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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s246605963
|
p03469
|
u462214100
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 33
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
s = input()
print('2018' + s[:4])
|
s180620982
|
Accepted
| 17
| 2,940
| 38
|
s = input()
print('2018/01/' + s[-2:])
|
s028590881
|
p02608
|
u699944218
| 2,000
| 1,048,576
|
Wrong Answer
| 2,206
| 8,796
| 307
|
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
import math
N = int(input())
n = int(math.sqrt(N))
ans = [0 for _ in range(N)]
for i in range(N):
for x in range(n):
for y in range(n):
for z in range(n):
if x ** 2 + y ** 2 + z ** 2 + x * y + y * z + z * y == N:
ans[i] += 1
for i in range(N):
print(ans[i])
|
s494238630
|
Accepted
| 800
| 9,244
| 279
|
N = int(input())
ans = [0 for i in range(10100)]
for x in range(1,100):
for y in range(1,100):
for z in range(1,100):
n = x ** 2 + y ** 2 + z ** 2 + x * y + y * z + z * x
if n < 10100:
ans[n] += 1
for n in range(N):
print(ans[n+1])
|
s069904330
|
p02612
|
u854294899
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 9,128
| 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)
|
s956955132
|
Accepted
| 31
| 9,148
| 75
|
n = int(input())
res = n % 1000
if (res > 0):
res = 1000 - res
print(res)
|
s037953869
|
p03377
|
u439392790
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 80
|
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<=B:
print('Yes')
else:
print('No')
|
s815269560
|
Accepted
| 17
| 2,940
| 90
|
A,B,X=map(int,input().split())
if X-A<=B and A<=X:
print('YES')
else:
print('NO')
|
s527550948
|
p03369
|
u486302338
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 68
|
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()
cnt = 0
for s in S:
cnt += 1
print(700 + 100 * cnt)
|
s572596373
|
Accepted
| 17
| 2,940
| 83
|
S = input()
cnt = 0
for s in S:
if s == 'o':
cnt += 1
print(700 + 100 * cnt)
|
s753290467
|
p03469
|
u735008991
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 28
|
On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in `yyyy/mm/dd` format. For example, January 23, 2018 should be written as `2018/01/23`. After finishing the document, she noticed that she had mistakenly wrote `2017` at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to `2018` and prints it.
|
print('2018' + input()[:4])
|
s723290732
|
Accepted
| 17
| 2,940
| 28
|
print('2018' + input()[4:])
|
s499864404
|
p03657
|
u674934587
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,016
| 95
|
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)%2==0:
print("Possible")
else:
print("Impossible")
|
s203520606
|
Accepted
| 27
| 9,076
| 111
|
a,b=map(int,input().split())
if a%3==0 or b%3==0 or (a+b)%3==0:
print("Possible")
else:
print("Impossible")
|
s622538910
|
p03845
|
u617515020
| 2,000
| 262,144
|
Wrong Answer
| 24
| 9,200
| 266
|
Joisino is about to compete in the final round of a certain programming competition. In this contest, there are N problems, numbered 1 through N. Joisino knows that it takes her T_i seconds to solve problem i(1≦i≦N). Also, there are M kinds of drinks offered to the contestants, numbered 1 through M. If Joisino takes drink i(1≦i≦M), her brain will be stimulated and the time it takes for her to solve problem P_i will become X_i seconds. It does not affect the time to solve the other problems. A contestant is allowed to take exactly one of the drinks before the start of the contest. For each drink, Joisino wants to know how many seconds it takes her to solve all the problems if she takes that drink. Here, assume that the time it takes her to solve all the problems is equal to the sum of the time it takes for her to solve individual problems. Your task is to write a program to calculate it instead of her.
|
n=int(input())
t=list(map(int,input().split()))
m=int(input())
px=[]
for _ in range(m):
p,x=map(int,input().split())
px.append((p-1,x-1))
for i in range(m):
ans=0
p,x=px[i]
for j in range(n):
if j==p:
ans+=x
else:
ans+=t[j]
print(ans)
|
s451315039
|
Accepted
| 23
| 9,148
| 264
|
n=int(input())
t=list(map(int,input().split()))
m=int(input())
px=[]
for _ in range(m):
p,x=map(int,input().split())
px.append((p-1,x))
for i in range(m):
ans=0
p,x=px[i]
for j in range(n):
if j==p:
ans+=x
else:
ans+=t[j]
print(ans)
|
s939907523
|
p03080
|
u482157295
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 187
|
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.
|
import math
N = int(input())
s = input()
count = 0
for i in s:
if i == "R":
count = count + 1
if count > math.floor(N//2):
print("yes")
else:
print("No")
|
s444420214
|
Accepted
| 17
| 2,940
| 175
|
import math
N = int(input())
s = input()
count = 0
for i in s:
if i == "R":
count = count + 1
if count > N//2:
print("Yes")
else:
print("No")
|
s570408474
|
p03854
|
u448659077
| 2,000
| 262,144
|
Wrong Answer
| 82
| 3,672
| 295
|
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
#ABC049
s = input()
while len(s) != 0:
if s[-5:] == "dream":
s = s[:-5]
elif s[-7:] == "dreamer":
s = s[:-7]
elif s[-5:] == "erase":
s = s[:-5]
elif s[-6:] == "eraser":
s = s[:-6]
else:
print("NO")
break
print("YES")
|
s284831868
|
Accepted
| 68
| 3,188
| 301
|
#ABC049
s = input()
while len(s) != 0:
if s[-5:] == "dream":
s = s[:-5]
elif s[-7:] == "dreamer":
s = s[:-7]
elif s[-5:] == "erase":
s = s[:-5]
elif s[-6:] == "eraser":
s = s[:-6]
else:
print("NO")
break
else:
print("YES")
|
s725559517
|
p03474
|
u977389981
| 2,000
| 262,144
|
Wrong Answer
| 21
| 2,940
| 217
|
The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen `-`, and the other characters are digits from `0` through `9`. You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.
|
a, b = map(int, input().split())
S = input()
ans = 'Yes'
if S[a + 1] == '-':
S = S.replace('-', '', 1)
for i in S:
if not i in '1234567890':
ans = 'No'
else:
ans = 'No'
print(ans)
|
s613266311
|
Accepted
| 17
| 3,064
| 166
|
a, b = map(int, input().split())
s = input()
ans = 'No'
if s[a] == '-':
if s[: a].isdecimal() and s[a + 1 :].isdecimal():
ans = 'Yes'
print(ans)
|
s645778926
|
p03693
|
u253011685
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 85
|
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=(input().split())
N=int(r+g+b)
if(N%4==0):
print('Yes')
else:
print('No')
|
s253159569
|
Accepted
| 17
| 2,940
| 85
|
r,g,b=(input().split())
N=int(r+g+b)
if(N%4==0):
print('YES')
else:
print('NO')
|
s487232506
|
p03909
|
u021916304
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 230
|
There is a grid with H rows and W columns. The square at the i-th row and j-th column contains a string S_{i,j} of length 5. The rows are labeled with the numbers from 1 through H, and the columns are labeled with the uppercase English letters from `A` through the W-th letter of the alphabet. Exactly one of the squares in the grid contains the string `snuke`. Find this square and report its location. For example, the square at the 6-th row and 8-th column should be reported as `H6`.
|
alpha = 'abcdefghijklmnopqrstuvwxyz'
h,w = map(int,input().split())
for i in range(h):
A = list(map(str,input().split()))
for j in range(w):
if A[j] == 'snuke':
print(alpha[j]+str(i))
exit()
|
s080426480
|
Accepted
| 17
| 2,940
| 223
|
alpha = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
h,w = map(int,input().split())
for i in range(h):
A = list(map(str,input().split()))
for j in range(w):
if A[j] == 'snuke':
ans = alpha[j]+str(i+1)
print(ans)
|
s726621717
|
p03611
|
u686036872
| 2,000
| 262,144
|
Wrong Answer
| 2,104
| 13,964
| 153
|
You are given an integer sequence of length N, a_1,a_2,...,a_N. For each 1≤i≤N, you have three choices: add 1 to a_i, subtract 1 from a_i or do nothing. After these operations, you select an integer X and count the number of i such that a_i=X. Maximize this count by making optimal choices.
|
N = input()
A = list(map(int, input().split()))
ans = 0
for i in range(min(A), max(A)+1):
ans = A.count(i-1) + A.count(i) + A.count(i+1)
print(ans)
|
s351934367
|
Accepted
| 161
| 14,564
| 192
|
import collections
N = input()
A = list(map(int, input().split()))
c = collections.Counter(A)
ans = 0
for i in range(min(A), max(A)+1):
ans = max(c[i-1] + c[i] + c[i+1], ans)
print(ans)
|
s218838958
|
p03644
|
u189397279
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 200
|
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())
if 1 <= N <2:
print(0)
if 2 <= N <4:
print(1)
if 4 <= N <8:
print(2)
if 8 <= N <16:
print(3)
if 16 <= N <32:
print(4)
if 32 <= N <64:
print(5)
if 64 <= N <100:
print(6)
|
s746916741
|
Accepted
| 17
| 3,060
| 204
|
N = int(input())
if 1 <= N <2:
print(1)
if 2 <= N <4:
print(2)
if 4 <= N <8:
print(4)
if 8 <= N <16:
print(8)
if 16 <= N <32:
print(16)
if 32 <= N <64:
print(32)
if 64 <= N <= 100:
print(64)
|
s448971574
|
p03997
|
u453500284
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 57
|
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a, b, h = [int(input()) for i in range(3)]
print(a*b*h/2)
|
s656603039
|
Accepted
| 17
| 2,940
| 48
|
x = lambda:int(input())
print((x()+x())*x()//2)
|
s685076989
|
p03730
|
u640603056
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 449
|
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`.
