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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s655817869 | p03544 | u011872685 | 2,000 | 262,144 | Wrong Answer | 2,205 | 9,120 | 150 | 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())
def ryuka(n):
if n==0:
return 2
elif n==1:
return 1
else:
return ryuka(n-1)+ryuka(n-2)
ryuka(N) | s986350194 | Accepted | 29 | 9,088 | 126 | #79B
N=int(input())
data=[2]
data.append(1)
for i in range(2,N+1):
data.append(data[i-1]+data[i-2])
print(data[N]) |
s984960646 | p02747 | u180926680 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 117 | A Hitachi string is a concatenation of one or more copies of the string `hi`. For example, `hi` and `hihi` are Hitachi strings, while `ha` and `hii` are not. Given a string S, determine whether S is a Hitachi string. | S = str(input())
S_size = len(S)
hi_size = S.count('hi')
if S_size == hi_size * 2:
print('yes')
else:
print('no') | s534963942 | Accepted | 17 | 2,940 | 117 | S = str(input())
S_size = len(S)
hi_size = S.count('hi')
if S_size == hi_size * 2:
print('Yes')
else:
print('No') |
s694720756 | p02401 | u671553883 | 1,000 | 131,072 | Wrong Answer | 40 | 7,556 | 310 | Write a program which reads two integers a, b and an operator op, and then prints the value of a op b. The operator op is '+', '-', '*' or '/' (sum, difference, product or quotient). The division should truncate any fractional part. | while True:
a, op, b = input().split()
a = int(a)
b = int(b)
if op == '?':
break
if op == '+':
print(a + b)
if op == '-':
print(a - b)
if op == '*':
print(a * b)
if op == '%':
print(a % b) | s085230414 | Accepted | 20 | 7,652 | 317 | while True:
a, op, b = input().split()
a = int(a)
b = int(b)
if op == '?':
break
if op == '+':
print(a + b)
if op == '-':
print(a - b)
if op == '*':
print(a * b)
if op == '/':
print(a // b)
|
s225072615 | p03494 | u627530854 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 209 | There are N positive integers written on a blackboard: A_1, ..., A_N. Snuke can perform the following operation when all integers on the blackboard are even: * Replace each integer X on the blackboard by X divided by 2. Find the maximum possible number of operations that Snuke can perform. | def count_factors(num, fact):
res = 0
while num % fact == 0:
res += 1
num //= fact
return res
nums = [int(tok) for tok in input().split()]
print(min(map((lambda n : count_factors(n, 2)), nums))) | s859151162 | Accepted | 18 | 3,060 | 220 | def count_factors(num, fact):
if num % fact != 0:
return 0
return 1 + count_factors(num // fact, fact)
input()
nums = [int(tok) for tok in input().split()]
print(min(map((lambda n : count_factors(n, 2)), nums))) |
s854396881 | p03997 | u121732701 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 70 | You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid. | a = int(input())
b = int(input())
h = int(input())
print((a+b)*h/2) | s645641843 | Accepted | 17 | 2,940 | 75 | a = int(input())
b = int(input())
h = int(input())
print(int((a+b)*h/2)) |
s176996513 | p02928 | u802191176 | 2,000 | 1,048,576 | Wrong Answer | 2,104 | 3,188 | 308 | We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of o... | N,K=(input().split())
N,K=int(N),int(K)
seq=list(input().split())
seq=[int(x) for x in seq]
print(seq)
cou=0
C=0
inc=0
for i in range(len(seq)):
for j in range(i+1,len(seq)):
if seq[i]>seq[j]: cou+=1
if seq[i]<seq[j]: inc+=1
for m in range(K):
C+=((K-m)*cou+m*inc)
print(C%(10**9+7))
| s566825342 | Accepted | 1,050 | 3,188 | 333 | N,K=(input().split())
N,K=int(N),int(K)
seq=list(input().split())
seq=[int(x) for x in seq]
cou=0
C=0
inc=0
for i in range(len(seq)):
for j in range(i+1,len(seq)):
if seq[i]>seq[j]: cou+=1
for i in range(len(seq)):
for j in range(len(seq)):
if seq[i]>seq[j]: inc+=1
C=(K*(K-1)//2)*inc+K*cou
print... |
s392890543 | p03024 | u874549552 | 2,000 | 1,048,576 | Wrong Answer | 18 | 2,940 | 136 | Takahashi is competing in a sumo tournament. The tournament lasts for 15 days, during which he performs in one match per day. If he wins 8 or more matches, he can also participate in the next tournament. The matches for the first k days have finished. You are given the results of Takahashi's matches as a string S cons... | s=list(input())
maru = 0
for i in range(len(s)):
if s[i] == "o":
maru = maru + 1
if maru >= 8:
print("YES")
else:
print("NO") | s632455723 | Accepted | 17 | 3,060 | 192 | s=list(input())
maru = 0
batu = 0
for i in range(len(s)):
if s[i] == "o":
maru = maru + 1
else:
batu = batu + 1
if 15 - maru - batu >= 8 - maru:
print("YES")
else:
print("NO") |
s333903618 | p02972 | u941645670 | 2,000 | 1,048,576 | Wrong Answer | 49 | 6,488 | 65 | There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following con... | #d
n = int(input())
a = list(map(int, input().split()))
print(-1) | s318931862 | Accepted | 220 | 14,776 | 294 | #d
n = int(input())
a = list(map(int, input().split()))
ans = [0]*len(a)
for i in range(len(a),0,-1):
if sum(ans[i-1::i])%2 != a[i-1]:
ans[i-1] = 1
print(sum(ans))
if sum(ans) > 0:
ans_n = [str(i+1) for i, x in enumerate(ans) if x==1]
ans_n=" ".join(ans_n)
print(ans_n) |
s006423500 | p02393 | u553058997 | 1,000 | 131,072 | Wrong Answer | 20 | 7,468 | 52 | Write a program which reads three integers, and prints them in ascending order. | a = list(map(int, input().split()))
print(sorted(a)) | s393211473 | Accepted | 30 | 7,576 | 78 | a = list(map(int, input().split()))
print(' '.join(list(map(str, sorted(a))))) |
s234174167 | p00036 | u647694976 | 1,000 | 131,072 | Time Limit Exceeded | 9,990 | 5,560 | 730 | 縦 8、横 8 のマスからなる図 1 のような平面があります。 □| □| □| □| □| □| □| □ ---|---|---|---|---|---|---|--- □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ □| □| □| □| □| □| □| □ 図１ --- この平面上に、以下の A から G の図形のどれかが一つだけ置かれて... | def trim():
num=[]
N=[input() for _ in range(8)]
for x in range(8):
for y in range(8):
if N[y][x]=="1":
if not num:
refer=(x,y)
num.append((x,y))
num=[(x-refer[0],y-refer[1]) for x,y in num]
return num
def point(point):
if p... | s036040581 | Accepted | 20 | 5,584 | 749 | def trim(N):
num=[]
for x in range(8):
for y in range(8):
if N[y][x]=="1":
if not num:
refer=(x,y)
num.append((x,y))
num=[(x-refer[0],y-refer[1]) for x,y in num]
return num
def points(point):
if point==[(0,0),(0,1),(1,0),(1,1)]:... |
s766100289 | p03469 | u931938233 | 2,000 | 262,144 | Wrong Answer | 29 | 9,020 | 36 | 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 col... | print('2018/01/',input()[8:],end='') | s841890999 | Accepted | 28 | 8,876 | 29 | print('2018/01/'+input()[8:]) |
s877027835 | p03385 | u361826811 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 266 | You are given a string S of length 3 consisting of `a`, `b` and `c`. Determine if S can be obtained by permuting `abc`. |
import sys
# import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
S = readline().decode('utf8')
print('Yes' if set('abc') == set(S) else 'No')
| s776035220 | Accepted | 17 | 2,940 | 275 |
import sys
# import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
S = readline().decode('utf8').rstrip()
print('Yes' if set('abc') == set(S) else 'No')
|
s558116860 | p03548 | u107039373 | 2,000 | 262,144 | Wrong Answer | 30 | 2,940 | 86 | We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two... | a,b,c = map(int,input().split())
x = 1
while a >= b*x + c*(x+1):
x += 1
print(x) | s185467341 | Accepted | 29 | 2,940 | 89 | a,b,c = map(int,input().split())
x = 1
while a >= b*x + c*(x+1):
x += 1
print(x-1)
|
