|
u802772880
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 79
|
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a=map(int,input().split())
ans='No'
if 2*max(a)==sum(a):
ans='Yes'
print(ans)
|
s915806697
|
Accepted
| 17
| 2,940
| 86
|
a=list(map(int,input().split()))
ans='No'
if 2*max(a)==sum(a):
ans='Yes'
print(ans)
|
s430278593
|
p02385
|
u073709667
| 1,000
| 131,072
|
Wrong Answer
| 40
| 11,828
| 959
|
Write a program which reads the two dices constructed in the same way as [Dice I](description.jsp?id=ITP1_11_A), and determines whether these two dices are identical. You can roll a dice in the same way as [Dice I](description.jsp?id=ITP1_11_A), and if all integers observed from the six directions are the same as that of another dice, these dices can be considered as identical.
|
import random
def N():
num=Dice[0]
Dice[0]=Dice[1]
Dice[1]=Dice[5]
Dice[5]=Dice[4]
Dice[4]=num
def E():
num=Dice[0]
Dice[0]=Dice[3]
Dice[3]=Dice[5]
Dice[5]=Dice[2]
Dice[2]=num
def W():
num=Dice[0]
Dice[0]=Dice[2]
Dice[2]=Dice[5]
Dice[5]=Dice[3]
Dice[3]=num
def S():
num=Dice[0]
Dice[0]=Dice[4]
Dice[4]=Dice[5]
Dice[5]=Dice[1]
Dice[1]=num
Dice=[int(i) for i in input().split()]
Dice2=[int(i) for i in input().split()]
if (Dice[0] in Dice2)==False:
print('No')
else:
while True:
rand=random.randint(1,4)
if rand==1:
N()
elif rand==2:
E()
elif rand==3:
W()
else:
S()
if Dice[0]==Dice2[0]:
break
for i in range(3):
N()
W()
S()
if Dice==Dice2:
break
if Dice==Dice2:
print('Yes')
else:
print('No')
|
s309460822
|
Accepted
| 50
| 7,800
| 861
|
def N():
num=Dice[0]
Dice[0]=Dice[1]
Dice[1]=Dice[5]
Dice[5]=Dice[4]
Dice[4]=num
def E():
num=Dice[0]
Dice[0]=Dice[3]
Dice[3]=Dice[5]
Dice[5]=Dice[2]
Dice[2]=num
def W():
num=Dice[0]
Dice[0]=Dice[2]
Dice[2]=Dice[5]
Dice[5]=Dice[3]
Dice[3]=num
def S():
num=Dice[0]
Dice[0]=Dice[4]
Dice[4]=Dice[5]
Dice[5]=Dice[1]
Dice[1]=num
Dice=[int(i) for i in input().split()]
Dice2=[int(i) for i in input().split()]
if (Dice[0] in Dice2)==False:
print('No')
else:
for j in range(6):
if j%2==0:
N()
else:
W()
for i in range(4):
N()
W()
S()
if Dice==Dice2:
break
if Dice==Dice2:
break
if Dice==Dice2:
print('Yes')
else:
print('No')
|
s869239068
|
p03836
|
u680851063
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,188
| 246
|
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
a, b, c, d = map(int,input().split())
x = c - a
y = d - b
z = []
z += y * 'U'
z += x * 'R'
z += y * 'D'
z += (x + 1) * 'L'
z += (y + 1) * 'U'
z += (1 + x) * 'R'
z += 'U'
z += 'R'
z += (y + 1) * 'U'
z += (1 + x) *'L'
z += 'U'
print(''.join(z))
|
s635400595
|
Accepted
| 18
| 3,188
| 245
|
a, b, c, d = map(int,input().split())
x = c - a
y = d - b
z = []
z += y * 'U'
z += x * 'R'
z += y * 'D'
z += (x + 1) * 'L'
z += (y + 1) * 'U'
z += (1 + x) * 'R'
z += 'D'
z += 'R'
z += (y + 1) * 'D'
z += (1 + x) *'L'
z += 'U'
print(''.join(z))
|
s628400545
|
p02255
|
u672443148
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 212
|
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
N=int(input())
arr=list(map(int,input().split()))
for key in range(1,len(arr)):
temp=arr[key]
i=key-1
while i>=0 and arr[i]>temp:
arr[i+1]=arr[i]
i-=1
arr[i+1]=temp
print(*arr)
|
s580120172
|
Accepted
| 30
| 5,980
| 242
|
N=int(input())
arr=list(map(int,input().split()))
print(' '.join(map(str,arr)))
for key in range(1,len(arr)):
temp=arr[key]
i=key-1
while i>=0 and arr[i]>temp:
arr[i+1]=arr[i]
i-=1
arr[i+1]=temp
print(*arr)
|
s231161493
|
p03605
|
u626228246
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 66
|
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
N = list(input())
print("Yes" if N[0] == 9 or N[1] == 9 else "No")
|
s572290902
|
Accepted
| 17
| 2,940
| 71
|
N = list(input())
print("Yes" if N[0] == "9" or N[1] == "9" else "No")
|
s179802832
|
p03047
|
u273006189
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 34
|
Snuke has N integers: 1,2,\ldots,N. He will choose K of them and give those to Takahashi. How many ways are there to choose K consecutive integers?
|
N, K = input().split(" ")
print(N)
|
s152221574
|
Accepted
| 30
| 3,060
| 218
|
line = input().split(" ")
N = int(line[0])
M = int(line[1])
count = 0
for i in range(1, N+1):
for j in range(M):
if i+j > N:
break
if j == M-1:
count += 1
print(count)
|
s751556367
|
p03139
|
u497883442
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 69
|
We conducted a survey on newspaper subscriptions. More specifically, we asked each of the N respondents the following two questions: * Question 1: Are you subscribing to Newspaper X? * Question 2: Are you subscribing to Newspaper Y? As the result, A respondents answered "yes" to Question 1, and B respondents answered "yes" to Question 2. What are the maximum possible number and the minimum possible number of respondents subscribing to both newspapers X and Y? Write a program to answer this question.
|
n,a,b = list(map(int, input().split()))
print(max(a,b), max(a+b-n,0))
|
s352748436
|
Accepted
| 17
| 2,940
| 69
|
n,a,b = list(map(int, input().split()))
print(min(a,b), max(a+b-n,0))
|
s734073608
|
p03067
|
u167307651
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 101
|
There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise.
|
a,b,c = map(int, input().split())
if ((a<b)&(b<c))|((a>b)&(b>c)):
print ("Yes")
else:
print("No")
|
s712353416
|
Accepted
| 17
| 2,940
| 101
|
a,b,c = map(int, input().split())
if ((a<c)&(c<b))|((a>c)&(c>b)):
print ("Yes")
else:
print("No")
|
s187813940
|
p03567
|
u814986259
| 2,000
| 262,144
|
Wrong Answer
| 28
| 8,992
| 59
|
Snuke built an online judge to hold a programming contest. When a program is submitted to the judge, the judge returns a verdict, which is a two-character string that appears in the string S as a contiguous substring. (The judge can return any two-character substring of S.) Determine whether the judge can return the string `AC` as the verdict to a program.