|
import sys
def ILI(): return list(map(int, sys.stdin.readline().rstrip().split()))
def ISI(): return map(int, sys.stdin.readline().rstrip().split())
def II(): return int(sys.stdin.readline().rstrip())
def ISS(): return sys.stdin.readline().rstrip().split()
def IS(): return sys.stdin.readline().rstrip()
A, B, C = ISI()
for i in range(1, B+1):
if A*i%B == C:
print("Yes")
break
else:
print("No")
|
s889426149
|
Accepted
| 18
| 3,064
| 449
|
import sys
def ILI(): return list(map(int, sys.stdin.readline().rstrip().split()))
def ISI(): return map(int, sys.stdin.readline().rstrip().split())
def II(): return int(sys.stdin.readline().rstrip())
def ISS(): return sys.stdin.readline().rstrip().split()
def IS(): return sys.stdin.readline().rstrip()
A, B, C = ISI()
for i in range(1, B+1):
if A*i%B == C:
print("YES")
break
else:
print("NO")
|
s076207287
|
p02833
|
u487594898
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 155
|
For an integer n not less than 0, let us define f(n) as follows: * f(n) = 1 (if n < 2) * f(n) = n f(n-2) (if n \geq 2) Given is an integer N. Find the number of trailing zeros in the decimal notation of f(N).
|
N = int(input())
cnt = 0
if N%2==1:
print(0)
else :
for i in range(30):
cnt +=N//5**i
if N < 5**i:
break
print(cnt)
|
s446105776
|
Accepted
| 17
| 3,060
| 159
|
import math
N = int(input())
cnt = 0
if N%2==1:
print(0)
else :
for i in range(1,30):
n = N//2
cnt +=math.floor(n//5**i)
print(cnt)
|
s216590920
|
p04043
|
u088974156
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 71
|
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.
|
n=input().split()
print("YES" if n.count("5")==2 and ("7")==1 else"NO")
|
s757693433
|
Accepted
| 17
| 2,940
| 79
|
a=input().split()
print("YES" if a.count("5")==2 and a.count("7")==1 else "NO")
|
s885495423
|
p02613
|
u439370642
| 2,000
| 1,048,576
|
Wrong Answer
| 139
| 16,260
| 387
|
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.
|
def main():
n = int(input())
mylist = []
for i in range(1, n+1):
s = input()
mylist.append(s)
ac=mylist.count('AC')
wa=mylist.count('WA')
tle=mylist.count('TLE')
re=mylist.count('RE')
print('AC × ' + str(ac))
print('WA × ' + str(wa))
print('TLE × ' + str(tle))
print('RE × ' + str(re))
if __name__ == "__main__":
main()
|
s736979450
|
Accepted
| 152
| 16,548
| 351
|
def main():
import collections
n = int(input())
mylist = []
for i in range(n):
s = input()
mylist.append(s)
c = collections.Counter(mylist)
print('AC x ' + str(c['AC']))
print('WA x ' + str(c['WA']))
print('TLE x ' + str(c['TLE']))
print('RE x ' + str(c['RE']))
if __name__ == "__main__":
main()
|
s234343526
|
p03997
|
u296150111
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 62
|
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)//2*h)
|
s277397240
|
Accepted
| 17
| 2,940
| 66
|
a=int(input())
b=int(input())
h=int(input())
print(int((a+b)/2*h))
|
s422888436
|
p03494
|
u364642100
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 2,940
| 140
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n = input()
a = list(map(int, input().split()))
cnt = 0
while all(i%2 == 0 for i in a):
a = [i % 2 for i in a]
cnt += 1
print(cnt)
|
s341613361
|
Accepted
| 17
| 3,060
| 164
|
import math
n = input()
a = list(map(int, input().split()))
ans = float("inf")
for i in a:
ans = min(ans, len(bin(i)) - bin(i).rfind("1") - 1)
print(round(ans))
|
s330465746
|
p02694
|
u578093902
| 2,000
| 1,048,576
|
Wrong Answer
| 24
| 9,432
| 164
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
import math
x = int(input())
n = int((math.log10(x)-2)/math.log10(1.01))
total = 100*(1.01**n)
while total < x:
n += 1
total = 100*(1.01**n)
print(total,n)
|
s082477085
|
Accepted
| 63
| 9,300
| 196
|
import math
x = int(input())
n = int((math.log10(x)-2)/math.log10(1.01))-1
total = 100
while total < x:
n += 1
total = 100
for _ in range(n):
total += int(total*0.01)
print(n)
|
s976696154
|
p03352
|
u893063840
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 141
|
You are given a positive integer X. Find the largest _perfect power_ that is at most X. Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.
|
x = int(input())
ans = 1
for i in range(2, x + 1):
tmp = i
while tmp <= x:
tmp *= i
ans = max(ans, tmp / i)
print(ans)
|
s326539711
|
Accepted
| 18
| 2,940
| 176
|
x = int(input())
import math
ans = 1
for i in range(2, math.ceil(math.sqrt(x)) + 1):
tmp = i
while tmp <= x:
tmp *= i
ans = max(ans, tmp // i)
print(ans)
|
s332357804
|
p02407
|
u613534067
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,572
| 68
|
Write a program which reads a sequence and prints it in the reverse order.
|
a = list(map(int, input().split()))
a.sort(reverse = True)
print(a)
|
s769170094
|
Accepted
| 20
| 5,584
| 85
|
input()
a = list(map(int, input().split()))
a.reverse()
print(" ".join(map(str, a)))
|
s988565866
|
p03852
|
u064246852
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 282
|
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
s = input()[::-1]
flag = 1
i = 0
while i < len(s):
if s[i:i+5] == "maerd" or s[i:i+5] == "esare":
i += 5
elif s[i:i+6] == "resare":
i += 6
elif s[i:i+7] == "remaerd":
i += 7
else:
flag = 0
break
print("YES" if flag else "NO")
|
s879820437
|
Accepted
| 17
| 2,940
| 46
|
print(input()in"aeiou"and"vowel"or"consonant")
|
s733336970
|
p02390
|
u229478139
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,588
| 160
|
Write a program which reads an integer $S$ [second] and converts it to $h:m:s$ where $h$, $m$, $s$ denote hours, minutes (less than 60) and seconds (less than 60) respectively.
|
temp = int(input("Your number:"))
hour = temp//3600
min = (temp%3600)//60
sec = (temp%3600)%60
answer = str(hour)+":"+str(min)+":"+str(sec)
print(answer)
|
s802571414
|
Accepted
| 20
| 5,588
| 117
|
s = int(input())
h = str(s // 3600)
m = str((s % 3600) // 60)
s = str((s % 3600) % 60)
print(h + ':' + m + ':' + s)
|
s023682111
|
p03399
|
u695079172
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 101
|
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.
|
s = int(input())
sm = sum([int(c) for c in str(s)])
if s%sm == 0:
print("Yes")
else:
print("No")
|
s623224253
|
Accepted
| 17
| 2,940
| 114
|
a,b,c,d = [int(input()) for _ in range(4)]
train = a if a <= b else b
bus = c if c <= d else d
print(train + bus)
|
s137762267
|
p02975
|
u861141787
| 2,000
| 1,048,576
|
Wrong Answer
| 72
| 14,116
| 194
|
Snuke has N hats. The i-th hat has an integer a_i written on it. There are N camels standing in a circle. Snuke will put one of his hats on each of these camels. If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print `Yes`; otherwise, print `No`. * The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself. What is XOR? The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non- negative integers x_1, x_2, \ldots, x_n is defined as follows: - When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even. For example, 3 \oplus 5 = 6.
|
n = int(input())
a = list(map(int, input().split()))
a = [a[-1]] + a + [a[0]]
print(a)
for i in range(1, n-2):
if a[i] != a[i-1] ^ a[i+1]:
print("No")
exit()
print("Yes")
|
s403483774
|
Accepted
| 49
| 14,212
| 316
|
n = int(input())
a = list(map(int, input().split()))
b = list(set(a))
if a.count(0) == n:
print("Yes")
elif len(b) == 2 and a.count(0) == n / 3:
print("Yes")
elif len(b) == 3 and b[0] ^ b[1] ^ b[2] == 0 and a.count(b[0]) == n / 3 and a.count(b[0]) == n / 3:
print("Yes")
else:
print("No")
|
s511728715
|
p02259
|
u150984829
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 166
|
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.