s064108868 | p03696 | u886747123 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 245 | You are given a string S of length N consisting of `(` and `)`. Your task is to insert some number of `(` and `)` into S to obtain a _correct bracket sequence_. Here, a correct bracket sequence is defined as follows: * `()` is a correct bracket sequence. * If X is a correct bracket sequence, the concatenation of... | # D - Insertion
N = int(input())
S = input().split("()")
ans = []
for x in S:
tmp = x
for _ in range(x.count(")")):
tmp = "(" + tmp
for _ in range(x.count("(")):
tmp += ")"
ans.append(tmp)
print("()".join(ans)) | s671230393 | Accepted | 20 | 3,060 | 162 | # D - Insertion
N = int(input())
S = input()
s = S
for _ in range(50):
s = s.replace("()", "")
ans = "(" * s.count(")") + S + ")" * s.count("(")
print(ans) |
s951429354 | p03635 | u725324053 | 2,000 | 262,144 | Wrong Answer | 31 | 9,036 | 69 | In _K-city_ , there are n streets running east-west, and m streets running north-south. Each street running east-west and each street running north-south cross each other. We will call the smallest area that is surrounded by four streets a block. How many blocks there are in K-city? | n,m = input().split()
n = int(n)
m = int(m)
print ((n + 1) * (m + 1)) | s239446398 | Accepted | 27 | 9,096 | 69 | n,m = input().split()
n = int(n)
m = int(m)
print ((n - 1) * (m - 1)) |
s763226394 | p03643 | u281303342 | 2,000 | 262,144 | Wrong Answer | 19 | 3,060 | 143 | 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 = input()
# print("ABC"+N)
N = int(input())
print("{:03d}".format(N)) | s133467807 | Accepted | 19 | 3,060 | 149 |
# N = input()
# print("ABC"+N)
N = int(input())
print("ABC"+"{:03d}".format(N)) |
s094746478 | p03861 | u163791883 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 55 | 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())
print((B - A) // X) | s294987802 | Accepted | 17 | 2,940 | 110 | A, B, X = map(int, input().split())
if A % X == 0:
print(B // X - A // X + 1)
else:
print(B // X - A // X) |
s782071755 | p03371 | u731448038 | 2,000 | 262,144 | Wrong Answer | 56 | 3,060 | 204 | "Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the curr... | a, b, c, x, y = map(int, input().split())
na, nb, nc = x, y, 0
mx = 10**18
for i in range(max(x,y)):
if i%2==1: continue
total = a*(na-i//2) + b*(nb-i//2) + c*i
if total<mx:
mx=total
print(mx) | s784302472 | Accepted | 128 | 3,060 | 274 | a, b, c, x, y = map(int, input().split())
mx = 10**18
for i in range(0, max(x,y)*2+1):
if i%2==1: continue
total = max(a*(x-i//2), 0) + max(b*(y-i//2), 0) + c*i
if total<mx:
mx=total
print(mx)
|
s996372070 | p03378 | u733608212 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 114 | There are N + 1 squares arranged in a row, numbered 0, 1, ..., N from left to right. Initially, you are in Square X. You can freely travel between adjacent squares. Your goal is to reach Square 0 or Square N. However, for each i = 1, 2, ..., M, there is a toll gate in Square A_i, and traveling to Square A_i incurs a c... | n, m, x = map(int, input().split())
li = list(map(int, input().split()))
print(min(len(li[:x]), len(li[x+1:])))
| s690738107 | Accepted | 17 | 2,940 | 164 | n, m, x = map(int, input().split())
li = list(map(int, input().split()))
li = [1 if i in li else 0 for i in range(1, n+1)]
print(min(sum(li[:x-1]), sum(li[x:])))
|
s184954047 | p04029 | u498975813 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 63 | There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total? | n = int(input())
ans=0
for i in range(n):
n+=1+i
print(ans) | s750966995 | Accepted | 17 | 2,940 | 65 | n = int(input())
ans=0
for i in range(n):
ans+=1+i
print(ans) |
s726473923 | p03760 | u768896740 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 211 | Snuke signed up for a new website which holds programming competitions. He worried that he might forget his password, and he took notes of it. Since directly recording his password would cause him trouble if stolen, he took two notes: one contains the characters at the odd-numbered positions, and the other contains the... | o = list(input())
e = list(input())
password = []
for i in range(len(e)):
password.append(o[i])
password.append(e[i])
if o > e:
password.append(o[-1])
password = ''.join(password)
print(password) | s194805096 | Accepted | 18 | 3,060 | 221 | o = list(input())
e = list(input())
password = []
for i in range(len(e)):
password.append(o[i])
password.append(e[i])
if len(o) > len(e):
password.append(o[-1])
password = ''.join(password)
print(password) |
s560475610 | p02678 | u711238850 | 2,000 | 1,048,576 | Wrong Answer | 1,069 | 110,256 | 4,884 | There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in th... | import heapq
from collections import deque
class Graph:
def __init__(self,v,edgelist,w_v = None,directed = False):
super().__init__()
self.v = v
self.w_e = [{} for _ in [0]*self.v]
self.neighbor = [[] for _ in [0]*self.v]
self.w_v = w_v
self.directed = directed
... | s804628671 | Accepted | 932 | 109,852 | 5,030 | import heapq
from collections import deque
class Graph:
def __init__(self,v,edgelist,w_v = None,directed = False):
super().__init__()
self.v = v
self.w_e = [{} for _ in [0]*self.v]
self.neighbor = [[] for _ in [0]*self.v]
self.w_v = w_v
self.directed = directed
... |
s518696487 | p02468 | u659034691 | 1,000 | 131,072 | Wrong Answer | 20 | 7,624 | 189 | For given integers m and n, compute mn (mod 1,000,000,007). Here, A (mod M) is the remainder when A is divided by M. | m,n=(int(i) for i in input().split())
m2=1
while n>0:
m1=m
l=1
while l<n:
m1*=m1
l*=2
n-=l
m2*=m1
if m2>1000000007:
m2//=1000000007
print(m2) | s564131461 | Accepted | 40 | 7,800 | 300 | # your code goes here
m,n=(int(i) for i in input().split())
m2=1
while n>0:
m1=m
l=1
while l*2<=n:
m1*=m1
if m1>1000000007:
m1%=1000000007
# print(m1)
l*=2
n-=l
m2*=m1
if m2>1000000007:
m2%=1000000007
# print(m2)
print(m2) |
s028324223 | p02612 | u033893324 | 2,000 | 1,048,576 | Wrong Answer | 26 | 9,144 | 30 | 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. | a=int(input())
print(a % 1000) | s526174085 | Accepted | 25 | 9,148 | 68 | a=int(input())
b=a % 1000
if b>>0:
print(1000-b)
else:
print(0)
|
s965364501 | p02613 | u107091170 | 2,000 | 1,048,576 | Wrong Answer | 160 | 9,196 | 239 | 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`,... | N=int(input())
a = 0
w = 0
t = 0
r = 0
for i in range(N):
S=input();
if S[0] == "A":
a+=1
elif S[0] == "W":
w+=1
elif S[0] == "T":
t+=1
else:
r+=1
print("AC x",a)
print("WC x",w)
print("TLE x",t)
print("RE x",r)
| s018581087 | Accepted | 150 | 9,136 | 239 | N=int(input())
a = 0
w = 0
t = 0
r = 0
for i in range(N):
S=input();
if S[0] == "A":
a+=1
elif S[0] == "W":
w+=1
elif S[0] == "T":
t+=1
else:
r+=1
print("AC x",a)
print("WA x",w)
print("TLE x",t)
print("RE x",r)
|
s005007880 | p03090 | u016128476 | 2,000 | 1,048,576 | Wrong Answer | 38 | 3,976 | 885 | You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved... | # input
n = int(input())
# 1. make validate graph fragment (all index are includede completely)
fragments = []
if n % 2 == 0:
fragments = [(i + 1, n - i) for i in range(n // 2)]
else:
fragments = [(i + 1, n - i - 1) for i in range(n // 2)]
fragments.append((n,))
# 2. each vertex connect all vert... | s192179119 | Accepted | 371 | 3,976 | 898 | # input
n = int(input())
# 1. make validate graph fragment (all index are includede completely)
fragments = []
if n % 2 == 0:
fragments = [(i + 1, n - i) for i in range(n // 2)]
else:
fragments = [(i + 1, n - i - 1) for i in range(n // 2)]
fragments.append((n,))
# 2. each vertex connect all vert... |
s604374273 | p03007 | u046432236 | 2,000 | 1,048,576 | Wrong Answer | 64 | 14,060 | 256 | 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... | n=int(input())