|
S = input()
if 'AC' in S:
print("AC")
else:
print("NO")
|
s578574767
|
Accepted
| 29
| 9,024
| 61
|
S = input()
if 'AC' in S:
print("Yes")
else:
print("No")
|
s331701529
|
p03478
|
u931462344
| 2,000
| 262,144
|
Wrong Answer
| 32
| 2,940
| 188
|
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())
sum = 0
for i in range(1, n+1):
tmpi = 0
for c in str(i):
tmpi += int(c)
if tmpi >= a and b >= tmpi:
sum += tmpi
print(sum)
|
s429324412
|
Accepted
| 32
| 2,940
| 185
|
n, a, b = map(int, input().split())
sum = 0
for i in range(1, n+1):
tmpi = 0
for c in str(i):
tmpi += int(c)
if tmpi >= a and b >= tmpi:
sum += i
print(sum)
|
s767832266
|
p03623
|
u059262067
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 103
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
a = 0
a, b, c = (abs(int(i) - a) for i in input().split())
if b > c:
print("B")
else:
print("A")
|
s025000218
|
Accepted
| 18
| 2,940
| 100
|
x, a, b = (int(i) for i in input().split())
if abs(a-x) > abs(b-x):
print("B")
else:
print("A")
|
s117752560
|
p03149
|
u657454538
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 215
|
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".
|
s=input()
n = len(s)
for i in range(n):
for j in range(i,n): #
tmp = s[:i]+s[j:]
if tmp == "keyence":
print("YES")
quit()
print("NO")
|
s397515221
|
Accepted
| 17
| 2,940
| 286
|
a = list(map(int,input().split()))
if 1 in a:
if 9 in a:
if 7 in a:
if 4 in a:
print("YES")
else :
print("NO")
else :
print("NO")
else :
print("NO")
else :
print("NO")
|
s181753340
|
p03433
|
u819663617
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,124
| 104
|
E869120 has A 1-yen coins and infinitely many 500-yen coins. Determine if he can pay exactly N yen using only these coins.
|
N = int(input())
A = int(input())
amari = N % 500
if amari <= A:
print("YES")
else:
print("NO")
|
s529536184
|
Accepted
| 25
| 9,072
| 101
|
N = int(input())
A = int(input())
amari = N % 500
if amari <= A:
print("Yes")
else:
print("No")
|
s532134926
|
p02406
|
u602702913
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,560
| 97
|
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
a=int(input())
for i in range(3,a+1):
if i% 3==0:
print(" "+str(i),end=" ")
|
s770304647
|
Accepted
| 30
| 7,948
| 197
|
n = int(input())
for i in range(1, n + 1):
if i % 3 == 0:
print(' ' + str(i), end='')
else:
s = str(i)
if '3' in s:
print(' ' + str(i), end='')
print('')
|
s660319596
|
p03795
|
u458608788
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 37
|
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
a=int(input())
print(a*800-200*a//15)
|
s752952835
|
Accepted
| 17
| 2,940
| 40
|
a=int(input())
print(a*800-200*(a//15))
|
s837584360
|
p03659
|
u970197315
| 2,000
| 262,144
|
Wrong Answer
| 233
| 24,832
| 196
|
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it. They will share these cards. First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards. Here, both Snuke and Raccoon have to take at least one card. Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively. They would like to minimize |x-y|. Find the minimum possible value of |x-y|.
|
N = int(input())
A = list(map(int,input().split()))
X = [0]*N
S = sum(A)
ans = 10**11
for i in range(1,N):
X[i] = X[i-1] + A[i]
t = abs(X[i]+X[i]-S)
ans = min(ans,t)
print(ans)
|
s354074473
|
Accepted
| 190
| 24,636
| 252
|
n=int(input())
a=list(map(int,input().split()))
from itertools import accumulate
s=list(accumulate(a))
a_sum=sum(a)
ans=10**18
if n==2:
print(abs(a[0]-a[1]))
exit(0)
for i in range(1,n-1):
x=s[i]
y=a_sum-x
ans=min(ans,abs(x-y))
print(ans)
|
s875668735
|
p03568
|
u432042540
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 163
|
We will say that two integer sequences of length N, x_1, x_2, ..., x_N and y_1, y_2, ..., y_N, are _similar_ when |x_i - y_i| \leq 1 holds for all i (1 \leq i \leq N). In particular, any integer sequence is similar to itself. You are given an integer N and an integer sequence of length N, A_1, A_2, ..., A_N. How many integer sequences b_1, b_2, ..., b_N are there such that b_1, b_2, ..., b_N is similar to A and the product of all elements, b_1 b_2 ... b_N, is even?
|
n = int(input())
a = [int(i) for i in input().split()]
print(a)
cnt = 0
for i in range(n):
if a[i] % 2 == 0:
cnt += 1
x = 3 ** n - 2 ** cnt
print(x)
|
s424470166
|
Accepted
| 17
| 2,940
| 155
|
n = int(input())
a = [int(i) for i in input().split()]
cnt = 0
for i in range(n):
if a[i] % 2 == 0:
cnt += 1
x = 3 ** n - 2 ** cnt
print(x)
|
s302758286
|
p03828
|
u521992338
| 2,000
| 262,144
|
Wrong Answer
| 44
| 3,064
| 265
|
You are given an integer N. Find the number of the positive divisors of N!, modulo 10^9+7.
|
n = int(input())
ps = [0] * (n + 1)
for i in range(2, n + 1):
x = i
p = 2
while x > 1:
while x % p == 0:
ps[p] += 1
x //= p
p += 1
print (ps)
ans = 1
for c in ps:
ans = (ans * (c + 1)) % 1000000007
print (ans)
|
s669180090
|
Accepted
| 45
| 3,060
| 254
|
n = int(input())
ps = [0] * (n + 1)
for i in range(2, n + 1):
x = i
p = 2
while x > 1:
while x % p == 0:
ps[p] += 1
x //= p
p += 1
ans = 1
for c in ps:
ans = (ans * (c + 1)) % 1000000007
print (ans)
|
s503034009
|
p03852
|
u063052907
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 64
|
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
#coding: utf-8
print(["vowel", "consonant"][input() in "aiueo"])
|
s138403976
|
Accepted
| 17
| 2,940
| 63
|
#coding: utf-8
print(["consonant","vowel"][input() in "aiueo"])
|
s826165820
|
p02678
|
u461454424
| 2,000
| 1,048,576
|
Wrong Answer
| 1,008
| 104,580
| 940
|
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.