|
n=int(input())
a=list(map(int,input().split()))
c=0
for i in range(n-2):
for j in range(n-1,i,-1):
if a[j-1]>a[j]:a[j-1],a[j]=a[j],a[j-1]
c+=1
print(*a)
print(c)
|
s613778855
|
Accepted
| 20
| 5,604
| 159
|
n=int(input())
a=list(map(int,input().split()))
c=0
for i in range(n-1):
for j in range(n-i-1):
if a[j]>a[j+1]:a[j:j+2]=a[j+1],a[j];c+=1
print(*a)
print(c)
|
s426715306
|
p03944
|
u148471096
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,064
| 381
|
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
w, h, n = map(int, input().split())
wmin = 0
hmin = 0
wmax = w
hmax = h
for i in range(n):
x, y, a = map(int, input().split())
if a==1:
wmax = max(wmax,x)
elif a==2:
wmin = min(wmin,x)
elif a==3:
hmax = max(hmax,y)
elif a==4:
hmin = min(hmin,y)
if wmax-wmin<0 or hmax-hmin<0:
print(0)
else:
print((wmax-wmin)*(hmax-hmin))
|
s590709969
|
Accepted
| 17
| 3,064
| 386
|
w, h, n = map(int, input().split())
wmin = w
hmin = h
wmax = 0
hmax = 0
for i in range(n):
x, y, a = map(int, input().split())
if a==1:
wmax = max(wmax,x)
elif a==2:
wmin = min(wmin,x)
elif a==3:
hmax = max(hmax,y)
elif a==4:
hmin = min(hmin,y)
if wmin-wmax<0 or hmin-hmax<0:
print(0)
else:
print((wmin-wmax)*(hmin-hmax))
|
s786253088
|
p03565
|
u519130434
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 779
|
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 = list(input())
T = list(input())
if len(S) < len(T):
print("UNRESTORABLE")
else:
solved = False
ind = -1
for i in reversed(range(len(S) - len(T) + 1)):
s = S[i:i+len(T)]
yes = True
for j in range(len(s)):
print(T[j], s[j])
if s[j] == '?' or s[j] == T[j]:
continue
else:
yes = False
break
if yes:
ind = i
solved = True
break
if not solved:
print("UNRESTORABLE")
else:
for i in range(len(S)):
if i < ind:
if S[i] == '?':
S[i] = 'a'
elif ind<= i < ind+len(T):
S[i] = T[i-ind]
print(''.join(S))
|
s133622653
|
Accepted
| 17
| 3,064
| 960
|
S = list(input())
T = list(input())
def solve():
global S, T
if len(S) < len(T):
print("UNRESTORABLE")
return
inds = []
for i in reversed(range(len(S) - len(T) + 1)):
s = S[i:i+len(T)]
# print(i, S[i], S[i:i+len(T)])
yes = True
for j in range(len(s)):
# print(T[j], s[j])
if s[j] == '?' or s[j] == T[j]:
continue
else:
yes = False
break
if yes:
inds.append(i)
# print(inds)
if len(inds) == 0:
print("UNRESTORABLE")
return
ans = []
for ind in inds:
ss = S[:]
for i in range(len(S)):
if ind<= i < ind+len(T):
ss[i] = T[i-ind]
elif ss[i] == '?':
ss[i] = 'a'
ans.append(ss)
# print(ans)
ans = list(map(lambda x: ''.join(x), ans))
print(min(ans))
solve()
|
s538107098
|
p03679
|
u391675400
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 215
|
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
x,a,b = map(int,(input().split()))
hantei = b - a
if hantei < 0:
print("delicious")
elif x > hantei:
print("sefe")
else:
print("dangerous")
|
s462809695
|
Accepted
| 17
| 3,068
| 174
|
x,a,b = map(int,(input().split()))
hantei = b - a
if hantei <= 0:
print("delicious")
elif x >= hantei:
print("safe")
elif (x + 1) <= hantei:
print("dangerous")
|
s149757621
|
p03494
|
u769538311
| 2,000
| 262,144
|
Wrong Answer
| 21
| 3,060
| 249
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
N = int(input())
a = list(map(int, input().split()))
c = 0
bool = True
while (bool):
for i in range(N):
if a[i] % 2 == 0:
a[i] = a[i] / 2
else:
bool = False
break
else:
c += 1
|
s221218887
|
Accepted
| 23
| 9,188
| 223
|
N=int(input())
A=list(map(int,input().split()))
ans_list = []
for n in A:
temp = n
count = 0
while temp % 2 == 0:
temp /= 2
count = count + 1
ans_list.append(count)
print(min(ans_list))
|
s033010983
|
p03624
|
u941753895
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,188
| 200
|
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=input()
l=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
for x in l:
if x in S:
print(x)
exit()
print('None')
|
s000185918
|
Accepted
| 17
| 3,188
| 204
|
S=input()
l=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
for x in l:
if x not in S:
print(x)
exit()
print('None')
|
s348250039
|
p03699
|
u416773418
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 213
|
You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?
|
n = int(input())
s=[int(input()) for i in range(n)]
ans=sum(s)
if ans % 10 != 0:
print(ans)
else:
for i in range(len(s)):
if s[i] % 10 != 0:
print(ans-s[i])
else:
print(0)
|
s096247032
|
Accepted
| 18
| 3,060
| 239
|
n = int(input())
s=[int(input()) for i in range(n)]
s.sort()
ans=sum(s)
if ans % 10 != 0:
print(ans)
else:
for i in range(len(s)):
if s[i] % 10 != 0:
print(ans-s[i])
break
else:
print(0)
|
s804483959
|
p04043
|
u343977188
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 101
|
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.
|
S=list(map(int,input().split()))
if S.count(5)==2 and S.count(7)==1:
print("Yes")
else:print("No")
|
s990479910
|
Accepted
| 17
| 2,940
| 101
|
S=list(map(int,input().split()))
if S.count(5)==2 and S.count(7)==1:
print("YES")
else:print("NO")
|
s492180851
|
p00007
|
u885889402
| 1,000
| 131,072
|
Wrong Answer
| 20
| 6,748
| 162
|
Your friend who lives in undisclosed country is involved in debt. He is borrowing 100,000-yen from a loan shark. The loan shark adds 5% interest of the debt and rounds it to the nearest 1,000 above week by week. Write a program which computes the amount of the debt in n weeks.
|
n = input()
a = 100000
for i in range(1,int(n)+1):
a = int(a*1.05)
print(a)
if(a%1000 > 0):
a = 1000*int(a/1000) + 1000
print(a)
print(a)
|
s951388262
|
Accepted
| 30
| 6,748
| 136
|
n = input()
a = 100000
for i in range(1,int(n)+1):
a = int(a*1.05)
if(a%1000 > 0):
a = 1000*int(a/1000) + 1000
print(a)
|
s676347162
|
p03351
|
u103657515
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 2,940
| 167
|
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())
AB = abs(A-B)
BC = abs(B-C)
ans = 0
if AB <= D and BC <= D:
ans = "yes"
else:
ans = "no"
print(ans)
|
s532720236
|
Accepted
| 17
| 2,940
| 185
|
A,B,C,D = map(int, input().split())
AB = abs(A-B)
BC = abs(B-C)
AC = abs(A-C)
ans = 0
if AB <= D and BC <= D or AC <= D:
ans = "Yes"
else:
ans = "No"
print(ans)
|
s619893848
|
p03377
|
u304630225
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 109
|
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,C=input().split()
A=int(A)
B=int(B)
C=int(C)
if C>=A and C<=A+B :
print("Yes")
else :
print("No")
|
s586260084
|
Accepted
| 17
| 3,060
| 109
|
A,B,C=input().split()
A=int(A)
B=int(B)
C=int(C)
if C>=A and C<=A+B :
print("YES")
else :
print("NO")
|
s807894658
|
p03007
|
u305422591
| 2,000
| 1,048,576
|
Wrong Answer
| 402
| 21,532
| 453
|
There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.
|
import heapq
N = int(input())
A = [int(item) for item in input().split()]
# N = 10**5
# A = [item for item in range(N)]
heapq.heapify(A)
res = []
while len(A) > 2:
min_ = A[0]
heapq.heappop(A)
min_2 = A[0]
heapq.heappop(A)
new = min_ - min_2
res.append([min_, min_2])
heapq.heappush(A, new)
min_ = A[0]
heapq.heappop(A)
min_2 = A[0]
heapq.heappop(A)
res.append([min_, min_2])
print(min_2 - min_)
for item in res:
print (item[0], item[1])
|
s496226942
|
Accepted
| 308
| 18,760
| 361
|
import heapq
N = int(input())
A = [int(item) for item in input().split()]
# N = 10**5
# A = [item for item in range(N)]
A.sort()
M = A.pop()
heapq.heapify(A)
m = heapq.heappop(A)
res = []
while A:
a = heapq.heappop(A)
if a < 0:
res.append((M, a))
M -= a
else:
res.append((m, a))
m -= a
print(M-m)
for x, y in res:
print(x, y)
print(M, m)
|
s761488839
|
p03385
|
u357050210
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
s = input()
s = sorted(s)
if s == "abc":
print("Yes")
else:
print("No")
|
s388198839
|
Accepted
| 20
| 2,940
| 90
|
s = input()
s = "".join(sorted(s))
if s == "abc":
print("Yes")
else:
print("No")
|
s117181265
|
p02608
|
u992541367
| 2,000
| 1,048,576
|
Wrong Answer
| 60
| 9,436
| 546
|
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())
answers = [0 for i in range(10**4+1)]
for x in range(1,50):
for y in range(x,50):
for z in range(y,50):
n = x**2+y**2+z**2+x*y+y*z+z*x
if 0 <= n <= 10000:
#print(x,y,z,n)
c = len(set([x,y,z]))
if c == 1:
answers[n] += 1
elif c == 2:
answers[n] += 3
else:
answers[n] += 6
#print(answers[:30])
for ans in answers[1:]:
print(ans)
|
s984590958
|
Accepted
| 154
| 9,092
| 589
|
N = int(input())
answers = [0 for i in range(10**4+1)]
for x in range(1,42):
for y in range(x,100):
for z in range(y,100):
n = x**2+y**2+z**2+x*y+y*z+z*x
if 0 <= n <= 10000:
#if n == 9803:
# print(x,y,z,n)
c = len(set([x,y,z]))
if c == 1:
answers[n] += 1
elif c == 2:
answers[n] += 3
else:
answers[n] += 6
#print(answers[-100:])
for ans in answers[1:N+1]:
print(ans)
|
s001940341
|
p02612
|
u023762741
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,140
| 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.