newlist = list(map(int, input().split()))
switch=0
mazu=newlist[0]
for nl in newlist:
if mazu!=nl:
switch=1
if switch==1:
wa=0
for nl in newlist:
wa+=abs(nl)
print(wa)
else:
print(mazu*(len(newlist)-2)) | s711404190 | Accepted | 363 | 22,264 | 3,068 | n=int(input())
newlist = list(map(int, input().split()))
switch=0
counter=0
mazu=newlist[0]
if newlist[0]>=0:
seihu=1
else:
seihu=-1
konzai=0
for nl in newlist:
if mazu!=nl:
switch=1
if switch==1:
wa=0
list2=[]
for nl in range(n):
list2.append([newlist[nl], abs(newlist[nl])])
... |
s133832989 | p00074 | u150984829 | 1,000 | 131,072 | Wrong Answer | 20 | 5,636 | 190 | 標準録画で 120 分のビデオテープがあります。テープを完全に巻き戻した状態でビデオデッキのカウンタを 00:00:00 にし、標準録画モードで録画したところ、あるカウンタ値になりました。このカウンタ値（時、分、秒）を入力し、残りのテープの長さ（録画可能時間）を求め、時：分：秒の形式で出力するプログラムを作成して下さい。 ただし、2 時間（120分）以内の入力とします。なお、テープ残量は標準録画モードと 3 倍録画モードの場合の２通りを計算し、出力例のように時、分、秒とも 2 桁ずつ出力します。また "05" のように 10 の位が 0 の場合は、"0" をつけてください。 | while 1:
t,h,s=map(int,input().split())
if t<0:break
d,e=7200,t*3600+h*60+s
a,b=d-e,d-e/3
print(f'{a//3600:02}:{a//60%60:02}:{a%60:02}')
print(f'{b//3600:02}:{b//60%60:02}:{b%60:02}')
| s773462136 | Accepted | 20 | 5,612 | 152 | p=lambda x:print(f'{x//3600:02}:{x//60%60:02}:{x%60:02}')
for e in iter(input,'-1 -1 -1'):
h,m,s=map(int,e.split())
d=7200-h*3600-m*60-s
p(d);p(d*3)
|
s250146373 | p02409 | u130834228 | 1,000 | 131,072 | Wrong Answer | 20 | 7,612 | 356 | You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth fl... | import sys
n = int(input())
list_room = [0 for i in range(120)]
for i in range(n):
b, f, r, v = map(int, input().split())
room = ((b-1)*3+(f-1))*10 + r-1
list_room[room] += v
#print(list_room)
for i in range(4):
for j in range(3):
line = ""
for k in range(10):
line += " "+str(list_room[i*30 + j*10 + k]) ... | s183475542 | Accepted | 20 | 7,632 | 368 | import sys
n = int(input())
list_room = [0 for i in range(120)]
for i in range(n):
b, f, r, v = map(int, input().split())
room = ((b-1)*3+(f-1))*10 + r-1
list_room[room] += v
#print(list_room)
for i in range(4):
for j in range(3):
line = ""
for k in range(10):
line += " "+str(list_room[i*30 + j*10 + k]) ... |
s050303320 | p03456 | u559103167 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 86 | AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number. | a = int("".join(list(input().split())))
print("YES" if (int(a**.5))**2 == a else "NO") | s287156774 | Accepted | 17 | 2,940 | 86 | a = int("".join(list(input().split())))
print("Yes" if (int(a**.5))**2 == a else "No") |
s411988946 | p03407 | u533679935 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 76 | An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan. | A,B,C=map(int,input().split())
if A+B>=C:
print('No')
else:
print('Yes') | s517958643 | Accepted | 17 | 2,940 | 84 | A,B,C=map(int,input().split())
if A+B>=C:
print('Yes')
elif A+B<=C:
print('No')
|
s508992981 | p03795 | u802772880 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 35 | Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant ... | n=int(input())
print(n*800-40*n//3) | s305777190 | Accepted | 17 | 2,940 | 39 | n=int(input())
print(n*800-200*(n//15)) |
s582141658 | p03796 | u354126779 | 2,000 | 262,144 | Wrong Answer | 42 | 2,940 | 69 | Snuke loves working out. He is now exercising N times. Before he starts exercising, his _power_ is 1. After he exercises for the i-th time, his power gets multiplied by i. Find Snuke's power after he exercises N times. Since the answer can be extremely large, print the answer modulo 10^{9}+7. | n=int(input())
p=1
for i in range(1,n+1):
p=p*i
p=p%(10**9+7) | s348471338 | Accepted | 42 | 2,940 | 78 | n=int(input())
p=1
for i in range(1,n+1):
p=p*i
p=p%(10**9+7)
print(p) |
s052465087 | p03672 | u384124931 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 130 | 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 del... | s = input()
for i in range(1, len(s), 2):
a = s[:-i]
if a[:len(a)//2]==a[len(a)//2:]:
print(len(a))
exit() | s422427660 | Accepted | 17 | 2,940 | 130 | s = input()
for i in range(2, len(s), 2):
a = s[:-i]
if a[:len(a)//2]==a[len(a)//2:]:
print(len(a))
exit() |
s071127273 | p03730 | u686036872 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 154 | 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 objecti... | A, B, C = map(int, input().split())
for i in range(1, 1000):
if B*i%A==0:
count=1
else:
count=0
print("No" if count==0 else "Yes") | s146787372 | Accepted | 19 | 2,940 | 175 | A, B, C = map(int, input().split())
for i in range(1, 10000):
if ((B*i)+C)%A==0:
count=1
break
else:
count=0
print("YES" if count==1 else "NO") |
s470023407 | p03474 | u244836567 | 2,000 | 262,144 | Wrong Answer | 26 | 9,076 | 188 | 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=input().split()
c=input()
a=int(a)
b=int(b)
d=0
if c[a]=="-":
for i in range(a):
if c[i]=="-":
d=d+1
if d==1:
print("Yes")
else:
print("No")
else:
print("No") | s430667625 | Accepted | 27 | 9,188 | 192 | a,b=input().split()
c=input()
a=int(a)
b=int(b)
d=0
if c[a]=="-":
for i in range(a+b+1):
if c[i]=="-":
d=d+1
if d==1:
print("Yes")
else:
print("No")
else:
print("No") |
s923649451 | p03435 | u214344212 | 2,000 | 262,144 | Wrong Answer | 18 | 3,064 | 373 | We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left. According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) ... | a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=list(map(int,input().split()))
true=0
while true==0:
if b[0]-a[0]==b[1]-a[1]==b[2]-a[2]:
true+=0
else:
true+=1
if c[0]-b[0]==c[1]-b[1]==c[2]-b[2]:
true+=0
break
else:
true+=1
break
... | s370314305 | Accepted | 17 | 3,064 | 443 | c1=list(map(int,input().split()))
c2=list(map(int,input().split()))
c3=list(map(int,input().split()))
num=0
if c1[1]-c1[0]==c2[1]-c2[0]==c3[1]-c3[0]:
num+=0
else:
num+=1
if c1[2]-c1[1]==c2[2]-c2[1]==c3[2]-c3[1]:
num+=0
else:
num+=1
if c2[0]-c1[0]==c2[1]-c1[1]==c2[2]-c1[2]:
num+=0
else:
num+=1
i... |
s456132965 | p04045 | u463836923 | 2,000 | 262,144 | Wrong Answer | 40 | 3,192 | 614 | Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K. She is shopping, and now paying at the cashier. Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change). However, as mentioned before, she is very ... | #coding=UTF-8
mojir=input()
hyo=mojir.split(' ')
N=int(hyo[0])
K=int(hyo[1])
mojir=input()
hyo=mojir.split(' ')
D=[int(mono) for mono in hyo]
kouho=N
def usecheck(gaku):
gakustr=str(gaku)
keta=len(gakustr)
dame=None
for idx in range(0,keta,1):
if gakustr[idx] in hyo:
dame=idx
... | s085766481 | Accepted | 40 | 3,064 | 615 | #coding=UTF-8
mojir=input()
hyo=mojir.split(' ')
N=int(hyo[0])
K=int(hyo[1])
mojir=input()
hyo=mojir.split(' ')
D=[int(mono) for mono in hyo]
kouho=N
def usecheck(gaku):
gakustr=str(gaku)
keta=len(gakustr)
dame=None
for idx in range(0,keta,1):
if gakustr[idx] in hyo:
dame=idx
... |
s454697086 | p03698 | u464912173 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 116 | You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different. | s = list(input().split())
s.sort()
for i in range(len(s)-1):
if s[i]==s[i+1]:
print('no')
else: print('yes') | s923230694 | Accepted | 17 | 2,940 | 72 | a = input()
if len(a)==len(set(a)):
print('yes')
else: print('no') |
s005032379 | p02409 | u587193722 | 1,000 | 131,072 | Wrong Answer | 20 | 7,736 | 339 | You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth fl... | date = [
[[0 for r in range(10)] for f in range(3)] for b in range(4)
]
count = int(input())
for c in range(count):
b,f,r,v= [int(i) for i in input().split()]
date[b-1][f-1][r-1] +=v
for b in date:
for f in b:
for r in f:
print(' {0}'.format(r), end='')