|
#atcoder template
def main():
import sys
imput = sys.stdin.readline
#input
N, M = map(int, input().split())
AB = [map(int, input().split()) for _ in range(M)]
C = [[] for _ in range(N+1)]
for a, b in AB:
C[a].append(b)
C[b].append(a)
#output
from collections import deque
search_stack = deque([1])
signpost = [-1] * (N+1)
while search_stack:
next_position = search_stack.popleft()
for node in C[next_position]:
if signpost[node] == -1:
signpost[node] = next_position
search_stack.append(node)
print("Yes")
for i in range(2, N):
print(signpost[i])
if __name__ == "__main__":
main()
|
s869842255
|
Accepted
| 941
| 104,484
| 942
|
#atcoder template
def main():
import sys
imput = sys.stdin.readline
#input
N, M = map(int, input().split())
AB = [map(int, input().split()) for _ in range(M)]
C = [[] for _ in range(N+1)]
for a, b in AB:
C[a].append(b)
C[b].append(a)
#output
from collections import deque
search_stack = deque([1])
signpost = [-1] * (N+1)
while search_stack:
next_position = search_stack.popleft()
for node in C[next_position]:
if signpost[node] == -1:
signpost[node] = next_position
search_stack.append(node)
print("Yes")
for i in range(2, N+1):
print(signpost[i])
if __name__ == "__main__":
main()
|
s022013178
|
p04044
|
u843768197
| 2,000
| 262,144
|
Wrong Answer
| 27
| 9,088
| 85
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
n, l = map(int, input().split())
a = list(input() for i in range(n))
print(sorted(a))
|
s035717837
|
Accepted
| 27
| 9,112
| 119
|
n, l = map(int, input().split())
a = list(input() for i in range(n))
ans = ''
for i in sorted(a):
ans += i
print(ans)
|
s907511674
|
p01137
|
u124909914
| 8,000
| 131,072
|
Wrong Answer
| 210
| 6,724
| 494
|
English text is not available in this practice contest. ケン・マリンブルーは，宇宙ヤシガニを求めて全銀河を旅するスペースハンターである．宇宙ヤシガニは，宇宙最大とされる甲殻類であり，成長後の体長は 400 メートル以上，足を広げれば 1,000 メートル以上にもなると言われている．既に多数の人々が宇宙ヤシガニを目撃しているが，誰一人として捕らえることに成功していない． ケンは，長期間の調査によって，宇宙ヤシガニの生態に関する重要な事実を解明した．宇宙ヤシガニは，驚くべきことに，相転移航法と呼ばれる最新のワープ技術と同等のことを行い，通常空間と超空間の間を往来しながら生きていた．さらに，宇宙ヤシガニが超空間から通常空間にワープアウトするまでには長い時間がかかり，またワープアウトしてからしばらくは超空間に移動できないことも突き止めた． そこで，ケンはついに宇宙ヤシガニの捕獲に乗り出すことにした．戦略は次のとおりである．はじめに，宇宙ヤシガニが通常空間から超空間に突入する際のエネルギーを観測する．このエネルギーを _e_ とするとき，宇宙ヤシガニが超空間からワープアウトする座標 ( _x_ , _y_ , _z_ ) は以下の条件を満たすことがわかっている． * _x_ , _y_ , _z_ はいずれも非負の整数である． * _x_ \+ _y_ 2 \+ _z_ 3 = _e_ である． * 上記の条件の下で _x_ \+ _y_ \+ _z_ の値を最小にする． これらの条件だけでは座標が一意に決まるとは限らないが， _x_ \+ _y_ \+ _z_ の最小値を _m_ としたときに，ワープアウトする座標が平面 _x_ \+ _y_ \+ _z_ = _m_ 上にあることは確かである．そこで，この平面上に十分な大きさのバリアを張る．すると，宇宙ヤシガニはバリアの張られたところにワープアウトすることになる．バリアの影響を受けた宇宙ヤシガニは身動きがとれなくなる．そこをケンの操作する最新鋭宇宙船であるウェポン・ブレーカー号で捕獲しようという段取りである． バリアは一度しか張ることができないため，失敗するわけにはいかない．そこでケンは，任務の遂行にあたって計算機の助けを借りることにした．あなたの仕事は，宇宙ヤシガニが超空間に突入する際のエネルギーが与えられたときに，バリアを張るべき平面 _x_ \+ _y_ \+ _z_ = _m_ を求めるプログラムを書くことである．用意されたテストケースの全てに対して正しい結果を出力したとき，あなたのプログラムは受け入れられるであろう．
|
cubics = [ x * x * x for x in range(0, 100)]
squares = [ x * x for x in range(0, 1000) if x * x < 100 ** 3 - 99 ** 3]
while True:
x, y, z = 0, 0, 0
e = int(input())
if e == 0: break
for i in range(len(cubics)):
if e - cubics[i] < 0:
e -= cubics[i - 1]
z = i - 1
break
for i in range(len(squares)):
if e - squares[i] < 0:
e -= squares[i -1]
y = i - 1
break
x = e
print(x + y + z)
|
s062249509
|
Accepted
| 540
| 6,816
| 371
|
from math import floor, sqrt
cubics = [ x * x * x for x in range(101)]
squares = [ x * x for x in range(1000)]
while True:
m = 1000000
e = int(input())
if e == 0: break
for z in range(len(cubics)):
if e < cubics[z]: break
tmp = e - cubics[z]
y = floor(sqrt(tmp))
x = tmp - y ** 2
m = min(m, x + y + z)
print(m)
|
s685719575
|
p03557
|
u790710233
| 2,000
| 262,144
|
Wrong Answer
| 536
| 23,360
| 300
|
The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower. He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i. To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar. How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.
|
import bisect
n = int(input())
A = sorted(map(int, input().split()))
B = sorted(map(int, input().split()))
C = sorted(map(int, input().split()))
ans = 0
for i in range(n):
b = B[i]
a = bisect.bisect_left(A, b)
c = n-bisect.bisect_right(C, b)
print(a, b, c)
ans += a*c
print(ans)
|
s138808544
|
Accepted
| 332
| 23,232
| 280
|
import bisect
n = int(input())
A = sorted(map(int, input().split()))
B = sorted(map(int, input().split()))
C = sorted(map(int, input().split()))
ans = 0
for i in range(n):
b = B[i]
a = bisect.bisect_left(A, b)
c = n-bisect.bisect_right(C, b)
ans += a*c
print(ans)
|
s807969227
|
p03759
|
u278143034
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,024
| 166
|
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a,b,c = map(int,input().split())
if b - a == b - c:
print("YES")
else:
print("NO")
|
s077426944
|
Accepted
| 31
| 9,160
| 166
|
a,b,c = map(int,input().split())
if b - a == c - b:
print("YES")
else:
print("NO")
|
s543982215
|
p02613
|
u975243484
| 2,000
| 1,048,576
|
Wrong Answer
| 150
| 16,300
| 224
|
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n = int(input())
s = []
for _ in range(n):
s1 = input()
s.append(s1)
print('AC × ' + str(s.count('AC')))
print('WA × ' + str(s.count('WA')))
print('TLE × ' + str(s.count('TLE')))
print('RE × ' + str(s.count('RE')))
|
s727058582
|
Accepted
| 151
| 16,220
| 220
|
n = int(input())
s = []
for _ in range(n):
s1 = input()
s.append(s1)
print('AC x ' + str(s.count('AC')))
print('WA x ' + str(s.count('WA')))
print('TLE x ' + str(s.count('TLE')))
print('RE x ' + str(s.count('RE')))
|
s594799882
|
p03377
|
u505420467
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 53
|
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());print("YNEOS"[a+b<=x])
|
s334559959
|
Accepted
| 18
| 2,940
| 63
|
a,b,x=map(int,input().split());print("YNEOS"[a+b<x or a>x::2])
|
s095770283
|
p02613
|
u635252313
| 2,000
| 1,048,576
|
Wrong Answer
| 151
| 16,120
| 144
|
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())
lst=[]
all=["AC","WA","TLE","RE"]
for _ in range(n):
a=input()
lst.append(a)
for i in all:
print(i,"×",lst.count(i))
|
s658535596
|
Accepted
| 139
| 16,224
| 132
|
N = int(input())
lst = [input() for _ in range(N)]
V=["AC","WA","TLE","RE"]
for v in V:
print("{} x {}".format(v, lst.count(v)))
|
s322121717
|
p03796
|
u248670337
| 2,000
| 262,144
|
Wrong Answer
| 32
| 2,940
| 74
|
Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7.
|
n=int(input())
ans=1
for i in range(n):
ans*=i
ans%=10**9+7
print(ans)
|
s861060077
|
Accepted
| 44
| 2,940
| 78
|
n=int(input())
ans=1
for i in range(1,n+1):
ans*=i
ans%=10**9+7
print(ans)
|
s177170439
|
p03473
|
u477696265
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,060
| 37
|
How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?