|
s = int(input())
print(s % 1000)
|
s747515341
|
Accepted
| 28
| 9,144
| 76
|
s = int(input()) % 1000
if s == 0:
print(0)
else:
print(1000 - s )
|
s187635615
|
p03730
|
u532087742
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 191
|
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`.
|
# -*- coding: utf-8 -*-
import sys
(A,B,C) = map(int,input().split(" "))
x = A % B
print(C,x)
for i in range(B):
if (A*i) % B == C:
print('YES')
sys.exit()
print('NO')
|
s445791277
|
Accepted
| 17
| 2,940
| 168
|
# -*- coding: utf-8 -*-
import sys
(A,B,C) = map(int,input().split(" "))
for i in range(B):
if (A*i) % B == C:
print('YES')
sys.exit()
print('NO')
|
s782105938
|
p03168
|
u686230543
| 2,000
| 1,048,576
|
Wrong Answer
| 202
| 13,048
| 192
|
Let N be a positive odd number. There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i. Taro has tossed all the N coins. Find the probability of having more heads than tails.
|
import numpy as np
n = int(input())
dp = np.zeros(n)
dp[0] = 1.
for p in map(float, input().split()):
tmp = (1. - p) * dp
tmp[1:] += p * dp[:-1]
dp = tmp
print(dp[n // 2 + 1:].sum())
|
s902694024
|
Accepted
| 202
| 12,468
| 196
|
import numpy as np
n = int(input())
dp = np.zeros(n + 1)
dp[0] = 1.
for p in map(float, input().split()):
tmp = (1. - p) * dp
tmp[1:] += p * dp[:-1]
dp = tmp
print(dp[n // 2 + 1:].sum())
|
s966435054
|
p02742
|
u996564551
| 2,000
| 1,048,576
|
Wrong Answer
| 26
| 9,136
| 238
|
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
H, W = input().split(' ')
H = int(H)
W = int(W)
if (H//2)==0 and (W//2)==0:
print(int((H/2) * W))
elif (H//2)==1 and (W//2)==0:
print(int((W/2) * H))
elif (H//2)==0 and (W//2)==1:
print(int((H/2) * W))
else:
print(int((H*W/2)+1))
|
s154356596
|
Accepted
| 25
| 8,984
| 286
|
H, W = input().split(' ')
H = int(H)
W = int(W)
if H==1 or W ==1:
print(1)
elif (H%2)==0 and (W%2)==0:
print(int((H/2) * W))
elif (H%2)==1 and (W%2)==0:
print(int((W/2) * H))
elif (H%2)==0 and (W%2)==1:
print(int((H/2) * W))
elif (H%2)==1 and (W%2)==1:
print(int((H*W//2)+1))
|
s060288280
|
p02394
|
u386861275
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,528
| 105
|
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
W,H,x,y,r= map(int, input().split())
if W>=x+r & r<=x & r<=y & H>=y+r:
print ("Yes")
else:
print ("No")
|
s302704652
|
Accepted
| 20
| 7,668
| 111
|
W,H,x,y,r= map(int, input().split())
if W>=x+r and r<=x and r<=y and H>=y+r:
print ("Yes")
else:
print ("No")
|
s388734798
|
p03853
|
u533084327
| 2,000
| 262,144
|
Wrong Answer
| 148
| 12,424
| 189
|
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 numpy as np
H,W = list(map(int, input().split()))
print (H)
C = []
for i in range(H):
C.append(input())
print (C)
for i in range(H):
print (str(C[i]))
print (str(C[i]))
|
s885561666
|
Accepted
| 148
| 12,424
| 191
|
import numpy as np
H,W = list(map(int, input().split()))
#print (H)
C = []
for i in range(H):
C.append(input())
#print (C)
for i in range(H):
print (str(C[i]))
print (str(C[i]))
|
s243236906
|
p04044
|
u506858457
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 88
|
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())
for i in range(N):
s=input()
s=sorted(s)
print(*s,sep='')
|
s655592652
|
Accepted
| 17
| 3,060
| 93
|
n,l=map(int,input().split())
s=sorted([input() for i in range(n)])
#print(s)
print(*s,sep="")
|
s905897742
|
p04011
|
u011872685
| 2,000
| 262,144
|
Wrong Answer
| 26
| 9,004
| 75
|
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
N=int(input())
K=int(input())
X=int(input())
Y=int(input())
a=X*K+Y*(N-K)
a
|
s486032863
|
Accepted
| 27
| 9,016
| 113
|
N=int(input())
K=int(input())
X=int(input())
Y=int(input())
if K<N:
print(X*K+Y*(N-K))
else:
print(X*N)
|
s545660124
|
p00060
|
u546285759
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,612
| 415
|
「1」から「10」までの数字が書かれたカードが各 1 枚、全部で 10 枚あります。このカードは、表側には数字が書かれ、裏側には何も書かれていません。このカードを使って、あなたと相手の 2 名で以下のルールでゲー ムを行います。 1. あなたと相手には、表を上にして 1 枚、裏を上にして 1 枚、計 2 枚のカードが配られています。あなたは相手の表のカードの数字を見ることができますが、裏のカードの数字は見えません。 2. 配られたカードの数字の合計が 20 以下で、かつ相手の数字の合計より大きいときに勝ちとなります。 例えば、あなたのカードが「7」「8」 (合計 15) 、相手のカードが「9」「10」 (合計 19) のときは、相手の勝ちです。 3. あなたと相手は最大であと 1 枚カードを引くことができます。それを引かなくても構いません。 ここで、あと 1 枚のカードを引くかどうかを決定する目安として、カードを引いたときに合計が 20 以下に なる確率を考え、その確率が 50% 以上のときはカードを引くこととしましょう。この確率を計算するときには、あなたの 2 枚のカードと相手の表のカードの計3枚のカードの情報を利用することができます。つまり、各カードは 1 枚ずつしかないので、それらのカードを引くことはないことになります。 あなたの 2 枚のカードと相手の表のカードを読み込んで、それぞれについて、あと 1 枚引いたときに合計が 20 以下になる確率が 50% 以上のときは YES、そうでないなら NO を出力するプログラムを作成してください。
|
while True:
try:
C1, C2, C3 = map(int, input().split())
except:
break
cards = [v for v in range(1, 11)]
p = []
for v in [C1, C2, C3]:
cards.remove(v)
for i in cards:
_card = cards[:]
_card.remove(i)
for j in _card:
if C1+C2+j <= 20:
p.append(1)
else:
p.append(0)
print(sum(p)/len(p))
|
s064093273
|
Accepted
| 30
| 7,700
| 441
|
while True:
try:
C1, C2, C3 = map(int, input().split())
except:
break
cards = [v for v in range(1, 11)]
p = []
for v in [C1, C2, C3]:
cards.remove(v)
for i in cards:
_card = cards[:]
_card.remove(i)
for j in _card:
if C1+C2+j <= 20:
p.append(1)
else:
p.append(0)
print("YES" if sum(p)/len(p) >= 0.5 else "NO")
|
s593984774
|
p03997
|
u941277791
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 107
|
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.
|
top=int(input())
bottom=int(input())
height=int(input())
num=top+bottom
ans=num*height/2
print(str(ans))
|
s586483263
|
Accepted
| 17
| 2,940
| 115
|
top=int(input())
bottom=int(input())
height=int(input())
num=top+bottom
ans=int(num*height/2)
print(str(ans))
|
s440041195
|
p03563
|
u945419374
| 2,000
| 262,144
|
Wrong Answer
| 19
| 2,940
| 97
|
Takahashi is a user of a site that hosts programming contests. When a user competes in a contest, the _rating_ of the user (not necessarily an integer) changes according to the _performance_ of the user, as follows: * Let the current rating of the user be a. * Suppose that the performance of the user in the contest is b. * Then, the new rating of the user will be the avarage of a and b. For example, if a user with rating 1 competes in a contest and gives performance 1000, his/her new rating will be 500.5, the average of 1 and 1000. Takahashi's current rating is R, and he wants his rating to be exactly G after the next contest. Find the performance required to achieve it.
|
N=int(input())
K=int(input())
x=1
for i in range(N):
if x<=K:
x=x*2
else:
x=x+K
print(x)
|
s174452985
|
Accepted
| 17
| 2,940
| 51
|
R=int(input())
G=int(input())
print(2*G-R)
|
s691722777
|
p02613
|
u316055872
| 2,000
| 1,048,576
|
Wrong Answer
| 144
| 16,272
| 236
|
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):
S.append(input())
a = S.count('AC')
b = S.count('WA')
c = S.count('TLE')
d = S.count('RE')
print ('AC ×' + str(a))
print('WA ×' + str(b))
print('TLE ×' + str(c))
print('RE ×' + str(d))
|
s974832859
|
Accepted
| 147
| 16,224
| 252
|
N = int(input())
S = []
for _ in range(N):
S.append(input())
a = S.count('AC')
b = S.count('WA')
c = S.count('TLE')
d = S.count('RE')
print ('AC x' +' ' + str(a))
print('WA x' + ' '+ str(b))
print('TLE x' +' ' + str(c))
print('RE x' + ' '+ str(d))
|
s555646320
|
p03434
|
u375695365
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 134
|
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()))
a.sort()
ab=list(a[::2])
ac=list(a[1::2])
print(sum(ab)-sum(ac))
#print(ab)
#print(ac)
|
s774912095
|
Accepted
| 17
| 3,060
| 146
|
n=int(input())
a=list(map(int,input().split()))
a.sort(reverse=True)
ab=list(a[::2])
ac=list(a[1::2])
print(sum(ab)-sum(ac))
#print(ab)
#print(ac)
|
s035667541
|
p03455
|
u559313689
| 2,000
| 262,144
|
Wrong Answer
| 25
| 9,088
| 109
|
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())
answer = (a * b)% 2
if answer == 0:
print('even')
else:
print('odd')
|
s109490959
|
Accepted
| 31
| 9,024
| 109
|
a, b = map(int, input().split())
answer = (a * b)% 2
if answer == 0:
print('Even')
else:
print('Odd')
|
s158771199
|
p02421
|
u395334793
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,500
| 328
|
Taro and Hanako are playing card games. They have n cards each, and they compete n turns. At each turn Taro and Hanako respectively puts out a card. The name of the animal consisting of alphabetical letters is written on each card, and the bigger one in lexicographical order becomes the winner of that turn. The winner obtains 3 points. In the case of a draw, they obtain 1 point each. Write a program which reads a sequence of cards Taro and Hanako have and reports the final scores of the game.