print('#' * 20)... | s540783846 | Accepted | 20 | 7,640 | 377 | data = [
[[0 for r in range(10)] for f in range(3)] for b in range(4)
]
count = int(input())
for c in range(count):
b, f, r, v = [int(i) for i in input().split()]
data[b - 1][f - 1][r - 1] += v
for bi, b in enumerate(data):
for f in b:
for r in f:
print(' {0}'.format(r), end=''... |
s924754135 | p03861 | u393512980 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 59 | 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())
print((b - a + 1) // x) | s807135676 | Accepted | 17 | 2,940 | 65 | a, b, x = map(int, input().split())
print(b // x - (a - 1) // x)
|
s945786732 | p02697 | u402629484 | 2,000 | 1,048,576 | Wrong Answer | 84 | 10,060 | 2,093 | You are going to hold a competition of one-to-one game called AtCoder Janken. _(Janken is the Japanese name for Rock-paper-scissors.)_ N players will participate in this competition, and they are given distinct integers from 1 through N. The arena has M playing fields for two players. You need to assign each playing fi... | import sys
sys.setrecursionlimit(1000000000)
import math
from math import gcd
def lcm(a, b): return a * b // gcd(a, b)
from itertools import count, permutations, chain
from functools import lru_cache
from collections import deque, defaultdict
from pprint import pprint
ii = lambda: int(input())
mis = lambda: map(int, in... | s470243184 | Accepted | 115 | 29,584 | 1,437 |
from collections import deque
def main():
N, M = map(int, input().split())
if N&1:
q = deque(range(M*2))
while q:
print(q.popleft()+1, q.pop()+1)
else:
q1 = deque(range(N))
q2 = deque(range(N-1))
p1 = []
while q1:
p... |
s629117997 | p03415 | u736729525 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 39 | We have a 3×3 square grid, where each square contains a lowercase English letters. The letter in the square at the i-th row from the top and j-th column from the left is c_{ij}. Print the string of length 3 that can be obtained by concatenating the letters in the squares on the diagonal connecting the top-left and bot... | print(input()[0],input()[1],input()[2]) | s998213218 | Accepted | 17 | 2,940 | 46 | print(input()[0],input()[1],input()[2],sep="") |
s841496863 | p03386 | u640303028 | 2,000 | 262,144 | Wrong Answer | 34 | 3,444 | 306 | Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers. | a,b,k = [int(i) for i in input().split()]
if a + b - 1 <= b:
for i in range(a,a+k):
print(i)
if b-a+1>=2*k:
for i in range(b-k+1,b+1):
print(i)
else:
for i in range(a+k,b+1):
print(i)
else:
for i in range(a,b+1):
print(i)
| s514101140 | Accepted | 17 | 3,060 | 293 | a,b,k = [int(i) for i in input().split()]
if a + k - 1 <= b:
for i in range(a,a+k):
print(i)
if b-a+1>=2*k:
for i in range(b-k+1,b+1):
print(i)
else:
for i in range(a+k,b+1):
print(i)
else:
for i in range(a,b+1):
print(i) |
s017952561 | p03637 | u879870653 | 2,000 | 262,144 | Wrong Answer | 65 | 14,252 | 290 | We have a sequence of length N, a = (a_1, a_2, ..., a_N). Each a_i is a positive integer. Snuke's objective is to permute the element in a so that the following condition is satisfied: * For each 1 ≤ i ≤ N - 1, the product of a_i and a_{i + 1} is a multiple of 4. Determine whether Snuke can achieve his objective. | N = int(input())
A = list(map(int,input().split()))
odd = 0
even2 = 0
even4 = 0
for i in A :
if not i % 4 :
even4 += 1
elif not i % 2 :
even2 += 1
else :
odd += 1
print(odd,even2,even4)
if odd <= even4 :
ans = "Yes"
else :
ans = "No"
print(ans)
| s758680984 | Accepted | 64 | 14,252 | 243 | N = int(input())
A = list(map(int,input().split()))
odd = 0
even4 = 0
for i in A :
if not i % 4 :
even4 += 1
elif i % 2 :
odd += 1
if (odd <= even4) or (N//2 <= even4) :
ans = "Yes"
else :
ans = "No"
print(ans)
|
s832071052 | p03657 | u861141787 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 132 | 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 ca... | A, B = map(int, input().split())
if A + B % 3 == 0 or A % 3 == 0 or B % 3 == 0:
print("Possible")
else:
print("Impossible") | s046044290 | Accepted | 17 | 2,940 | 139 | A, B = map(int, input().split())
if (A + B) % 3 == 0 or A % 3 == 0 or B % 3 == 0:
print("Possible")
else:
print("Impossible")
|
s041111801 | p04011 | u495677768 | 2,000 | 262,144 | Wrong Answer | 41 | 3,064 | 367 | 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. | '''
Created on 2016/08/28
@author: diyosko7
'''
import sys
if __name__ == '__main__':
n = 4
input_lines = [int(input()) for i in range(n)]
N = input_lines[0]
K = input_lines[1]
X = input_lines[2]
Y = input_lines[3]
SUM = 0
for i in range(0,N):
if i <= K:
SUM += X
... | s793761268 | Accepted | 40 | 3,316 | 383 | '''
Created on 2016/08/28
@author: diyosko7
'''
if __name__ == '__main__':
n = 4
input_lines = [int(input()) for i in range(n)]
N = int(input_lines[0])
K = int(input_lines[1])
X = int(input_lines[2])
Y = int(input_lines[3])
SUM = 0
for i in range(0,N):
if i < K:
... |
s208337679 | p02409 | u447630054 | 1,000 | 131,072 | Wrong Answer | 30 | 6,724 | 356 | You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth fl... | data = [[[0 for r in range(10)] for f in range(3)] for b in range(4)]
n = int(input())
for nc in range(n):
(b,f,r,v) = [int(i) for i in input().split()]
data[b - 1][f - 1][r - 1] += v
for b in range(4):
for f in range(3):
for r in range(10):
print('{0}'.format(data[b][f][r]), end='')... | s913369963 | Accepted | 30 | 6,724 | 378 | data = [[[0 for r in range(10)] for f in range(3)] for b in range(4)]
n = int(input())
for nc in range(n):
(b,f,r,v) = [int(i) for i in input().split()]
data[b - 1][f - 1][r - 1] += v
for b in range(4):
for f in range(3):
for r in range(10):
print('',data[b][f][r], end='')
p... |
s231533093 | p03447 | u477320129 | 2,000 | 262,144 | Wrong Answer | 27 | 9,088 | 69 | You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping? | X = int(input())
A = int(input())
B = int(input())
print((X-A) // B)
| s743515578 | Accepted | 28 | 9,152 | 68 | X = int(input())
A = int(input())
B = int(input())
print((X-A) % B)
|
s410463700 | p03399 | u864197622 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 48 | 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 minimu... | print(min(input(),input())+min(input(),input())) | s665169521 | Accepted | 17 | 2,940 | 61 | def A(): return int(input())
print(min(A(),A())+min(A(),A())) |
s895034672 | p03814 | u243572357 | 2,000 | 262,144 | Wrong Answer | 17 | 3,512 | 65 | Snuke has decided to construct a string that starts with `A` and ends with `Z`, by taking out a substring of a string s (that is, a consecutive part of s). Find the greatest length of the string Snuke can construct. Here, the test set guarantees that there always exists a substring of s that starts with `A` and ends w... | s = input()
print(len(s) - 1 - s[::-1].index('Z') - s.index('A')) | s068856934 | Accepted | 18 | 3,500 | 67 | a = input()
l = len(a) - a.index('A') - a[::-1].index('Z')
print(l) |
s434614387 | p03997 | u822179469 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 71 | 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 = map(int,[input() for i in range(3)])
s = (a+b)*h/2
print(s) | s508323287 | Accepted | 17 | 2,940 | 76 | a, b, h = map(int,[input() for i in range(3)])
s = (a+b)*h/2
print(int(s)) |
s133208636 | p03739 | u513081876 | 2,000 | 262,144 | Wrong Answer | 2,105 | 64,352 | 550 | You are given an integer sequence of length N. The i-th term in the sequence is a_i. In one operation, you can select a term and either increment or decrement it by one. At least how many operations are necessary to satisfy the following conditions? * For every i (1≤i≤n), the sum of the terms from the 1-st through ... | N = int(input())