|
a = input()
b = int(a) + 24
print(b)
|
s319977549
|
Accepted
| 27
| 9,040
| 49
|
a = input()
b = 24 - int(a)
c = 24 + b
print(c)
|
s310850676
|
p04011
|
u502149531
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 127
|
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
a = int(input())
b = int(input())
c = int(input())
d = int(input())
if a >= b :
print(c*a)
else :
print((a-b)*c + b*d)
|
s010712724
|
Accepted
| 17
| 2,940
| 128
|
a = int(input())
b = int(input())
c = int(input())
d = int(input())
if a <= b :
print(c*a)
else :
print(b*c + (a-b)*d)
|
s458261638
|
p03089
|
u987164499
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 256
|
Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
|
from sys import stdin
n = int(stdin.readline().rstrip())
li = list(map(int,stdin.readline().rstrip().split()))
lin = []
for i in range(n):
if i+1 >= li[i]:
lin.append(li[i])
else:
print(-1)
exit()
for i in lin:
print(i)
|
s070701489
|
Accepted
| 18
| 3,064
| 376
|
from sys import stdin
from itertools import groupby
n = int(stdin.readline().rstrip())
li = list(map(int,stdin.readline().rstrip().split()))
lin = []
while li:
now = 10**10
for i,j in enumerate(li,1):
if i == j:
now = i
if now == 10**10:
print(-1)
exit()
lin.append(now)
li.pop(now-1)
for i in lin[::-1]:
print(i)
|
s737846063
|
p03623
|
u953379577
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 62
|
Snuke lives at position x on a number line. On this line, there are two stores A and B, respectively at position a and b, that offer food for delivery. Snuke decided to get food delivery from the closer of stores A and B. Find out which store is closer to Snuke's residence. Here, the distance between two points s and t on a number line is represented by |s-t|.
|
x,a,b = map(int,input().split())
print(min(abs(a-x),abs(b-x)))
|
s801055822
|
Accepted
| 17
| 2,940
| 91
|
x,a,b = map(int,input().split())
if abs(a-x)<abs(b-x):
print("A")
else:
print("B")
|
s472806071
|
p03860
|
u425448230
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,064
| 37
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
# A
s = input()
print("A", s[8], "C")
|
s299841647
|
Accepted
| 22
| 3,064
| 37
|
# A
s = input()
print("A"+ s[8]+ "C")
|
s717487102
|
p03693
|
u626467464
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 107
|
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())
num = 100*r + 10*g + b
if num % 4 == 0:
print("Yes")
else:
print("No")
|
s205911693
|
Accepted
| 18
| 2,940
| 107
|
r,g,b = map(int,input().split())
num = 100*r + 10*g + b
if num % 4 == 0:
print("YES")
else:
print("NO")
|
s215503984
|
p00101
|
u555040407
| 1,000
| 131,072
|
Wrong Answer
| 70
| 5,780
| 906
|
An English booklet has been created for publicizing Aizu to the world. When you read it carefully, you found a misnomer (an error in writing) on the last name of Masayuki Hoshina, the lord of the Aizu domain. The booklet says "Hoshino" not "Hoshina". Your task is to write a program which replace all the words "Hoshino" with "Hoshina". You can assume that the number of characters in a text is less than or equal to 1000.
|
n = int(input())
for __ in range(n):
sentence = input()
length = 0
after = ""
while True:
try:
sentence[length]
length += 1
except:
break
for i in range(length-6):
if sentence[i] == "H":
if sentence[i+1] == "o":
if sentence[i+2] == "s":
if sentence[i+3] == "h":
if sentence[i+4] == "i":
if sentence[i+5] == "n":
if sentence[i+6] == "o":
for j in range(i+6):
after += sentence[j]
after += "a"
for j in range(i+7,length):
after += sentence[j]
print(after)
|
s395848632
|
Accepted
| 70
| 5,604
| 869
|
def get_length(sentence):
length = 0
while True:
try:
if sentence[length] is not None:
length += 1
except:
break
return length
n = int(input())
target = "Hoshino"
target_right = "Hoshina"
length_target = get_length(target)
for __ in range(n):
sentence = input()
length = get_length(sentence)
for i in range(length - length_target +1):
correct = True
for j in range(length_target):
if sentence[i+j] != target[j]:
correct = False
break
if correct:
after = ""
for j in range(i):
after += sentence[j]
after += target_right
for j in range(i+length_target, length):
after += sentence[j]
sentence = after
print(sentence)
|
s010916811
|
p03110
|
u616025987
| 2,000
| 1,048,576
|
Wrong Answer
| 19
| 3,188
| 202
|
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
import re
otoshidama = 0
N = int(input())
for i in range(0,N):
ni = re.split('\s+', input())
if(ni[1] == 'BTC'):
ni[0] = float(ni[0]) * 38000
otoshidama += float(ni[0])
print(otoshidama)
|
s254547533
|
Accepted
| 19
| 3,188
| 222
|
import re
otoshidama = 0
N = int(input())
for i in range(0,N):
ni = re.split('\s+', input())
if(ni[1] == 'BTC'):
otoshidama += (float(ni[0]) * 380000.00)
else:
otoshidama += int(ni[0])
print(otoshidama)
|
s627540413
|
p03815
|
u095969144
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 54
|
Snuke has decided to play with a six-sided die. Each of its six sides shows an integer 1 through 6, and two numbers on opposite sides always add up to 7. Snuke will first put the die on the table with an arbitrary side facing upward, then repeatedly perform the following operation: * Operation: Rotate the die 90° toward one of the following directions: left, right, front (the die will come closer) and back (the die will go farther). Then, obtain y points where y is the number written in the side facing upward. For example, let us consider the situation where the side showing 1 faces upward, the near side shows 5 and the right side shows 4, as illustrated in the figure. If the die is rotated toward the right as shown in the figure, the side showing 3 will face upward. Besides, the side showing 4 will face upward if the die is rotated toward the left, the side showing 2 will face upward if the die is rotated toward the front, and the side showing 5 will face upward if the die is rotated toward the back. Find the minimum number of operation Snuke needs to perform in order to score at least x points in total.
|
import math
i = int(input())
print(math.ceil(i / 11))
|
s186175014
|
Accepted
| 17
| 2,940
| 120
|
i = int(input())
x = i // 11
a = x * 2
b = i % 11
if b == 0:
pass
elif b <= 6:
a += 1
else:
a += 2
print(a)
|
s030154348
|
p03455
|
u393558653
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 168
|
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a,b = map(int,input().split())
c = a * b % 2
if (c == 0):
print("even")
else:
print("odd")
|
s849458843
|
Accepted
| 17
| 2,940
| 168
|
a,b = map(int,input().split())
c = a * b % 2
if (c == 0):
print("Even")
else:
print("Odd")
|
s097539616
|
p03456
|
u371763408
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 121
|
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
import math
a,b = map(int,input().split())
num = int(a+b)
if math.sqrt(num) % 1 == 0:
print('Yes')
else:
print('No')
|
s792658509
|
Accepted
| 20
| 3,316
| 113
|
import math
a,b = input().split()
num = int(a+b)
if math.sqrt(num) % 1 == 0:
print('Yes')
else:
print('No')
|
s547418674
|
p02613
|
u498531348
| 2,000
| 1,048,576
|
Wrong Answer
| 165
| 16,324
| 479
|
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.