|
count = int(input())
Taro_an = []
Hanako_an = []
Ts = 0
Hs = 0
for i in range(count) :
animal = input().split(" ")
Taro_an = animal[0]
Hanako_an = animal[1]
if Taro_an > Hanako_an :
Ts += 3
else :
if Hanako_an > Taro_an :
Hs += 3
else :
++Ts, ++Hs
print(Ts,Hs)
|
s840082101
|
Accepted
| 20
| 7,488
| 334
|
count = int(input())
Taro_an = []
Hanako_an = []
Ts = 0
Hs = 0
for i in range(count) :
animal = input().split(" ")
Taro_an = animal[0]
Hanako_an = animal[1]
if Taro_an > Hanako_an :
Ts += 3
else :
if Hanako_an > Taro_an :
Hs += 3
else :
Ts += 1; Hs += 1
print(Ts,Hs)
|
s631277623
|
p03167
|
u954774382
| 2,000
| 1,048,576
|
Wrong Answer
| 600
| 51,968
| 1,576
|
There is a grid with H horizontal rows and W vertical columns. Let (i, j) denote the square at the i-th row from the top and the j-th column from the left. For each i and j (1 \leq i \leq H, 1 \leq j \leq W), Square (i, j) is described by a character a_{i, j}. If a_{i, j} is `.`, Square (i, j) is an empty square; if a_{i, j} is `#`, Square (i, j) is a wall square. It is guaranteed that Squares (1, 1) and (H, W) are empty squares. Taro will start from Square (1, 1) and reach (H, W) by repeatedly moving right or down to an adjacent empty square. Find the number of Taro's paths from Square (1, 1) to (H, W). As the answer can be extremely large, find the count modulo 10^9 + 7.
|
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
# import numpy as np
starttime = time.time()
# import numpy as np
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
sys.setrecursionlimit(int(pow(10,6)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
n,m=L()
A=[input() for i in range(n)]
dp=[[0 for i in range(m)] for j in range(n)]
for i in range(n):
for j in range(m):
if i==0 and j==0:
dp[i][j]=1
continue
if i==0:
dp[i][j]=dp[i][j-1]
elif j==0:
dp[i][j]=dp[i-1][j]
else:
if A[i][j]==".":
dp[i][j]=(dp[i-1][j]+dp[i][j-1])%mod
else:
dp[i][j]=0
print(dp[-1][-1])
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
|
s900921109
|
Accepted
| 609
| 52,164
| 1,662
|
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
# import numpy as np
starttime = time.time()
# import numpy as np
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
sys.setrecursionlimit(int(pow(10,6)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
n,m=L()
A=[input() for i in range(n)]
dp=[[0 for i in range(m)] for j in range(n)]
for i in range(n):
for j in range(m):
if i==0 and j==0:
dp[i][j]=1
continue
if i==0:
if A[i][j]==".":
dp[i][j]=dp[i][j-1]
elif j==0:
if A[i][j]==".":
dp[i][j]=dp[i-1][j]
else:
if A[i][j]==".":
dp[i][j]=(dp[i-1][j]+dp[i][j-1])%mod
else:
dp[i][j]=0
# print(*dp[i])
print(dp[-1][-1])
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
|
s383481435
|
p02239
|
u797673668
| 1,000
| 131,072
|
Wrong Answer
| 30
| 8,012
| 524
|
Write a program which reads an directed graph $G = (V, E)$, and finds the shortest distance from vertex $1$ to each vertex (the number of edges in the shortest path). Vertices are identified by IDs $1, 2, ... n$.
|
from collections import deque
n = int(input())
g = [None] * n
visited = [0] * n
dist = [-1] * n
queue = deque()
while n:
l = list(map(int, input().split()))
connected = [i - 1 for i in l[2:]]
g[l[0] - 1] = [i - 1 for i in l[2:]]
n -= 1
queue.append((0, 0))
visited[0] = 1
while queue:
print(queue)
i, d = queue.popleft()
dist[i] = d
for c in g[i]:
if not visited[c]:
visited[c] = 1
queue.append((c, d + 1))
for i, d in enumerate(dist):
print(i + 1, d)
|
s839252610
|
Accepted
| 40
| 8,068
| 507
|
from collections import deque
n = int(input())
g = [None] * n
visited = [0] * n
dist = [-1] * n
queue = deque()
while n:
l = list(map(int, input().split()))
connected = [i - 1 for i in l[2:]]
g[l[0] - 1] = [i - 1 for i in l[2:]]
n -= 1
queue.append((0, 0))
visited[0] = 1
while queue:
i, d = queue.popleft()
dist[i] = d
for c in g[i]:
if not visited[c]:
visited[c] = 1
queue.append((c, d + 1))
for i, d in enumerate(dist):
print(i + 1, d)
|
s913878293
|
p04030
|
u637918426
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 222
|
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
l = list(input())
print(l)
list = []
for i in range(len(l)):
if l[i] == 'B':
if not list:
continue
else:
del list[-1]
else:
list.append(l[i])
print(''.join(list))
|
s084368643
|
Accepted
| 17
| 2,940
| 213
|
l = list(input())
list = []
for i in range(len(l)):
if l[i] == 'B':
if not list:
continue
else:
del list[-1]
else:
list.append(l[i])
print(''.join(list))
|
s138798006
|
p03672
|
u585819925
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 223
|
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()
while True:
if len(s) % 2 == 0:
if s[:len(s)//2] == s[len(s)//2:]:
print(len(s))
break
s = s[:-1]
|
s880522154
|
Accepted
| 17
| 2,940
| 228
|
s=input()[:-1]
while True:
if len(s) % 2 == 0:
if s[:len(s)//2] == s[len(s)//2:]:
print(len(s))
break
s = s[:-1]
|
s502129929
|
p03861
|
u525423408
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 87
|
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,x=map(int,input().split())
if a==0:
print(b//x +1)
else:
print(b//x - a//x)
|
s399466686
|
Accepted
| 17
| 3,064
| 93
|
a,b,x=map(int,input().split())
if a % x == 0:
print(b//x - a//x +1)
else:print(b//x-a//x)
|
s106703183
|
p02936
|
u869919400
| 2,000
| 1,048,576
|
Wrong Answer
| 2,118
| 236,564
| 390
|
Given is a rooted tree with N vertices numbered 1 to N. The root is Vertex 1, and the i-th edge (1 \leq i \leq N - 1) connects Vertex a_i and b_i. Each of the vertices has a counter installed. Initially, the counters on all the vertices have the value 0. Now, the following Q operations will be performed: * Operation j (1 \leq j \leq Q): Increment by x_j the counter on every vertex contained in the subtree rooted at Vertex p_j. Find the value of the counter on each vertex after all operations.
|
N, Q = map(int, input().split())
rs = [map(int, input().split()) for _ in range(N-1)]
ps = [map(int, input().split()) for _ in range(Q)]
cs = [0] * N
nodes = [set() for _ in range(N)]
for n1, n2 in rs:
nodes[n1-1].add(n2-1)
def f(n, p):
cs[n] += p
if not nodes[n]:
return
for _n in nodes[n]:
f(_n, p)
while ps:
n, p = ps.pop()
f(n-1, p)
print(cs)
|
s885451193
|
Accepted
| 1,419
| 234,516
| 532
|
import sys
sys.setrecursionlimit(500000)
N, Q = map(int, input().split())
tree = [[] for _ in range(N)]
for _ in range(N-1):
a, b = map(int, input().split())
tree[a-1].append(b-1)
tree[b-1].append(a-1)
scores = [0] * N
for _ in range(Q):
p, x = map(int, input().split())
scores[p-1] += x
ans = [0] * N
visited = set()
def dfs(n: int, score: int):
score += scores[n]
ans[n] = score
visited.add(n)
for v in tree[n]:
if v not in visited:
dfs(v, score)
dfs(0, 0)
print(*ans)
|
s205067731
|
p03149
|
u732963817
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 101
|
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".