a = [int(i) for i in input().split()]
def pura(A):
ans = 0
check = 0
if A[0] <= 0:
ans += abs(1-A[0])
A[0] = 1
check += A[0]
for i in range(1, N):
if i % 2 != 0:
if check + A[i] >= 0:
ans += abs(A[i] + 1 + check)
... | s009439359 | Accepted | 159 | 14,468 | 884 | n = int(input())
a = [int(i) for i in input().split()]
ans1 = 0
ans2 = 0
summ = 0
summ2 = 0
a2 = a[:]
for i in range(n):
if i % 2 == 0:
if summ + a[i] <= 0:
ans1 += abs(summ + a[i]) + 1
a[i] = -summ + 1
summ = 1
else:
summ += a[i]
else:
i... |
s727062062 | p02578 | u382639013 | 2,000 | 1,048,576 | Wrong Answer | 136 | 32,256 | 148 | N persons are standing in a row. The height of the i-th person from the front is A_i. We want to have each person stand on a stool of some heights - at least zero - so that the following condition is satisfied for every person: Condition: Nobody in front of the person is taller than the person. Here, the height of a ... | n = int(input())
a = list(map(int, input().split()))
ans = 0
for i in range(len(a)-1):
if a[i]>a[i+1]:
ans += a[i] - a[i+1]
print(ans) | s286446351 | Accepted | 141 | 32,064 | 207 | n = int(input())
a = list(map(int, input().split()))
ans = 0
max_a = a[0]
for i in range(len(a)-1):
if max_a<a[i+1]:
max_a = a[i+1]
if max_a>a[i+1]:
ans += max_a - a[i+1]
print(ans) |
s950784354 | p03448 | u158778550 | 2,000 | 262,144 | Wrong Answer | 54 | 3,060 | 237 | 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 co... | A = int(input())
B = int(input())
C = int(input())
X = int(input())
counter = 0
for a in range(A):
for b in range(B):
for c in range(C):
x = 500*a + 100*b + 50*c
if X == x:
counter += 1
print(counter) | s630946437 | Accepted | 56 | 3,060 | 243 | A = int(input())
B = int(input())
C = int(input())
X = int(input())
counter = 0
for a in range(A+1):
for b in range(B+1):
for c in range(C+1):
x = 500*a + 100*b + 50*c
if X == x:
counter += 1
print(counter) |
s718935747 | p03777 | u792078574 | 2,000 | 262,144 | Wrong Answer | 21 | 8,920 | 65 | 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`, AtCo... | a, b = input().split()
if a == b:
print('D')
else:
print('H') | s020010662 | Accepted | 27 | 9,020 | 65 | a, b = input().split()
if a == b:
print('H')
else:
print('D') |
s220036909 | p03549 | u038408819 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 95 | Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`. When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases. Then, he rewrote the code to correctly solve each of th... | N, M = map(int, input().split())
ms = 1900 * M + 100 * (N - M)
p = (1 / 2) ** M
print(ms*p**-1) | s339220118 | Accepted | 17 | 2,940 | 101 | N, M = map(int, input().split())
ms = 1900 * M + 100 * (N - M)
p = (1 / 2) ** M
print(int(ms*p**-1))
|
s986739500 | p03377 | u483945851 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 83 | There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals. | A, B, X=map(int,input().split())
if A<=X<=(A+B):
print("Yes")
else:
print("No") | s748050344 | Accepted | 17 | 2,940 | 83 | A, B, X=map(int,input().split())
if A<=X<=(A+B):
print("YES")
else:
print("NO") |
s571769628 | p02694 | u505181116 | 2,000 | 1,048,576 | Wrong Answer | 24 | 9,160 | 112 | 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 abo... | x = int(input())
i = 0
money = 100
while money < x:
money *= 1.01
money = int(money)
i += 1
print(i,money) | s735455111 | Accepted | 25 | 9,160 | 107 | x = int(input())
i = 0
money = 100
while money < x:
money *= 1.01
money = int(money)
i += 1
print(i)
|
s171758388 | p03555 | u155251346 | 2,000 | 262,144 | Wrong Answer | 31 | 8,960 | 181 | You are given a grid with 2 rows and 3 columns of squares. The color of the square at the i-th row and j-th column is represented by the character C_{ij}. Write a program that prints `YES` if this grid remains the same when rotated 180 degrees, and prints `NO` otherwise. | word1 = input()
word2 = input()
up = list(word1)
down = list(word2)
if word1[0]==word2[2] and word1[1]== word2[1] and word1[2] == word2[0]:
print("Yes")
else:
print("No")
| s915803881 | Accepted | 26 | 9,008 | 220 |
A = input()
B = input()
if A[0]==B[2] and A[1]==B[1] and A[2]==B[0]:
print("YES")
else:
print("NO")
|
s041344848 | p03251 | u662449766 | 2,000 | 1,048,576 | Wrong Answer | 19 | 3,064 | 403 | Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes incli... | import sys
input = sys.stdin.readline
def main():
n, m, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
war = "No War"
for z in range(X, Y + 1):
if X <= z <= Y and all([i < z for i in x]) and all([i >= z for i in y]):
wa... | s254675834 | Accepted | 19 | 3,064 | 392 | import sys
input = sys.stdin.readline
def main():
n, m, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
war = "War"
for z in range(X + 1, Y + 1):
if all([i < z for i in x]) and all([i >= z for i in y]):
war = "No War"
... |
s340727282 | p03545 | u225183661 | 2,000 | 262,144 | Wrong Answer | 28 | 9,120 | 550 | Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given in... | num=input()
A=int(num[0])
B=int(num[1])
C=int(num[2])
D=int(num[3])
for i in range(2):
for j in range(2):
for k in range(2):
b=B*(-1)**i
c=C*(-1)**j
d=D*(-1)**k
ans=A+b+c+d
print(ans)
if ans==7:
sgn=[i,j,k]
... | s401919552 | Accepted | 29 | 9,132 | 534 | num=open(0).read()
A=int(num[0])
B=int(num[1])
C=int(num[2])
D=int(num[3])
for i in range(2):
for j in range(2):
for k in range(2):
b=B*(-1)**i
c=C*(-1)**j
d=D*(-1)**k
ans=A+b+c+d
if ans==7:
sgn=[i,j,k]
for l in rang... |
s162084699 | p03067 | u420194640 | 2,000 | 1,048,576 | Wrong Answer | 325 | 21,400 | 130 | There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively. Print `Yes` if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print `No` otherwise. | import numpy as np
def test(A, B, C):
if (A<C) & (C<B):
print('Yes')
elif (B<C) & (C<A):
print('Yes')
else:
print('No')
| s352503962 | Accepted | 17 | 3,060 | 132 | A,B,C = input().split()
A=int(A)
B=int(B)
C=int(C)
AB=abs(A-B)
ACB=abs(A-C)+abs(C-B)
if AB==ACB:
print('Yes')
else:
print('No') |
s611813495 | p03450 | u368796742 | 2,000 | 262,144 | Wrong Answer | 2,109 | 91,012 | 684 | There are N people standing on the x-axis. Let the coordinate of Person i be x_i. For every i, x_i is an integer between 0 and 10^9 (inclusive). It is possible that more than one person is standing at the same coordinate. You will given M pieces of information regarding the positions of these people. The i-th piece of... | from collections import deque
n,m = map(int,input().split())
e = [[] for i in range(n)]
for i in range(m):
a,b,c = map(int,input().split())
a -= 1
b -= 1
e[a].append([b,c])
e[b].append([a,-c])
print(e)
dis = [False]*n
def search(x):
q = deque([])
dis[x] = 0
q.append(x)
while q:
... | s371893764 | Accepted | 1,349 | 78,888 | 917 | def main():
from collections import deque
import sys
input = sys.stdin.readline
n,m = map(int,input().split())
e = [[] for i in range(n)]
for i in range(m):
a,b,c = map(int,input().split())
a -= 1
b -= 1
e[a].append([b,c])
e[b].append([a,-c])
di... |
s533944334 | p02928 | u150787983 | 2,000 | 1,048,576 | Wrong Answer | 871 | 3,188 | 459 | We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of o... | N, K = list(map(int, input().split()))
A = list(map(int, input().split()))
def cal(A, N):
cnt1 = 0
cnt2 = 0
for i in range(N):
for j in range(i+1,N):
if A[N-1-j] > A[N-1-i]:
cnt1 += 1
for i in range(N):
for j in range(N):
if A[N-1-i] < A[j]:
... | s211008185 | Accepted | 873 | 3,188 | 482 | N, K = list(map(int, input().split()))
A = list(map(int, input().split()))
def cal(A, N):
cnt1 = 0
cnt2 = 0
for i in range(N):
for j in range(i+1,N):