|
IN = [input() for i in range(1)]
N = int(IN[0])
INP = [input() for i in range(int(N))]
AC = 0
WA = 0
TLE = 0
RE = 0
i = 0
while i < N:
if INP[i] == "AC":
AC = AC + 1
i = i + 1
elif INP[i] == "WA":
WA = WA + 1
i = i + 1
elif INP[i] == "TLE":
TLE = TLE + 1
i = i + 1
elif INP[i] == "RE":
RE = RE + 1
i = i + 1
else:
i = i + 1
print("AC × " + str(AC))
print("WA × " + str(WA))
print("TLE × " + str(TLE))
print("RE × " + str(RE))
|
s945963443
|
Accepted
| 160
| 15,964
| 475
|
IN = [input() for i in range(1)]
N = int(IN[0])
INP = [input() for i in range(int(N))]
AC = 0
WA = 0
TLE = 0
RE = 0
i = 0
while i < N:
if INP[i] == "AC":
AC = AC + 1
i = i + 1
elif INP[i] == "WA":
WA = WA + 1
i = i + 1
elif INP[i] == "TLE":
TLE = TLE + 1
i = i + 1
elif INP[i] == "RE":
RE = RE + 1
i = i + 1
else:
i = i + 1
print("AC x " + str(AC))
print("WA x " + str(WA))
print("TLE x " + str(TLE))
print("RE x " + str(RE))
|
s061673970
|
p03448
|
u078944864
| 2,000
| 262,144
|
Wrong Answer
| 984
| 3,064
| 319
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
a = int(input())
b = int(input())
c = int(input())
total = int(input())
max_a = total // 500
max_b = total // 100
max_c = total // 50
ans = 0
for i in range(0, max_a):
for j in range(0, max_b):
for k in range(0, max_c):
sum = 500 * i + 100 * j + 50 * k
if (sum == total) :
ans += 1
print(ans)
|
s290601951
|
Accepted
| 48
| 3,064
| 411
|
a = int(input())
b = int(input())
c = int(input())
total = int(input())
max_a = a if total // 500 > a else total // 500
max_b = b if total // 100 > b else total // 100
max_c = c if total // 50 > c else total // 50
ans = 0
for i in range(0, max_a + 1):
for j in range(0, max_b + 1):
for k in range(0, max_c + 1):
sum = 500 * i + 100 * j + 50 * k
if (sum == total) :
ans += 1
print(ans)
|
s846288686
|
p03470
|
u922449550
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 169
|
An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you have a 3-layered kagami mochi; if you put just one mochi, you have a 1-layered kagami mochi. Lunlun the dachshund has N round mochi, and the diameter of the i-th mochi is d_i centimeters. When we make a kagami mochi using some or all of them, at most how many layers can our kagami mochi have?
|
N = int(input())
D = list(map(int, input().split()))
D.sort(reverse=True)
ans = 0
pre_d = float('inf')
for d in D:
if d < pre_d:
ans += 1
pre_d = d
print(ans)
|
s638810674
|
Accepted
| 18
| 3,060
| 170
|
N = int(input())
D = [int(input()) for i in range(N)]
D.sort(reverse=True)
ans = 0
pre_d = float('inf')
for d in D:
if d < pre_d:
ans += 1
pre_d = d
print(ans)
|
s369128155
|
p03624
|
u432226259
| 2,000
| 262,144
|
Wrong Answer
| 81
| 6,376
| 255
|
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(map(str,input()))
S_new = sorted(S)
a = []
index = 0
print(S_new)
for i in range(0, len(S_new) - 1):
if ord(S_new[i + 1]) - ord(S_new[i]) > 1:
index = i
break
if index == 0:
print('None')
else:
ans = ord(S_new[index]) + 1
print(chr(ans))
|
s149736471
|
Accepted
| 73
| 5,092
| 387
|
def cal_word(n):
return n - 97
def cal_word_rev(n):
return n + 97
S = list(map(str,input()))
S_new = sorted(S)
index = -1
S_ans = []
for i in range(0, 26):
S_ans.append('false')
for i in range(0, len(S)):
S_ans[cal_word(ord(S[i]))] = 'true'
for i in range(0, 26):
if S_ans[i] == 'false':
index = i
break
if index == -1:
print('None')
else:
print(chr(cal_word_rev(index)))
|
s552491702
|
p04043
|
u874303957
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 96
|
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
ABC = list(map(int, input().split()))
print('Yes' if ABC[0] * ABC[1] * ABC[2] == 175 else 'No')
|
s320386710
|
Accepted
| 17
| 2,940
| 97
|
ABC = list(map(int, input().split()))
print('YES' if ABC[0] * ABC[1] * ABC[2] == 175 else 'NO')
|
s816379687
|
p03778
|
u451017206
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 107
|
AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W. If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle. Find the minimum distance it needs to be moved.
|
W, a, b = map(int, input().split())
if b > a + W:print(b - a + W)
elif b + W > a:print(b+W-a)
else:print(0)
|
s445790764
|
Accepted
| 17
| 2,940
| 107
|
W, a, b = map(int, input().split())
if b > a + W:print(b - a - W)
elif b + W < a:print(a-b-W)
else:print(0)
|
s829792448
|
p03645
|
u986254798
| 2,000
| 262,144
|
Wrong Answer
| 865
| 49,748
| 574
|
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.
|
from queue import Queue
n, m = map(int, input().split())
adj = [[] for i in range(n+1)]
for i in range(m):
a, b = map(int, input().split())
adj[a].append(b)
adj[b].append(a)
def bfs():
queue = Queue()
queue.put((1, 0))
visited = [False]*(n+1)
while not queue.empty():
u, step = queue.get()
for a in adj[u]:
if a == n:
print ("Possible")
return
if step + 1 < 2 and not visited[a]:
queue.put((a, step+1))
print ("IMPOSSIBLE")
return
bfs()
|
s151488085
|
Accepted
| 838
| 43,676
| 638
|
from queue import Queue
n, m = map(int, input().split())
adj = [[] for i in range(n+1)]
for i in range(m):
a, b = map(int, input().split())
adj[a].append(b)
adj[b].append(a)
def bfs():
queue = Queue()
queue.put(1)
distance = [10**6]*(n+1)
distance[1] = 0
while not queue.empty():
u = queue.get()
if u == n and distance[u] == 2:
print ("POSSIBLE")
return
for a in adj[u]:
if distance[u] + 1 <= 2 and distance[a] > distance[u] + 1:
distance[a] = distance[u] + 1
queue.put(a)
print ("IMPOSSIBLE")
return
bfs()
|
s346988796
|
p02612
|
u104005543
| 2,000
| 1,048,576
|
Wrong Answer
| 22
| 9,140
| 32
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
print(n % 1000)
|
s871540356
|
Accepted
| 29
| 9,056
| 34
|
n = int(input())
print(-n % 1000)
|
s144377275
|
p03697
|
u163320134
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 78
|
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())
if a+b>=10:
print('error')
else:
print('a+b')
|
s186044049
|
Accepted
| 18
| 2,940
| 76
|
a,b=map(int,input().split())
if a+b>=10:
print('error')
else:
print(a+b)
|
s653910509
|
p02407
|
u426534722
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,572
| 42
|
Write a program which reads a sequence and prints it in the reverse order.
|
print(*sorted(map(int, input().split())))
|
s093010712
|
Accepted
| 20
| 5,560
| 38
|
input()
print(*input().split()[::-1])
|
s829238966
|
p03993
|
u953379577
| 2,000
| 262,144
|
Wrong Answer
| 72
| 20,548
| 131
|
There are N rabbits, numbered 1 through N. The i-th (1≤i≤N) rabbit likes rabbit a_i. Note that no rabbit can like itself, that is, a_i≠i. For a pair of rabbits i and j (i＜j), we call the pair (i，j) a _friendly pair_ if the following condition is met. * Rabbit i likes rabbit j and rabbit j likes rabbit i. Calculate the number of the friendly pairs.