|
N = list(map(int, input().split()))
print("Yes" if 1 in N and 9 in N and 7 in N and 4 in N else "No")
|
s173583795
|
Accepted
| 17
| 2,940
| 122
|
N = list(map(int, input().split()))
print("YES" if len(set(N)) == 4 and 1 in N and 9 in N and 7 in N and 4 in N else "NO")
|
s490254191
|
p04030
|
u306412379
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,016
| 219
|
Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key. To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string: * The `0` key: a letter `0` will be inserted to the right of the string. * The `1` key: a letter `1` will be inserted to the right of the string. * The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted. Sig has launched the editor, and pressed these keys several times. You are given a string s, which is a record of his keystrokes in order. In this string, the letter `0` stands for the `0` key, the letter `1` stands for the `1` key and the letter `B` stands for the backspace key. What string is displayed in the editor now?
|
s = list(input())
x = len(s)
a = []
for i in range(x):
if s[i] == '0':
a.append(0)
elif s[i] == '1':
a.append(1)
elif s[i] == 'B':
if a != []:
a.remove(a[-1])
else:
continue
print(*a)
|
s326913698
|
Accepted
| 30
| 8,932
| 230
|
s = list(input())
x = len(s)
a = []
for i in range(x):
if s[i] == '0':
a.append('0')
elif s[i] == '1':
a.append('1')
elif s[i] == 'B':
if a == []:
continue
else:
a.pop(-1)
b = ''.join(a)
print(b)
|
s140832590
|
p03943
|
u402629484
| 2,000
| 262,144
|
Wrong Answer
| 22
| 3,064
| 48
|
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.
|
s = input()
print(s.count('WB') + s.count('BW'))
|
s330970007
|
Accepted
| 22
| 3,064
| 92
|
a,b,c = list(map(int, input().split()))
print("Yes" if a+b==c or b+c==a or c+a==b else "No")
|
s587889538
|
p02261
|
u760369990
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,728
| 1,571
|
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
def isStable(before_sort, after_sort):
num = len(before_sort)
for i in range(0, num-1):
for j in range(i+1, num):
for a in range(0, num-1):
for b in range(a+1, num):
if before_sort[i][1] == after_sort[j][1] and \
before_sort[i] == after_sort[b] and \
before_sort[j] == after_sort[a]:
print("i=%d / j=%d / a=%d / b=%d /before=%s / after=%s" % \
(i, j, a, b, before_sort[i][1], after_sort[j][1]))
return False
return True
def isStable2(sorted_data, bubble_sort_data):
for i in range(0, len(sorted_data)):
if sorted_data[i] != bubble_sort_data[i]:
return False
return True
def BubbleSort(c, n):
for i in range(0, n):
for j in range(n-1, i, -1):
if int(c[j][1]) < int(c[j-1][1]):
c[j], c[j-1] = c[j-1], c[j]
def SelectionSort(c, n):
for i in range(0, n):
minj = i
for j in range(i, n):
if int(c[j][1]) < int(c[minj][1]):
minj = j
c[i], c[minj] = c[minj], c[i]
num = int(input())
data = input().split(' ')
copy_data = data.copy()
BubbleSort(copy_data, num)
print(copy_data)
if isStable(copy_data, copy_data):
print("Stable")
else:
print("Not stable")
bubble_sort_data = data.copy()
SelectionSort(bubble_sort_data, num)
print(bubble_sort_data)
if isStable(bubble_sort_data, copy_data):
print("Stable")
else:
print("Not stable")
|
s843366925
|
Accepted
| 150
| 7,844
| 1,427
|
def isStable(before_sort, after_sort):
num = len(before_sort)
for i in range(0, num-1):
for j in range(i+1, num):
for a in range(0, num-1):
for b in range(a+1, num):
if before_sort[i][1] == after_sort[j][1] and \
before_sort[i] == after_sort[b] and \
before_sort[j] == after_sort[a]:
return False
return True
def isStable2(sorted_data, bubble_sort_data):
for i in range(0, len(sorted_data)):
if sorted_data[i] != bubble_sort_data[i]:
return False
return True
def BubbleSort(c, n):
for i in range(0, n):
for j in range(n-1, i, -1):
if int(c[j][1]) < int(c[j-1][1]):
c[j], c[j-1] = c[j-1], c[j]
def SelectionSort(c, n):
for i in range(0, n):
minj = i
for j in range(i, n):
if int(c[j][1]) < int(c[minj][1]):
minj = j
c[i], c[minj] = c[minj], c[i]
num = int(input())
data = input().split(' ')
copy_data = data.copy()
BubbleSort(copy_data, num)
print(' '.join(copy_data))
if isStable(copy_data, copy_data):
print("Stable")
else:
print("Not stable")
bubble_sort_data = data.copy()
SelectionSort(bubble_sort_data, num)
print(' '.join(bubble_sort_data))
if isStable(bubble_sort_data, copy_data):
print("Stable")
else:
print("Not stable")
|
s297105978
|
p02388
|
u538972897
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,572
| 28
|
Write a program which calculates the cube of a given integer x.
|
x = int(input())
print(x**x)
|
s335872983
|
Accepted
| 20
| 5,576
| 28
|
x = int(input())
print(x**3)
|
s651639743
|
p03044
|
u518042385
| 2,000
| 1,048,576
|
Wrong Answer
| 2,106
| 108,104
| 607
|
We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied: * For any two vertices painted in the same color, the distance between them is an even number. Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.
|
import sys
sys.setrecursionlimit(1000000)
n=int(input())
l=[[] for i in range(n+1)]
for i in range(n-1):
a,b,c=map(int,input().split())
l[a].append([b,c])
l[b].append([a,c])
color=[-1 for i in range(n+1)]
def dfs(s,c):
color[s]=c
for i in range(len(l[s])):
index=l[s][i][0]
dis=l[s][i][1]
if color[index]!=-1:
if dis%2==1:
if color[index]==c:
color[index]=1-c
dfs(index,1-c)
else:
if dis%2==1:
color[s]=1-c
dfs(index,1-c)
else:
color[s]=c
dfs(index,1-c)
dfs(1,1)
for i in range(1,n+1):
print(color[i])
|
s937372880
|
Accepted
| 857
| 42,228
| 499
|
from collections import deque
q=deque([1])
n=int(input())
l=[[] for i in range(n+1)]
for i in range(n-1):
a,b,c=map(int,input().split())
c=c%2
l[a].append([b,c])
l[b].append([a,c])
inf=10**10
d=[inf for i in range(n+1)]
d[1]=0
def dik(s):
q=deque([s])
while q:
a=q.popleft()
for i in l[a]:
index=i[0]
dis=i[1]
if d[index]>dis+d[a]:
d[index]=dis+d[a]
q.append(index)
dik(1)
for i in range(1,n+1):
if d[i]%2==0:
print(0)
else:
print(1)
|
s376503054
|
p02614
|
u137226361
| 1,000
| 1,048,576
|
Wrong Answer
| 128
| 9,348
| 969
|
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 sys, copy
input = sys.stdin.readline
def main():
h, w , k = map(int, input().split())
ls = []
for _ in range(h):
ms = []
t=str(input().rstrip())
for i in range(len(t)):
if t[i] =='.':
ms.append(0)
else:
ms.append(1)
ls.append(ms)
def num(x, y):
lt = copy.deepcopy(ls)
x2 = format(x, 'b')
y2 = format(y, 'b')
for i in range(len(x2)):
if x2[i]=='0':
lt[i]= [0]*len(y2)
for j in range(len(y2)):
if y2[j]=='0':
for t in range(len(x2)):
lt[t][j]= 0
re = 0
for i in lt:
re += sum(i)
return re
ans = 0
for i in range(2**h):
for j in range(2**w):
if num(i,j) == k:
ans +=1
print(i, j)
print(ans)
print(ls)
if __name__ == '__main__':
main()
|
s590259860
|
Accepted
| 130
| 9,168
| 957
|
import sys, copy
input = sys.stdin.readline
def main():
h, w , k = map(int, input().split())
ls = []
for _ in range(h):
ms = []
t=str(input().rstrip())
for i in range(len(t)):
if t[i] =='.':
ms.append(0)
else:
ms.append(1)
ls.append(ms)
def num(x, y):
lt = copy.deepcopy(ls)
x2 = format(x, 'b').rjust(h, '0')
y2 = format(y, 'b').rjust(w, '0')
for i in range(len(x2)):
if x2[i]=='0':
lt[i]= [0]*len(y2)
for j in range(len(y2)):
if y2[j]=='0':
for t in range(len(x2)):
lt[t][j]= 0
re = 0
for i in lt:
re += sum(i)
return re
ans = 0
for i in range(2**h):
for j in range(2**w):
if num(i,j) == k:
ans +=1
print(ans)
if __name__ == '__main__':
main()
|
s515501416
|
p02613
|
u047116039
| 2,000
| 1,048,576
|
Wrong Answer
| 161
| 9,220
| 314
|
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.