if A[N-1-j] > A[N-1-i]:
cnt1 += 1
for i in range(N):
for j in range(N):
if A[N-1-i] < A[j]:
... |
s515125164 | p03044 | u968404618 | 2,000 | 1,048,576 | Wrong Answer | 665 | 79,488 | 450 | 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... | import sys
sys.setrecursionlimit(10 ** 7)
def dfs(v, p, c):
color[v] = c
for vi, wi in g[v]:
if vi == p: continue
if wi % 2: dfs(vi, v, 0)
else: dfs(vi, v, 1)
return color
n = int(input())
g = [[] for _ in range(n)]
for _ in range(n-1):
u, v, w = map(int, input().split())
u... | s017750401 | Accepted | 717 | 79,492 | 454 | import sys
sys.setrecursionlimit(10 ** 7)
def dfs(v, p, c):
color[v] = c
for vi, wi in g[v]:
if vi == p: continue
if wi % 2: dfs(vi, v, 1-c)
else: dfs(vi, v, c)
return color
n = int(input())
g = [[] for _ in range(n)]
for _ in range(n-1):
u, v, w = map(int, input().split())
... |
s021061239 | p02846 | u780475861 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,064 | 398 | Takahashi and Aoki are training for long-distance races in an infinitely long straight course running from west to east. They start simultaneously at the same point and moves as follows **towards the east** : * Takahashi runs A_1 meters per minute for the first T_1 minutes, then runs at A_2 meters per minute for th... | T1, T2 = [int(i) for i in input().split()]
A1, A2 = [int(i) for i in input().split()]
B1, B2 = [int(i) for i in input().split()]
d1 = T1*(A1-B1)
d2 = T2*(A2-B2)
if d1 + d2 == 0:
print('infinity')
if (d1 > 0 and d2 > 0) or (d1 < 0 and d2 < 0):
print(0)
if d1 * (d2 - d1) > 0:
if d1 % (d2 - d1) != 0:
... | s272551309 | Accepted | 17 | 3,064 | 369 | T1, T2 = [int(i) for i in input().split()]
A1, A2 = [int(i) for i in input().split()]
B1, B2 = [int(i) for i in input().split()]
d1 = T1*(A1-B1)
d2 = T2*(B2-A2)
if d1 == d2:
print('infinity')
if d1 * (d2 - d1) < 0:
print(0)
if d1 * (d2 - d1) > 0:
if d1 % (d2 - d1) != 0:
print(d1 // (d2 - d1) * 2... |
s836088529 | p04029 | u951601135 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 56 | 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())
x=0
for i in range(N):
x+=i**2
print(x) | s024790188 | Accepted | 18 | 2,940 | 57 | N=int(input())
x=0
for i in range(N):
x+=(i+1)
print(x) |
s555269631 | p02613 | u533039576 | 2,000 | 1,048,576 | Wrong Answer | 152 | 9,180 | 225 | 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`,... | n = int(input())
cnt = {'AC': 0, 'WA': 0, 'TLE': 0, 'RE': 0}
for _ in range(n):
si = input()
cnt[si] += 1
print(f'AC x {cnt["AC"]}')
print(f'WA x {cnt["WA"]}')
print(f'TLT x {cnt["TLE"]}')
print(f'RE x {cnt["RE"]}')
| s950577017 | Accepted | 147 | 9,180 | 225 | n = int(input())
cnt = {'AC': 0, 'WA': 0, 'TLE': 0, 'RE': 0}
for _ in range(n):
si = input()
cnt[si] += 1
print(f'AC x {cnt["AC"]}')
print(f'WA x {cnt["WA"]}')
print(f'TLE x {cnt["TLE"]}')
print(f'RE x {cnt["RE"]}')
|
s473043318 | p03251 | u780475861 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 189 | Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes incli... | n,m,x,y = map(int,input().split())
x1 = max(map(int,input().split())) + 1
y1 = min(map(int,input().split()))
if x1 <= y1 and x1 <= y and y1 > x:
print('No war')
else:
print('War') | s911715305 | Accepted | 17 | 2,940 | 189 | n,m,x,y = map(int,input().split())
x1 = max(map(int,input().split())) + 1
y1 = min(map(int,input().split()))
if x1 <= y1 and x1 <= y and y1 > x:
print('No War')
else:
print('War') |
s856669578 | p02831 | u670961163 | 2,000 | 1,048,576 | Wrong Answer | 27 | 9,136 | 136 | Takahashi is organizing a party. At the party, each guest will receive one or more snack pieces. Takahashi predicts that the number of guests at this party will be A or B. Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted. We assume that a piece cannot be ... | def main():
a, b = map(int, input().split())
import math
print(math.gcd(a,b))
if __name__ == "__main__":
main()
| s428910792 | Accepted | 30 | 9,044 | 145 | def main():
a, b = map(int, input().split())
import math
print(int(a*b/math.gcd(a,b)))
if __name__ == "__main__":
main()
|
s433308337 | p02613 | u115877451 | 2,000 | 1,048,576 | Wrong Answer | 143 | 16,536 | 447 | 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`,... | import collections
a=int(input())
b=[input() for i in range(a)]
c=collections.Counter(b)
n,m=zip(*c.most_common())
n=list(n)
m=list(m)
result=[0]*4
for i in range(len(n)):
if n[i]=='AC':
result[0]=m[i]
if n[i]=='WA':
result[1]=m[i]
if n[i]=='TLE':
result[2]=m[i]
if n[i]=='RE':
... | s766730208 | Accepted | 142 | 16,520 | 444 | import collections
a=int(input())
b=[input() for i in range(a)]
c=collections.Counter(b)
n,m=zip(*c.most_common())
n=list(n)
m=list(m)
result=[0]*4
for i in range(len(n)):
if n[i]=='AC':
result[0]=m[i]
if n[i]=='WA':
result[1]=m[i]
if n[i]=='TLE':
result[2]=m[i]
if n[i]=='RE':
... |
s577205186 | p02613 | u225493896 | 2,000 | 1,048,576 | Wrong Answer | 146 | 9,204 | 290 | 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`,... | N = int(input())
C = [ 0 for i in range(4) ]
s2num = {"AC":0, "WA":1, "TLE":2, "RE":3}
num2s = {0:"AC", 1:"WA", 2:"TLE", 3:"RE"}
# read
for i in range(N):
num = s2num[ input() ]
C[num] += 1
#print(num)
# print
for i in range(4):
print("{} × {}".format( num2s[i], C[i] ))
| s511158312 | Accepted | 151 | 9,200 | 289 | N = int(input())
C = [ 0 for i in range(4) ]
s2num = {"AC":0, "WA":1, "TLE":2, "RE":3}
num2s = {0:"AC", 1:"WA", 2:"TLE", 3:"RE"}
# read
for i in range(N):
num = s2num[ input() ]
C[num] += 1
#print(num)
# print
for i in range(4):
print("{} x {}".format( num2s[i], C[i] ))
|
s046994498 | p02795 | u595952233 | 2,000 | 1,048,576 | Wrong Answer | 24 | 9,092 | 86 | We have a grid with H rows and W columns, where all the squares are initially white. You will perform some number of painting operations on the grid. In one operation, you can do one of the following two actions: * Choose one row, then paint all the squares in that row black. * Choose one column, then paint all t... | n, h, w = [int(input()) for i in range(3)]
k = max(h, w)
ans = (n+k-1)//k
print(ans) | s139698725 | Accepted | 30 | 9,096 | 83 | h, w, n= [int(input()) for i in range(3)]
k = max(h, w)
ans = (n+k-1)//k
print(ans) |
s006177676 | p02397 | u706023549 | 1,000 | 131,072 | Wrong Answer | 10 | 5,604 | 204 | Write a program which reads two integers x and y, and prints them in ascending order. | # coding:utf-8
n = list(map(int, input().split()))
print(n)
if n[0] > n[1]:
i = n[0]
n[0] = n[1]
n[1] = i
print (str(n[0]) + " " + str(n[1]))
else:
print (str(n[0]) + " " + str(n[1]))
| s201876173 | Accepted | 70 | 5,620 | 251 | n = []
i = 0
while i < 3000:
n = [int(i) for i in input().split()]
if n[0] == 0 and n[1] == 0:
break
m = 0
if n[0] > n[1]:
m = n[0]
n[0] = n[1]
n[1] = m
print(str(n[0]) + " " + str(n[1]))
i += 1
|
s497156850 | p03150 | u886747123 | 2,000 | 1,048,576 | Wrong Answer | 19 | 3,060 | 213 | A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string. | S = input()
if S == "keyence":
print("Yes")
exit()
for i in range(len(S)-1):
for j in range(i+1, len(S)):
if S[:i]+S[j:] == "keyence":
print("Yes")
exit()
print("No") | s061621093 | Accepted | 18 | 3,060 | 213 | S = input()
if S == "keyence":
print("YES")
exit()
for i in range(len(S)-1):
for j in range(i+1, len(S)):
if S[:i]+S[j:] == "keyence":
print("YES")
exit()
print("NO") |
s112872447 | p03644 | u045270305 | 2,000 | 262,144 | Wrong Answer | 18 | 2,940 | 138 | 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... | N = int(input())
val = 1
res = 0
while val <= N:
val *= 2
print('val = ', val)
mod_2 = divmod(val, 2)
mod_2 = mod_2[0]
print(mod_2)
| s576767592 | Accepted | 17 | 2,940 | 108 | N = int(input())