|
n = int(input())
s = list(map(int,input().split()))
ans = 0
for i in range(n):
if s[s[i]-1] == i+1:
ans += 1
print(ans)
|
s070471860
|
Accepted
| 70
| 20,508
| 135
|
n = int(input())
s = list(map(int,input().split()))
ans = 0
for i in range(n):
if s[s[i]-1] == i+1:
ans += 1
print(ans//2)
|
s443448327
|
p03494
|
u360842608
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 335
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
def answer(in0, in1):
N = int(in0)
A = [int(a) for a in in1.split()]
print(N)
print(A)
cnt = 0
while (True):
for i in range(N):
a = A[i]
if a % 2 != 0:
print(cnt)
return
A[i] = int(a / 2)
cnt += 1
answer(input(), input())
|
s058505069
|
Accepted
| 18
| 2,940
| 309
|
def answer(in0, in1):
N = int(in0)
A = [int(a) for a in in1.split()]
cnt = 0
while (True):
for i in range(N):
a = A[i]
if a % 2 != 0:
print(cnt)
return
A[i] = int(a / 2)
cnt += 1
answer(input(), input())
|
s132033728
|
p03543
|
u897436032
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 51
|
We call a 4-digit integer with three or more consecutive same digits, such as 1118, **good**. You are given a 4-digit integer N. Answer the question: Is N **good**?
|
input_ = input().split(",")
print(" ".join(input_))
|
s439821954
|
Accepted
| 17
| 2,940
| 75
|
a,b,c,d = input()
if a==b==c or b==c==d:
print("Yes")
else:
print("No")
|
s743114213
|
p04012
|
u053856575
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 233
|
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.
|
W=input()
count=1
w=[]
for j in range(len(W)):
w.append(W[j])
w.sort()
for i in range(len(w)-1):
if w[i]==w[i+1]:
count+=1
else:
if count %2 !=0:
print("No")
exit()
print("Yes")
|
s678045949
|
Accepted
| 18
| 3,064
| 281
|
W=input()
count=1
w=[]
for j in range(len(W)):
w.append(W[j])
w.sort()
w.append("")
for i in range(len(w)-1):
if w[i]==w[i+1]:
count+=1
else:
if count %2 !=0:
print("No")
exit()
else:
count=1
print("Yes")
|
s095017030
|
p02748
|
u857070771
| 2,000
| 1,048,576
|
Wrong Answer
| 354
| 18,608
| 267
|
You are visiting a large electronics store to buy a refrigerator and a microwave. The store sells A kinds of refrigerators and B kinds of microwaves. The i-th refrigerator ( 1 \le i \le A ) is sold at a_i yen (the currency of Japan), and the j-th microwave ( 1 \le j \le B ) is sold at b_j yen. You have M discount tickets. With the i-th ticket ( 1 \le i \le M ), you can get a discount of c_i yen from the total price when buying the x_i-th refrigerator and the y_i-th microwave together. Only one ticket can be used at a time. You are planning to buy one refrigerator and one microwave. Find the minimum amount of money required.
|
a,b,m=map(int,input().split())
*a,=map(int,input().split())
*b,=map(int,input().split())
mi=10**9
for i in range(m):
x,y,c=map(int,input().split())
if len(a)<=x and len(b)<=y:
res=a[x-1]+b[y-1]-c
mi=min(mi,res)
l=min(a)+min(b)
print(min(mi,l))
|
s115929483
|
Accepted
| 467
| 18,648
| 271
|
a,b,m=map(int,input().split())
*a,=map(int,input().split())
*b,=map(int,input().split())
ans=10**9
for i in range(m):
x,y,c=map(int,input().split())
if len(a)>=x and len(b)>=y:
res=a[x-1]+b[y-1]-c
ans=min(ans,res)
l=min(a)+min(b)
print(min(ans,l))
|
s296801527
|
p03680
|
u870297120
| 2,000
| 262,144
|
Wrong Answer
| 192
| 7,072
| 172
|
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())
l = list(int(input()) for _ in range(n))
num = 1
ans = -1
for i in range(n):
num = l[num-1]
if num == 2:
ans = i
break
print(ans)
|
s310318142
|
Accepted
| 195
| 7,072
| 172
|
n = int(input())
l = list(int(input()) for _ in range(n))
num = 1
ans = -1
for i in range(n):
if num == 2:
ans = i
break
num = l[num-1]
print(ans)
|
s702667021
|
p03827
|
u111365959
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 104
|
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())
a = input()
x = 0
for b in a:
x += int(b.replace('I','1').replace('D','-1'))
print(x)
|
s392478253
|
Accepted
| 17
| 2,940
| 134
|
input()
tmp = x = 0
for a in input():
x += int(a.replace('I','1').replace('D','-1'))
if tmp < x:
tmp = x
print(max([0,tmp]))
|
s055716435
|
p03448
|
u633914031
| 2,000
| 262,144
|
Wrong Answer
| 161
| 4,080
| 260
|
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
A=int(input())
B=int(input())
C=int(input())
X=int(input())
cnt=0
i=0
j=0
k=0
while i <= A:
while j <= B:
while k <= C:
sum=500*i+100*j+50*k
print(sum)
if sum== X:
cnt+=1
k+=1
j+=1
k=0
i+=1
j=0
k=0
print(cnt)
|
s648930712
|
Accepted
| 65
| 3,064
| 243
|
A=int(input())
B=int(input())
C=int(input())
X=int(input())
cnt=0
i=0
j=0
k=0
while i <= A:
while j <= B:
while k <= C:
sum=500*i+100*j+50*k
if sum== X:
cnt+=1
k+=1
j+=1
k=0
i+=1
j=0
k=0
print(cnt)
|
s905156835
|
p03605
|
u952130512
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 73
|
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
s=input()
if s[:1]==9 or s[-1:]==9:
print("Yes")
else:
print("No")
|
s453901488
|
Accepted
| 17
| 2,940
| 74
|
s=input()
if s[0]=="9" or s[1]=="9":
print("Yes")
else:
print("No")
|
s668086654
|
p03079
|
u233729690
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 105
|
You are given three integers A, B and C. Determine if there exists an equilateral triangle whose sides have lengths A, B and C.
|
A,B,C = map(int, input().split())
print(A,B,C)
if A == B and B == C:
print('Yes')
else:
print('No')
|
s415824197
|
Accepted
| 17
| 2,940
| 91
|
A,B,C = map(int, input().split())
if A == B and B == C:
print('Yes')
else:
print('No')
|
s461366973
|
p03399
|
u182249053
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 316
|
You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses.
|
A = int(input("A = "))
B = int(input("B = "))
C = int(input("C = "))
D = int(input("D = "))
if A > B:
if C > D:
print("最小値 = %d" % (B+D))
else:
print("最小値 = %d" % (B+C))
else:
if C > D:
print("最小値 = %d" % (A+D))
else:
print("最小値 = %d" % (A+C))
|
s201104538
|
Accepted
| 17
| 3,060
| 244
|
A = int(input())
B = int(input())
C = int(input())
D = int(input())
if A > B:
if C > D:
print("%d" % (B+D))
else:
print("%d" % (B+C))
else:
if C > D:
print("%d" % (A+D))
else:
print("%d" % (A+C))
|
s661746804
|
p03371
|
u863044225
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 127
|
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
a,b,c,x,y=map(int,input().split())
c=min(c*2,a+b)
ans=min(x,y)*c
if x>y:ans+=min(a,c)*(y-x)
else:ans+=min(b,c)*(x-y)
print(ans)
|
s781513678
|
Accepted
| 18
| 3,060
| 127
|
a,b,c,x,y=map(int,input().split())
c=min(c*2,a+b)
ans=min(x,y)*c
if x>y:ans+=min(a,c)*(x-y)
else:ans+=min(b,c)*(y-x)
print(ans)
|
s834367046
|
p02612
|
u274615057
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,084
| 94
|
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.