|
ac,wa,tle,re = 0,0,0,0
N=int(input())
for _ in range(N):
s=str(input())
if(s=="AC"):
ac+=1
elif(s=="WA"):
wa+=1
elif(s=="TLE"):
tle+=1
elif(s=="RE"):
re+=1
print("AC × "+str(ac)+"\n")
print("WA × "+str(wa)+"\n")
print("TLE × "+str(tle)+"\n")
print("RE × "+str(re)+"\n")
|
s292464654
|
Accepted
| 162
| 9,212
| 272
|
ac,wa,tle,re = 0,0,0,0
N=int(input())
for _ in range(N):
s=str(input())
if(s=="AC"):
ac+=1
elif(s=="WA"):
wa+=1
elif(s=="TLE"):
tle+=1
elif(s=="RE"):
re+=1
print("AC x "+str(ac))
print("WA x "+str(wa))
print("TLE x "+str(tle))
print("RE x "+str(re))
|
s363152954
|
p03379
|
u503901534
| 2,000
| 262,144
|
Wrong Answer
| 371
| 25,220
| 184
|
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
n = int(input())
x = list(map(int,input().split()))
xs = list(sorted(x))
for i in range(n):
if x[i] <= xs[n // 2]:
print(xs[n // 2])
else:
print(xs[n // 2 -1])
|
s776934597
|
Accepted
| 340
| 25,620
| 183
|
n = int(input())
x = list(map(int,input().split()))
xs = list(sorted(x))
for i in range(n):
if x[i] < xs[n // 2]:
print(xs[n // 2])
else:
print(xs[n // 2 -1])
|
s761590286
|
p03351
|
u077019541
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 92
|
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 (b-a)+(c-b)<=d:
print("Yes")
else:
print("No")
|
s902724379
|
Accepted
| 17
| 2,940
| 140
|
a,b,c,d = map(int,input().split())
if abs(b-a)<=d and abs(c-b)<=d:
print("Yes")
elif abs(c-a)<=d:
print("Yes")
else:
print("No")
|
s260406227
|
p03493
|
u181769094
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 61
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
list=list(map(int,input().split()))
a=list.count(1)
print(a)
|
s175006129
|
Accepted
| 18
| 2,940
| 52
|
list=list(map(int,input()))
a=list.count(1)
print(a)
|
s530095893
|
p04043
|
u594956556
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 131
|
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()))
abc.sort()
if (abc[0]==5) and (abc[0]==5) and (abc[0]==7):
print('YES')
else:
print('NO')
|
s157171132
|
Accepted
| 17
| 2,940
| 132
|
abc = list(map(int, input().split()))
abc.sort()
if (abc[0]==5) and (abc[1]==5) and (abc[2]==7):
print('YES')
else:
print('NO')
|
s095183229
|
p02694
|
u597626771
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 9,092
| 110
|
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())
money = 100
year = 0
while money <= X:
year += 1
money = int(money*1.01)
print(year)
|
s081712988
|
Accepted
| 24
| 9,084
| 108
|
X = int(input())
money = 100
year = 0
while money < X:
year += 1
money = int(money*1.01)
print(year)
|
s139992261
|
p03385
|
u569272329
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 85
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
S = str(input())
S = sorted(S)
if "abc" == S:
print("Yes")
else:
print("No")
|
s352562186
|
Accepted
| 17
| 2,940
| 122
|
S = str(input())
if not(S[0] == S[1]) and not(S[1] == S[2]) and not(S[0] == S[2]):
print("Yes")
else:
print("No")
|
s127137317
|
p03730
|
u739843002
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,124
| 126
|
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`.
|
import math
A, B, C = list(map(lambda n: int(n), input().split(" ")))
print("Yes") if C % math.gcd(A,B) == 0 else print("No")
|
s186275939
|
Accepted
| 25
| 9,200
| 127
|
import math
A, B, C = list(map(lambda n: int(n), input().split(" ")))
print("YES") if C % math.gcd(A,B) == 0 else print("NO")
|
s349994264
|
p03007
|
u583276018
| 2,000
| 1,048,576
|
Wrong Answer
| 2,104
| 13,964
| 386
|
There are N integers, A_1, A_2, ..., A_N, written on a blackboard. We will repeat the following operation N-1 times so that we have only one integer on the blackboard. * Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y. Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.
|
n = int(input())
num = [int(i) for i in input().split()]
anslist = []
for i in range(n-2):
anslist.append([min(num), max(num)])
a = min(num) - max(num)
num.remove(min(num))
num.remove(max(num))
num.append(a)
ans = max(num) - min(num)
print(ans)
for i in range(n-2):
print(str(anslist[i][0]) + " " + str(anslist[i][1]))
print(str(max(num)) + " " + str(min(num)))
|
s968275740
|
Accepted
| 1,800
| 23,240
| 1,469
|
n = int(input())
num = [int(i) for i in input().split()]
plus = []
minus = []
sumlist = []
for i in range(n):
if(num[i] >= 0):
plus.append(num[i])
else:
minus.append(num[i])
ans = []
if(len(plus) >= len(minus)):
last = min(num)
if(last >= 0):
plus.remove(min(plus))
else:
minus.remove(min(minus))
while(len(minus) > 0):
sumlist.append(plus[0] - minus[0])
ans.append([plus.pop(0), minus.pop(0)])
for i in range(len(plus)):
sumlist.append(plus[i])
for i in range(len(sumlist)-1):
tmp = last
last -= sumlist[0]
ans.append([tmp, sumlist.pop(0)])
ans.append([sumlist[0], last])
score = sumlist[0] - last
print(score)
for i in range(len(ans)):
print(str(ans[i][0]) + " " + str(ans[i][1]))
else:
last = max(num)
if(last >= 0):
plus.remove(max(plus))
else:
minus.remove(max(minus))
while(len(plus) > 0):
sumlist.append(minus[0] - plus[0])
ans.append([minus.pop(0), plus.pop(0)])
for i in range(len(minus)):
sumlist.append(minus[i])
for i in range(len(sumlist)-1):
tmp = last
last -= sumlist[0]
ans.append([tmp, sumlist.pop(0)])
ans.append([last, sumlist[0]])
score = last - sumlist[0]
print(score)
for i in range(len(ans)):
print(str(ans[i][0]) + " " + str(ans[i][1]))
|
s027355593
|
p03494
|
u727801592
| 2,000
| 262,144
|
Wrong Answer
| 23
| 3,064
| 140
|
There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform.
|
n=input()
div=30
count=0
for i in input().split():
while int(i)%2==0:
i=int(i)/2
count+=1
if count<div:
div=count
print(div)
|
s602536398
|
Accepted
| 23
| 2,940
| 150
|
n=input()
div=30
count=0
for i in input().split():
while int(i)%2==0:
i=int(i)/2
count+=1
if count<div:
div=count
count=0
print(div)
|
s702531526
|
p04029
|
u733321071
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 37
|
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((1 + n) / 2)
|
s970750236
|
Accepted
| 17
| 2,940
| 38
|
n = int(input())
print(int((n+1)*n/2))
|
s895715845
|
p03369
|
u010437136
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 191
|
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.
|
import sys
s = sys.stdin.readline()
l = [0,0,0]
for i in range(2):
if s[i] == "o":
l[i] = 100
elif s[i] == "x":
l[i] = 0
goukei = 700+l[0]+l[1]+l[2]
print(goukei)
|
s638812136
|
Accepted
| 17
| 2,940
| 191
|
import sys
s = sys.stdin.readline()
l = [0,0,0]
for i in range(3):
if s[i] == "o":
l[i] = 100
elif s[i] == "x":
l[i] = 0
goukei = 700+l[0]+l[1]+l[2]
print(goukei)
|
s568956910
|
p02608
|
u933129390
| 2,000
| 1,048,576
|
Wrong Answer
| 467
| 18,232
| 401
|
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())
xyz = [[[0 for i in range(101)] for j in range(101)] for k in range(101)]
ans = [0 for i in range(n+1)]
def gen_n(x, y, z):
return x*2 + y*2 + z*2 + x*y + y*z + z*x
for i in range(1, 101):
for j in range(1, 101):
for k in range(1, 101):
num = gen_n(i, j, k)
if n >= num:
ans[num] += 1
for i in range(1, n+1):
print(ans[i])
|
s743498723
|
Accepted
| 954
| 18,416
| 404
|
n = int(input())
xyz = [[[0 for i in range(101)] for j in range(101)] for k in range(101)]
ans = [0 for i in range(n+1)]
def gen_n(x, y, z):
return x**2 + y**2 + z**2 + x*y + y*z + z*x
for i in range(1, 101):
for j in range(1, 101):
for k in range(1, 101):
num = gen_n(i, j, k)
if n >= num:
ans[num] += 1
for i in range(1, n+1):
print(ans[i])
|
s052175139
|
p03385
|
u580362735
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 63
|
You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`.
|
S = input()
print('Yes') if sorted(S) == 'abc' else print('No')
|
s476651953
|
Accepted
| 18
| 2,940
| 77
|
S = list(input())
print('Yes') if sorted(S) == ['a','b','c'] else print('No')
|
s483654110
|
p03544
|
u170324846
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 108
|
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
N = int(input())
L = [0] * (N + 1)
L[0] = 2
L[1] = 1
for i in range(2,N):
L[i] = L[i-1]+L[i-2]
print(L[N])
|
s534582760
|
Accepted
| 17
| 2,940
| 111
|
N = int(input())
L = [0] * (N + 1)
L[0] = 2
L[1] = 1
for i in range(2,N+1):
L[i] = L[i-1]+L[i-2]
print(L[N])
|
s735063433
|
p03351
|
u468940068
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 3,316
| 167
|
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.
|
# coding: utf-8
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')
|
s970456600
|
Accepted
| 22
| 3,316
| 167
|
# coding: utf-8
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')
|
s658791342
|
p03160
|
u041401317
| 2,000
| 1,048,576
|
Wrong Answer
| 99
| 13,980
| 408
|
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
def main():
n = int(input())
hts = list(map(int,input().split()))
if n == 2:
print(abs(hts[0]-hts[1]))
return
dp = [float('inf')]*n
dp[0] = 0
dp[1] = abs(dp[1]-dp[0])
## dp[i]
for i in range(2,n):
dp[i] = min(dp[i-1]+abs(hts[i]-hts[i-1]),
dp[i-2]+abs(hts[i]-hts[i-2]))
print(dp[-1])
main()
|
s247054559
|
Accepted
| 102
| 13,928
| 426
|
def main():
n = int(input())
hts = list(map(int,input().split()))
if n == 2:
print(abs(hts[0]-hts[1]))
return
dp = [float('inf')]*n
dp[0] = 0
dp[1] = abs(hts[1]-hts[0])
## dp[i]
for i in range(2,n):
dp[i] = min(dp[i-1]+abs(hts[i]-hts[i-1]),
dp[i-2]+abs(hts[i]-hts[i-2]))
#print(dp)
print(dp[-1])
main()
|
s336333507
|
p02613
|
u950847221
| 2,000
| 1,048,576
|
Wrong Answer
| 29
| 9,068
| 267
|
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.