val = 1
while val <= N:
val *= 2
mod_2 = divmod(val, 2)
mod_2 = mod_2[0]
print(mod_2) |
s500882142 | p03606 | u773686010 | 2,000 | 262,144 | Wrong Answer | 27 | 9,044 | 164 | Joisino is working as a receptionist at a theater. The theater has 100000 seats, numbered from 1 to 100000. According to her memo, N groups of audiences have come so far, and the i-th group occupies the consecutive seats from Seat l_i to Seat r_i (inclusive). How many people are sitting at the theater now? |
N = int(input())
ans = 0
for i in range(N):
st,sp = map(int,input().split())
ans += abs(sp-st)
print(ans) | s151464428 | Accepted | 31 | 9,180 | 166 |
N = int(input())
ans = 0
for i in range(N):
st,sp = map(int,input().split())
ans += abs(sp-st+1)
print(ans) |
s423141495 | p02928 | u762420987 | 2,000 | 1,048,576 | Wrong Answer | 185 | 14,532 | 413 | We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}. Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2. Find the inversion number of B, modulo 10^9 + 7. Here the inversion number of B is defined as the number of o... | import numpy as np
N,K = map(int,input().split())
Alist = np.array(list(map(int,input().split())))
counter = 0
for i,A in enumerate(Alist,0):
fir_tenti = np.count_nonzero(A>Alist[i:])
print(fir_tenti)
counter += (K*(fir_tenti*(K+1)))/2
second_tenti = np.count_nonzero(A>Alist)
print(second_tenti)
... | s991005362 | Accepted | 945 | 3,188 | 564 | N, K = map(int, input().split())
Alist = list(map(int, input().split()))
mod = 10**9 + 7
def sum_n(n):
return (n * (n+1))//2
ans = 0
left = [0] * N
right = [0] * N
for i in range(N):
a = Alist[i]
in_left = True
for j in range(N):
if i == j:
in_left = False
continue
... |
s050338511 | p04043 | u692178082 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 238 | 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 ... |
example = ('5', '7', '5')
five = 0
seven = 0
ipt = []
s = input(">")
s = s.split()
for i in s:
if i == '5':
five += 1
elif i == '7':
seven += 1
if five == 2 and seven == 1:
print("YES")
else:
print("NO") | s921057498 | Accepted | 17 | 3,060 | 235 |
example = ('5', '7', '5')
five = 0
seven = 0
ipt = []
s = input()
s = s.split()
for i in s:
if i == '5':
five += 1
elif i == '7':
seven += 1
if five == 2 and seven == 1:
print("YES")
else:
print("NO") |
s250255447 | p02843 | u134520518 | 2,000 | 1,048,576 | Wrong Answer | 17 | 2,940 | 249 | AtCoder Mart sells 1000000 of each of the six items below: * Riceballs, priced at 100 yen (the currency of Japan) each * Sandwiches, priced at 101 yen each * Cookies, priced at 102 yen each * Cakes, priced at 103 yen each * Candies, priced at 104 yen each * Computers, priced at 105 yen each Takahashi want... | x = int(input())
nokori = x%100
max_n = x//100
print(nokori//5 + nokori%5//4 + nokori%5%4//3 + nokori%5%4%3//2 + nokori%5%4%3%2)
if max_n < nokori//5 + nokori%5//4 + nokori%5%4//3 + nokori%5%4%3//2 + nokori%5%4%3%2:
print(0)
else:
print(1)
| s674054461 | Accepted | 17 | 2,940 | 168 | x = int(input())
nokori = x%100
max_n = x//100
if max_n < nokori//5 + nokori%5//4 + nokori%5%4//3 + nokori%5%4%3//2 + nokori%5%4%3%2:
print(0)
else:
print(1)
|
s101784445 | p03447 | u917558625 | 2,000 | 262,144 | Wrong Answer | 29 | 9,044 | 57 | You went shopping to buy cakes and donuts with X yen (the currency of Japan). First, you bought one cake for A yen at a cake shop. Then, you bought as many donuts as possible for B yen each, at a donut shop. How much do you have left after shopping? | X=int(input())
A=int(input())
B=int(input())
print(X-A-B) | s403193399 | Accepted | 28 | 9,100 | 61 | X=int(input())
A=int(input())
B=int(input())
print((X-A-B)%B) |
s786919342 | p02396 | u229478139 | 1,000 | 131,072 | Wrong Answer | 20 | 5,504 | 64 | In the online judge system, a judge file may include multiple datasets to check whether the submitted program outputs a correct answer for each test case. This task is to practice solving a problem with multiple datasets. Write a program which reads an integer x and print it as is. Note that multiple datasets are give... | def testcase(x):
print("Case {}: {}".format(str(x),str(x)))
| s849089701 | Accepted | 140 | 5,568 | 116 | c = 0
while True:
n = input()
if n == '0':
break
c += 1
print('Case ' + str(c) + ': ' + n)
|
s969713677 | p03338 | u548545174 | 2,000 | 1,048,576 | Wrong Answer | 18 | 3,060 | 220 | You are given a string S of length N consisting of lowercase English letters. We will cut this string at one position into two strings X and Y. Here, we would like to maximize the number of different letters contained in both X and Y. Find the largest possible number of different letters contained in both X and Y when ... | N = int(input())
S = input()
ans = 0
for sakaime in range(N):
left = S[:sakaime]
right = S[sakaime:]
cnt = 0
for c in left:
if c in right:
cnt += 1
ans = max(ans, cnt)
print(ans) | s406513882 | Accepted | 25 | 3,772 | 294 | N = int(input())
S = input()
import string
abcs = string.ascii_lowercase
ans = 0
for sakaime in range(N):
cnt = 0
left = S[:sakaime]
right = S[sakaime:]
for alpha in abcs:
if alpha in left and alpha in right:
cnt += 1
ans = max(ans, cnt)
print(ans) |
s549995729 | p03409 | u934788990 | 2,000 | 262,144 | Wrong Answer | 25 | 3,064 | 512 | On a two-dimensional plane, there are N red points and N blue points. The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i). A red point and a blue point can form a _friendly pair_ when, the x-coordinate of the red point is smaller than that of the blue point, ... | n = int(input())
AB = []
for i in range(n):
a,b=map(int,input().split())
AB.append([a,b])
CD = []
for i in range(n):
c,d=map(int,input().split())
CD.append([c,d])
coutAB = []
coutCD = []
count = 0
for i in range(n):
for j in range(n):
if AB[i][0] <= CD[j][0] and AB[i][1] <= CD[j][1]:
... | s526477987 | Accepted | 18 | 3,064 | 293 | N = int(input())
x = sorted([list(map(int, input().split())) for _ in range(N)], key=lambda x: -x[1])
y = sorted([list(map(int, input().split())) for _ in range(N)])
cnt = 0
for c, d in y:
for a, b in x:
if a < c and b < d:
x.remove([a, b])
cnt += 1
break
print(cnt)
|
s327506620 | p03720 | u957872856 | 2,000 | 262,144 | Wrong Answer | 26 | 3,444 | 235 | There are N cities and M roads. The i-th road (1≤i≤M) connects two cities a_i and b_i (1≤a_i,b_i≤N) bidirectionally. There may be more than one road that connects the same pair of two cities. For each city, how many roads are connected to the city? | N, M = map(int, input().split())
dp = [[0 for i in range(N)] for j in range(N)]
for i in range(M):
a, b = map(int, input().split())
dp[a-1][b-1] += 1
dp[b-1][a-1] += 1
print(dp)
for i in range(N):
print(sum(dp[i])) | s725858129 | Accepted | 18 | 3,060 | 140 | n, m = map(int,input().split())
l = []
for i in range(m):
l += list(map(int, input().split()))
for i in range(1, n+1):
print(l.count(i)) |
s160531580 | p03943 | u367130284 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 74 | 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. Not... | a=map(int,input().split())
print("Yes") if sum(a)//2 in a else print("No") | s455000586 | Accepted | 18 | 2,940 | 85 | a=list(map(int,input().split()))
print("Yes") if sum(a)/2 in list(a) else print("No") |
s077417244 | p03251 | u874741582 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 190 | Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes incli... | n,m,x,y=map(int,input().split())
X= list(map(int,input().split()))
Y= list(map(int,input().split()))
maxx = max(X)
miny = min(Y)
if maxx +1 < miny:
print("No War")
else:
print("War") | s911364434 | Accepted | 17 | 3,064 | 280 | n,m,x,y=map(int,input().split())