|
def main():
n = int(input())
print(n % 1000)
if __name__ == "__main__":
main()
|
s968005646
|
Accepted
| 27
| 9,164
| 154
|
def main():
n = int(input())
if n % 1000 != 0:
print(1000 - n % 1000)
else:
print(0)
if __name__ == "__main__":
main()
|
s882482375
|
p03369
|
u811730180
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 97
|
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()
Ss = [S[0],S[1],S[2]]
Ss_extract = [x for x in Ss if x == "o"]
print(len(Ss_extract))
|
s770127936
|
Accepted
| 17
| 2,940
| 107
|
S = input()
Ss = [S[0],S[1],S[2]]
Ss_extract = [x for x in Ss if x == "o"]
print(700 + len(Ss_extract)*100)
|
s359500723
|
p02259
|
u825178626
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,484
| 404
|
Write a program of the Bubble Sort algorithm which sorts a sequence _A_ in ascending order. The algorithm should be based on the following pseudocode: BubbleSort(A) 1 for i = 0 to A.length-1 2 for j = A.length-1 downto i+1 3 if A[j] < A[j-1] 4 swap A[j] and A[j-1] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 4 of the pseudocode.
|
# coding: utf-8
# Here your code !
S = int(input())
N = list(map(int,input().split()))
count=0
def bubblesort(n,s):
global count
flag = 1
while flag:
flag = 0
for i in range(s-1,0,-1):
key = n[i]
if n[i]<n[i-1]:
n[i]=n[i-1]
n[i-1]=key
count+=1
flag=1
bubblesort(N,S)
print(count)
|
s398659786
|
Accepted
| 20
| 7,636
| 446
|
# coding: utf-8
# Here your code !
S = int(input())
N = list(map(int,input().split()))
count=0
def bubblesort(n,s):
global count
flag = 1
while flag:
flag = 0
for i in range(s-1,0,-1):
key = n[i]
if n[i]<n[i-1]:
n[i]=n[i-1]
n[i-1]=key
count+=1
flag=1
print(" ".join(list(map(str,n))))
print(count)
bubblesort(N,S)
|
s532059283
|
p03861
|
u854093727
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 61
|
You are given nonnegative integers a and b (a ≤ b), and a positive integer x. Among the integers between a and b, inclusive, how many are divisible by x?
|
a,b,c = map(int,input().split())
print(int(b/c - ((a-1)/c)))
|
s443981473
|
Accepted
| 17
| 2,940
| 63
|
a,b,c = map(int,input().split())
print(int(b//c - ((a-1)//c)))
|
s863313216
|
p03573
|
u550535134
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers.
|
A, B, C = map(int, input().split())
print("A" if B == C else "B" if A == C else "C")
|
s926205984
|
Accepted
| 18
| 2,940
| 79
|
A, B, C = map(int, input().split())
print(A if B == C else B if A == C else C)
|
s303395059
|
p02458
|
u847467233
| 4,000
| 262,144
|
Wrong Answer
| 30
| 5,628
| 628
|
For a set $S$ of integers, perform a sequence of the following operations. Note that _multiple elements can have equivalent values in $S$_. * insert($x$): Insert $x$ to $S$ and report the number of elements in $S$ after the operation. * find($x$): Report the number of $x$ in $S$. * delete($x$): Delete all $x$ from $S$. * dump($L$, $R$): Print elements $x$ in $S$ such that $L \leq x \leq R$.
|
# AOJ ITP2_7_D: Multi-Set
# Python3 2018.6.24 bal4u
from bisect import bisect_left, bisect_right, insort_left
dict = {}
keytbl = []
cnt = 0
q = int(input())
for i in range(q):
a = list(input().split())
ki = int(a[1])
if a[0] == '0':
if ki not in dict:
dict[ki] = 1
insort_left(keytbl, ki)
else: dict[ki] += 1
cnt += 1
print(cnt)
elif a[0] == '1': print(dict[ki] if ki in dict else 0)
elif a[0] == '2':
if ki in dict:
cnt -= dict[ki]
dict[ki] = 0
else:
L = bisect_left (keytbl, int(a[1]))
R = bisect_right(keytbl, int(a[2]), L)
for j in range(L, R):
if dict[keytbl[j]] > 0: print(keytbl[j])
|
s921684839
|
Accepted
| 3,420
| 17,908
| 660
|
# AOJ ITP2_7_D: Multi-Set
# Python3 2018.6.24 bal4u
from bisect import bisect_left, bisect_right, insort_left
dict = {}
keytbl = []
cnt = 0
q = int(input())
for i in range(q):
a = list(input().split())
ki = int(a[1])
if a[0] == '0':
if ki not in dict:
dict[ki] = 1
insort_left(keytbl, ki)
else: dict[ki] += 1
cnt += 1
print(cnt)
elif a[0] == '1': print(dict[ki] if ki in dict else 0)
elif a[0] == '2':
if ki in dict:
cnt -= dict[ki]
dict[ki] = 0
else:
L = bisect_left (keytbl, int(a[1]))
R = bisect_right(keytbl, int(a[2]), L)
r = []
for j in range(L, R): r += [keytbl[j]]*dict[keytbl[j]]
if r != []: print(*r, sep='\n')
|
s116682179
|
p03089
|
u797664206
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 311
|
Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
|
n = int(input())
l = [int(i) for i in input().split()]
a = [0]*n
o = 0
for i in l:
o+=1
if(i<=o):
if(i<=len(a)):
if(a[i-1]!=0):
a[a.index(0)] = i
else:
a[i-1] = i
if(a == l):
print("\n".join([str(i) for i in a]))
else:
print(-1)
|
s376249742
|
Accepted
| 18
| 3,060
| 299
|
n = int(input())
l = input().split()
res = []
for i in range(n):
for j in range(len(l)-1,-1,-1):
if(l[j] == str(j+1)):
res.append(l[j])
del(l[j])
break
if(l != []):
print(-1)
else:
print("\n".join(res[::-1]))
|
s087120649
|
p02600
|
u961683878
| 2,000
| 1,048,576
|
Time Limit Exceeded
| 2,222
| 520,056
| 1,123
|
M-kun is a competitor in AtCoder, whose highest rating is X. In this site, a competitor is given a _kyu_ (class) according to his/her highest rating. For ratings from 400 through 1999, the following kyus are given: * From 400 through 599: 8-kyu * From 600 through 799: 7-kyu * From 800 through 999: 6-kyu * From 1000 through 1199: 5-kyu * From 1200 through 1399: 4-kyu * From 1400 through 1599: 3-kyu * From 1600 through 1799: 2-kyu * From 1800 through 1999: 1-kyu What kyu does M-kun have?