|
tr=int(input())
word=input()
a=0
w=0
t=0
r=0
i=0
for i in range(1<=tr<=10*5):
i+=1
if word=='AC':
a+=1
elif word=='WA':
w+=1
elif word=='TLE':
t+=1
else:
r+=1
print('AC*a')
print('WA*w')
print('TLE*t')
print('RE*r')
|
s697658370
|
Accepted
| 145
| 9,156
| 288
|
tr=int(input())
a=0
w=0
t=0
r=0
for i in range(tr):
word=input()
if word=='AC':
a+=1
elif word=='WA':
w+=1
elif word=='TLE':
t+=1
else:
r+=1
print('AC x '+str(a))
print('WA x '+str(w))
print('TLE x '+str(t))
print('RE x '+str(r))
|
s689571325
|
p02612
|
u297089927
| 2,000
| 1,048,576
|
Wrong Answer
| 28
| 9,148
| 48
|
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.
|
import math
n=int(input())
print(math.ceil(n)-n)
|
s132134268
|
Accepted
| 30
| 8,960
| 29
|
n=int(input())
print(-n%1000)
|
s246371719
|
p03852
|
u960570220
| 2,000
| 262,144
|
Wrong Answer
| 29
| 9,100
| 121
|
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`.
|
c = str(input())
if c == 'a' and c == 'e' and c == 'i' and c == 'o' and c == 'u':
print('Yes')
else:
print('No')
|
s381211625
|
Accepted
| 26
| 9,028
| 126
|
c = str(input())
if c == 'a' or c == 'e' or c == 'i' or c == 'o' or c == 'u':
print('vowel')
else:
print('consonant')
|
s501318799
|
p03493
|
u497326082
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,316
| 110
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
a = input()
b = 1
c = 0
print(a)
print(a[b])
for b in range(3):
if int(a[b]) == 1:
c += 1
print(c)
|
s394313872
|
Accepted
| 17
| 2,940
| 86
|
s=input()
ans=0
if s[0]=="1":ans+=1
if s[1]=="1":ans+=1
if s[2]=="1":ans+=1
print(ans)
|
s993720128
|
p02293
|
u279605379
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,652
| 318
|
For given two lines s1 and s2, print "2" if they are parallel, "1" if they are orthogonal, or "0" otherwise. s1 crosses points p0 and p1, and s2 crosses points p2 and p3.
|
def dot(a,b):return a[0]*b[0] + a[1]*b[1]
def cross(a,b):return a[0]*b[1] - a[1]*b[0]
q = int(input())
for i in range(q):
x0,y0,x1,y1,x2,y2,x3,y3 = [int(i) for i in input().split()]
a = [x1-x0,y1-y0]
b = [x3-x2,y3-y1]
if dot(a,b) == 0 : print(1)
elif cross(a,b) == 0 : print(2)
else : print(0)
|
s033306097
|
Accepted
| 30
| 7,748
| 318
|
def dot(a,b):return a[0]*b[0] + a[1]*b[1]
def cross(a,b):return a[0]*b[1] - a[1]*b[0]
q = int(input())
for i in range(q):
x0,y0,x1,y1,x2,y2,x3,y3 = [int(i) for i in input().split()]
a = [x1-x0,y1-y0]
b = [x3-x2,y3-y2]
if dot(a,b) == 0 : print(1)
elif cross(a,b) == 0 : print(2)
else : print(0)
|
s925451197
|
p02233
|
u869924057
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 168
|
Write a program which prints $n$-th fibonacci number for a given integer $n$. The $n$-th fibonacci number is defined by the following recursive formula: \begin{equation*} fib(n)= \left \\{ \begin{array}{ll} 1 & (n = 0) \\\ 1 & (n = 1) \\\ fib(n - 1) + fib(n - 2) & \\\ \end{array} \right. \end{equation*}
|
memo = [0, 1]
def fib(n):
if (n in memo):
return memo[n]
ans = fib(n-1) + fib(n-2)
memo.append(ans)
return ans
n = int(input())
print(fib(n))
|
s391124287
|
Accepted
| 20
| 5,600
| 163
|
memo = [1, 1]
def fib(n):
if (n < len(memo)):
return memo[n]
memo.append(fib(n-1) + fib(n-2))
return memo[n]
N = int(input())
print(fib(N))
|
s084994995
|
p03777
|
u407106303
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 60
|
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()
print('H' if (a=='D')^(b=='D') else 'D')
|
s743618191
|
Accepted
| 17
| 2,940
| 60
|
a,b=input().split()
print('D' if (a=='H')^(b=='H') else 'H')
|
s676893764
|
p03643
|
u473023730
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 33
|
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.
|
n=str(input())
a="ABC"
print(n+a)
|
s511171602
|
Accepted
| 17
| 2,940
| 33
|
n=str(input())
a="ABC"
print(a+n)
|
s731663155
|
p02975
|
u377989038
| 2,000
| 1,048,576
|
Wrong Answer
| 86
| 22,976
| 311
|
Snuke has N hats. The i-th hat has an integer a_i written on it. There are N camels standing in a circle. Snuke will put one of his hats on each of these camels. If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print `Yes`; otherwise, print `No`. * The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself. What is XOR? The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non- negative integers x_1, x_2, \ldots, x_n is defined as follows: - When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even. For example, 3 \oplus 5 = 6.
|
from collections import Counter as ct
import sys
input = sys.stdin.readline
n = int(input())
a = input().split()
c = ct(a)
print(c)
if c["0"] == n:
print("Yes")
elif len(c) == 2 and c["0"] * 3 == n:
print("Yes")
elif len(c) == 3 and c[0] == c[1] and c[1] == c[2]:
print("Yes")
else:
print("No")
|
s783736507
|
Accepted
| 53
| 14,468
| 393
|
from collections import Counter as ct
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
c = ct(a)
if c[0] == n:
print("Yes")
elif len(c) == 2 and c[0] * 3 == n:
print("Yes")
elif len(c) == 3 and len(set(c.values())) == 1:
a, b, c = c.keys()
if (a ^ b ^ c) == 0:
print("Yes")
else:
print("No")
else:
print("No")
|
s380249676
|
p03998
|
u017719431
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 58
|
Alice, Bob and Charlie are playing _Card Game for Three_ , as below: * At first, each of the three players has a deck consisting of some number of cards. Each card has a letter `a`, `b` or `c` written on it. The orders of the cards in the decks cannot be rearranged. * The players take turns. Alice goes first. * If the current player's deck contains at least one card, discard the top card in the deck. Then, the player whose name begins with the letter on the discarded card, takes the next turn. (For example, if the card says `a`, Alice takes the next turn.) * If the current player's deck is empty, the game ends and the current player wins the game. You are given the initial decks of the players. More specifically, you are given three strings S_A, S_B and S_C. The i-th (1≦i≦|S_A|) letter in S_A is the letter on the i-th card in Alice's initial deck. S_B and S_C describes Bob's and Charlie's initial decks in the same way. Determine the winner of the game.
|
cards_list = [input() for _ in range(3)]
print(cards_list)
|
s111174100
|
Accepted
| 17
| 3,064
| 447
|
cards_list = [input() for _ in range(3)]
def battle(cards_list):
dict = {"a": 0, "b": 1, "c": 2}
current = "a"
while True:
index = dict[current] #0
current = cards_list[index][0] #b
if len(cards_list[index]) > 1:
cards_list[index] = cards_list[index][1:]
else:
cards_list[index] = []
index = dict[current]
if len(cards_list[index]) == 0:
return current.upper()
print(battle(cards_list))
|
s417725862
|
p03448
|
u502028059
| 2,000
| 262,144
|
Wrong Answer
| 49
| 3,060
| 234
|
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())
ans = 0
for i in range(a+1):
for x in range(b+1):
for z in range(c+1):
if 500 * i + 100 * x + 10 * z == x:
ans += 1
print(ans)
|
s327717570
|
Accepted
| 51
| 3,060
| 251
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
ans = 0
for i in range(a+1):
for y in range(b+1):
for z in range(c+1):
if 500 * i + 100 * y + 50 * z == x:
ans += 1
print(ans)
|
s041869300
|
p03813
|
u117629640
| 2,000
| 262,144
|
Wrong Answer
| 23
| 3,192
| 237
|
Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) otherwise. You are given Smeke's current rating, x. Print `ABC` if Smeke will participate in ABC, and print `ARC` otherwise.
|
# -*- coding: utf-8 -*-
def main():
x = int(input())
ans = (x // 11) * 2
x %= 11
if x > 0:
if x > 6:
ans += 2
else:
ans += 1
print(ans)
if __name__ == '__main__':
main()
|
s756192451
|
Accepted
| 24
| 3,064
| 144
|
# -*- coding: utf-8 -*-
def main():
a = int(input())
print('ABC') if a < 1200 else print('ARC')
if __name__ == '__main__':
main()
|
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