X= list(map(int,input().split()))
Y= list(map(int,input().split()))
maxx = max(X)
miny = min(Y)
frag = 0
for i in range(x+1,y+1):
if i >maxx and i <=miny :
frag = 1
break
if frag == 1:
print("No War")
else:
print("War") |
s422516451 | p03470 | u747005359 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 50 | An _X -layered kagami mochi_ (X ≥ 1) is a pile of X round mochi (rice cake) stacked vertically where each mochi (except the bottom one) has a smaller diameter than that of the mochi directly below it. For example, if you stack three mochi with diameters of 10, 8 and 6 centimeters from bottom to top in this order, you h... | input()
print(len(set(map(int, input().split())))) | s331431895 | Accepted | 17 | 2,940 | 67 | N = int(input())
d = [input() for i in range(N)]
print(len(set(d))) |
s417811338 | p02927 | u615820360 | 2,000 | 1,048,576 | Wrong Answer | 25 | 3,060 | 288 | Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq... | a,b=map(int,input().split())
x=0
for i in range(1,a+1):
for j in range(11,b):
lst = []
while j > 0:
lst.append(j%10)
j //= 10
lst.reverse()
if lst[0] >= 2 and lst[1] >= 2:
if i == lst[0]*lst[1]:
print(lst[0],lst[1])
x+=1
print(x)
| s670885486 | Accepted | 24 | 3,060 | 261 | a,b=map(int,input().split())
x=0
for i in range(1,a+1):
for j in range(11,b+1):
lst = []
while j > 0:
lst.append(j%10)
j //= 10
lst.reverse()
if lst[0] >= 2 and lst[1] >= 2:
if i == lst[0]*lst[1]:
x+=1
print(x)
|
s399094839 | p02833 | u387774811 | 2,000 | 1,048,576 | Wrong Answer | 17 | 3,060 | 108 | 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())
if N%2==1:
print(0)
else:
N/=2
a=0
i=0
while (5**i)<=N:
i+=1
a+=N//(5**i)
print(a) | s001168027 | Accepted | 17 | 2,940 | 114 | N= int(input())
if N%2==1:
print(0)
else:
a=0
i=0
while (5**i)*2<=N:
i+=1
b=(5**i)*2
a+=(N//b)
print(a) |
s863724498 | p02280 | u153665391 | 1,000 | 131,072 | Wrong Answer | 20 | 5,616 | 1,389 | A rooted binary tree is a tree with a root node in which every node has at most two children. Your task is to write a program which reads a rooted binary tree _T_ and prints the following information for each node _u_ of _T_ : * node ID of _u_ * parent of _u_ * sibling of _u_ * the number of children of _u_ ... | N = int(input())
binary_tree = [{"parent": -1, "sibling": -1} for _ in range(N)]
for _ in range(N):
node_input = input()
id, left, right = map(int, node_input.split())
binary_tree[id]["left"] = left
binary_tree[id]["right"] = right
degree = 0
if left != -1:
degree += 1
binary_... | s177008299 | Accepted | 20 | 5,624 | 1,947 | N = int(input())
class Node():
def __init__(self, parent = -1, left = -1, right = -1):
self.parent = parent
self.left = left
self.right = right
binary_tree = [Node() for _ in range(N)]
for _ in range(N):
node_input = input()
id, left, right = map(int, node_input.split())
binar... |
s295169836 | p03827 | u396266329 | 2,000 | 262,144 | Wrong Answer | 17 | 3,060 | 148 | You have an integer variable x. Initially, x=0. Some person gave you a string S of length N, and using the string you performed the following operation N times. In the i-th operation, you incremented the value of x by 1 if S_i=`I`, and decremented the value of x by 1 if S_i=`D`. Find the maximum value taken by x duri... | N = int(input())
S = input()
num = [0]
cu = 0
for i in S:
if i == "I":
cu += 1
num.append(cu)
else:
cu -= 1
num.sort()
print(num[0]) | s481663886 | Accepted | 17 | 3,060 | 149 | N = int(input())
S = input()
num = [0]
cu = 0
for i in S:
if i == "I":
cu += 1
num.append(cu)
else:
cu -= 1
num.sort()
print(num[-1]) |
s665421208 | p03712 | u492447501 | 2,000 | 262,144 | Wrong Answer | 20 | 3,700 | 234 | You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1. | H, W = map(int, input().split())
pic = ["#"]*(W+2)
pic_list = []
pic_list.append(pic)
for i in range(H):
pic = "#"+input()+"#"
pic_list.append(pic)
pic = ["#"]*(W+2)
pic_list.append(pic)
for i in pic_list:
print(*i)
| s562142926 | Accepted | 21 | 4,596 | 421 | H,W = map(int, input().split())
H_output = H+2
W_output = W+2
S = []
for _ in range(H):
s = list(input())
S.append(s)
count = 0
for i in range(H_output):
if i==0 or i==H+1:
row = ["#"]*(W+2)
print(*row,sep="")
continue
else:
row_left = ["#"]*1
row_right = ["#"]... |
s321890465 | p03449 | u062459048 | 2,000 | 262,144 | Wrong Answer | 18 | 3,060 | 224 | We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j). You are initially in the top-left square, (1, 1). You will travel to the bottom-right square, (2, N), by repeatedly moving right or down. The square (i, j) contains A_{i, j} candies. You ... | n = int(input())
a1list = list(map(int, input().split()))
a2list = list(map(int, input().split()))
countlist = []
for i in range(n):
p = sum(a1list[0:i]) + sum(a2list[i:n-1])
countlist.append(p)
print(max(countlist)) | s505627405 | Accepted | 17 | 3,060 | 225 | n = int(input())
a1list = list(map(int, input().split()))
a2list = list(map(int, input().split()))
countlist = []
for i in range(n):
p = sum(a1list[0:i+1]) + sum(a2list[i:n])
countlist.append(p)
print(max(countlist))
|
s119041529 | p03693 | u102242691 | 2,000 | 262,144 | Wrong Answer | 17 | 3,064 | 134 | 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 i... |
r,g,b = input().split()
r = int(r)
g = int(g)
b = int(b)
if (r * 100 + g * 10 * b) % 2 == 0:
print("Yes")
else:
print("No")
| s876044232 | Accepted | 18 | 2,940 | 147 | r,g,b = input().split()
r = int(r)
g = int(g)
b = int(b)
number = r * 100 + g * 10 + b
if number % 4 == 0:
print("YES")
else:
print("NO") |
s705507962 | p03150 | u288948615 | 2,000 | 1,048,576 | Wrong Answer | 20 | 3,060 | 104 | A string is called a KEYENCE string when it can be changed to `keyence` by removing its contiguous substring (possibly empty) only once. Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string. | S = input()
if S.startswith('keyence') or S.endswith('keyence'):
print('YES')
else:
print('NO') | s482306634 | Accepted | 18 | 3,060 | 256 | S = input()
KEYENCE = 'keyence'
patterns = []
for i in range(len(KEYENCE)):
patterns.append((KEYENCE[:i:], KEYENCE[i::]))
for stt, end in patterns:
if S.startswith(stt) and S.endswith(end):
print('YES')
break
else:
print('NO') |
s168542958 | p03624 | u923662841 | 2,000 | 262,144 | Wrong Answer | 26 | 3,832 | 147 | 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. | import string
import sys
a = string.ascii_lowercase
S =sys.stdin.readline()
a = sorted(set(a)^set(S))
if a:
print(a[0])
else:
print("None") | s681589753 | Accepted | 26 | 3,832 | 123 | import string
a = string.ascii_lowercase
S =input()
a = sorted(set(a)^set(S))
if a:
print(a[0])
else:
print("None") |
s899477287 | p03729 | u444722572 | 2,000 | 262,144 | Wrong Answer | 17 | 2,940 | 113 | You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `Y... | A,B,C=input().split()
a,b=len(A),len(B)
if(A[a-1]==B[0] and B[b-1]==C[0]):
print("Yes")
else:
print("No") | s199681540 | Accepted | 19 | 3,060 | 113 | A,B,C=input().split()
a,b=len(A),len(B)
if(A[a-1]==B[0] and B[b-1]==C[0]):
print("YES")
else:
print("NO") |
s489564777 | p03379 | u179169725 | 2,000 | 262,144 | Wrong Answer | 239 | 26,772 | 140 | 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, ..... | N = int(input())
X = list(map(int, input().split()))
s = N // 2
for _ in range(s):
print(X[s - 1])
for _ in range(s, N):
print(X[s]) | s203878248 | Accepted | 339 | 26,180 | 231 | N = int(input())
X = list(map(int, input().split()))
X_sort = X.copy()
X_sort.sort()
s = N // 2
median = (X_sort[s - 1] + X_sort[s]) / 2
for x in X:
if x < median:
print(X_sort[s])
else:
print(X_sort[s-1])
|
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