|
#! /usr/bin/env python3
import sys
int1 = lambda x: int(x) - 1
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(500000)
N = int(readline())
XYP = [list(map(int, readline().split())) for _ in range(N)]
x_cost = [[0 for _ in range(N)] for _ in range(1 << N)]
y_cost = [[0 for _ in range(N)] for _ in range(1 << N)]
for bit in range(1 << N):
for i in range(N):
x_cost[bit][i] = abs(XYP[i][0])
y_cost[bit][i] = abs(XYP[i][1])
for k in range(N):
if bit & (1 << k):
x_cost[bit][i] = min(x_cost[bit][i], abs(XYP[i][0] - XYP[k][0]))
y_cost[bit][i] = min(y_cost[bit][i], abs(XYP[i][1] - XYP[k][1]))
ans = [sys.maxsize for _ in range(N + 1)]
for bit in range(1 << N):
cnt = bin(bit).count('1')
bit_ = bit
while bit_ >= 0:
bit_ &= bit
cost = 0
for k in range(N):
cost += 1 * XYP[k][2] * min(x_cost[bit_][k], y_cost[bit - bit_][k])
ans[cnt] = min(ans[cnt], cost)
bit_ -= 1
for i in range(N + 1):
print(ans[i])
|
s258713616
|
Accepted
| 126
| 27,148
| 467
|
#! /usr/bin/env python3
import sys
import numpy as np
int1 = lambda x: int(x) - 1
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(500000)
X = int(readline())
if X <= 599:
print(8)
elif X <= 799:
print(7)
elif X <= 999:
print(6)
elif X <= 1199:
print(5)
elif X <= 1399:
print(4)
elif X <= 1599:
print(3)
elif X <= 1799:
print(2)
elif X <= 1999:
print(1)
|
s607780934
|
p02613
|
u202145645
| 2,000
| 1,048,576
|
Wrong Answer
| 155
| 16,640
| 266
|
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.
|
from collections import Counter
N = int(input())
list_s = []
for i in range(N):
S = input()
list_s.append(S)
D = Counter(list_s)
print("AC ｘ " + str(D["AC"]))
print("WA ｘ " + str(D["WA"]))
print("TLE ｘ " + str(D["TLE"]))
print("RE ｘ " + str(D["RE"]))
|
s618203079
|
Accepted
| 152
| 16,624
| 262
|
from collections import Counter
N = int(input())
list_s = []
for i in range(N):
S = input()
list_s.append(S)
D = Counter(list_s)
print("AC x " + str(D["AC"]))
print("WA x " + str(D["WA"]))
print("TLE x " + str(D["TLE"]))
print("RE x " + str(D["RE"]))
|
s911520655
|
p02972
|
u726615467
| 2,000
| 1,048,576
|
Wrong Answer
| 1,269
| 10,936
| 448
|
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.
|
import sys
# input = sys.stdin.readline
inf = 10 ** 18
P = 10 ** 9 + 7
N = int(input())
a = list(map(int, input().split()))
x = [None] * N
for i_inv in range(N):
i = (N - 1) - i_inv
j = 2
xi = a[i] % 2
while (i + 1) * j < N:
xi ^= x[(i + 1) * j - 1]
j += 1
x[i] = xi
print(sum(x))
ans = []
for i, xi in enumerate(x):
if xi == 1:
ans.append(i + 1)
if sum(x) < 0:
print(' '.join(map(str, ans)))
|
s344683596
|
Accepted
| 1,349
| 19,892
| 543
|
import sys
# input = sys.stdin.readline
inf = 10 ** 18
P = 10 ** 9 + 7
N = int(input())
a = list(map(int, input().split()))
x = [ai % 2 for ai in a]
for i_inv in range(N):
i = (N - 1) - i_inv
j = 2
xi = x[i]
# print("#", i, a[i])
while (i + 1) * j - 1 < N:
# print("##", (i + 1) * j - 1, x[(i + 1) * j - 1])
xi ^= x[(i + 1) * j - 1]
j += 1
x[i] = xi
print(sum(x))
ans = []
for i, xi in enumerate(x):
if xi == 1:
ans.append(i + 1)
if sum(x) > 0:
print(' '.join(map(str, ans)))
|
s557942492
|
p03698
|
u035210736
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,024
| 78
|
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
s = list(input())
if len(s) == len(set(s)):
print("YES")
else:
print("NO")
|
s446185683
|
Accepted
| 30
| 8,988
| 79
|
s = list(input())
if len(s) == len(set(s)):
print("yes")
else:
print("no")
|
s824053922
|
p02615
|
u616382321
| 2,000
| 1,048,576
|
Wrong Answer
| 131
| 31,432
| 110
|
Quickly after finishing the tutorial of the online game _ATChat_ , you have decided to visit a particular place with N-1 players who happen to be there. These N players, including you, are numbered 1 through N, and the **friendliness** of Player i is A_i. The N players will arrive at the place one by one in some order. To make sure nobody gets lost, you have set the following rule: players who have already arrived there should form a circle, and a player who has just arrived there should cut into the circle somewhere. When each player, except the first one to arrive, arrives at the place, the player gets **comfort** equal to the smaller of the friendliness of the clockwise adjacent player and that of the counter-clockwise adjacent player. The first player to arrive there gets the comfort of 0. What is the maximum total comfort the N players can get by optimally choosing the order of arrivals and the positions in the circle to cut into?
|
N = int(input())
A = list(map(int, input().split()))
A.sort(reverse=True)
print(A)
B = A[:-1]
print(sum(B))
|
s872179188
|
Accepted
| 106
| 31,428
| 180
|
N = int(input())
A = list(map(int, input().split()))
A.sort(reverse=True)
if N%2 == 0:
print(sum(A[:int(N/2)])*2 - A[0])
else:
print(sum(A[:N//2+1])*2 - A[0] - A[N//2])
|
s752391113
|
p03485
|
u923010184
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 58
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a,b = map(int,input().split())
print(int(a / 2 + b /2) +1)
|
s540811484
|
Accepted
| 17
| 2,940
| 55
|
a,b = map(int,input().split())
print(((a + b +1) // 2))
|
s686654952
|
p03657
|
u241159583
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 127
|
Snuke is giving cookies to his three goats. He has two cookie tins. One contains A cookies, and the other contains B cookies. He can thus give A cookies, B cookies or A+B cookies to his goats (he cannot open the tins). Your task is to determine whether Snuke can give cookies to his three goats so that each of them can have the same number of cookies.
|
a, b = map(int, input().split())
if a + b % 3 == 0 or a % 3 == 0 or b % 3 == 0:
print("Possible")
else:
print("Impossible")
|
s898167401
|
Accepted
| 18
| 2,940
| 101
|
a,b = map(int, input().split())
print("Possible" if a%3==0 or b%3==0 or (a+b)%3==0 else "Impossible")
|
s039914955
|
p03994
|
u100927237
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 109,836
| 363
|
Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times. * Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`. For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`. Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.
|
makeA = lambda c: (ord('z')-ord(c)+1) % 26
s = list(map(makeA,list(input())))
print(s)
K = int(input())
len_s = len(s)
print(len_s)
for i in range(len_s-1):
if s[i] <= K:
K -= s[i]
s[i] = 0
print(s)
if K > 0:
s[-1] = (s[-1] + 26 - (K % 26)) % 26
print(s)
makeS = lambda n: chr((26-n)%26+ord('a'))
print(''.join(list(map(makeS,s))))
|
s033365469
|
Accepted
| 97
| 4,836
| 369
|
makeA = lambda c: (ord('z')-ord(c)+1) % 26
s = list(map(makeA,list(input())))
#print(s)
K = int(input())
len_s = len(s)
#print(len_s)
for i in range(len_s-1):
if s[i] <= K:
K -= s[i]
s[i] = 0
#print(s)
if K > 0:
s[-1] = (s[-1] + 26 - (K % 26)) % 26
#print(s)
makeS = lambda n: chr((26-n)%26+ord('a'))
print(''.join(list(map(makeS,s))))
|
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