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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s006857839
|
p03563
|
u032599513
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 51 |
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.
|
R = float(input())
G = float(input())
print(2*G-R)
|
s712133969
|
Accepted
| 17 | 2,940 | 47 |
R = int(input())
G = int(input())
print(2*G-R)
|
s529154342
|
p03814
|
u827202523
| 2,000 | 262,144 |
Wrong Answer
| 69 | 3,500 | 166 |
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 with `Z`.
|
s = input()
aind = 0
for i , c in enumerate(s[::-1]):
if c == "A":
aind = i
zind = 0
for i,c in enumerate(s):
if c == "Z":
zind = i
print(zind-aind)
|
s326240630
|
Accepted
| 44 | 3,516 | 176 |
s = input()
aind = 0
for i , c in enumerate(s):
if c == "A":
aind = i
break
zind = 0
for i,c in enumerate(s):
if c == "Z":
zind = i
print(zind-aind + 1)
|
s847330160
|
p03370
|
u551692187
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 99 |
Akaki, a patissier, can make N kinds of doughnut using only a certain powder called "Okashi no Moto" (literally "material of pastry", simply called Moto below) as ingredient. These doughnuts are called Doughnut 1, Doughnut 2, ..., Doughnut N. In order to make one Doughnut i (1 ≤ i ≤ N), she needs to consume m_i grams of Moto. She cannot make a non-integer number of doughnuts, such as 0.5 doughnuts. Now, she has X grams of Moto. She decides to make as many doughnuts as possible for a party tonight. However, since the tastes of the guests differ, she will obey the following condition: * For each of the N kinds of doughnuts, make at least one doughnut of that kind. At most how many doughnuts can be made here? She does not necessarily need to consume all of her Moto. Also, under the constraints of this problem, it is always possible to obey the condition.
|
N,X = map(int, input().split())
m = [int(input()) for i in range(N)]
print(X-sum(m) // min(m) + N)
|
s477054242
|
Accepted
| 17 | 2,940 | 100 |
N,X = map(int, input().split())
m = [int(input()) for i in range(N)]
print((X-sum(m)) // min(m) + N)
|
s751437530
|
p03352
|
u629607744
| 2,000 | 1,048,576 |
Wrong Answer
| 60 | 8,964 | 150 |
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())
min = 1000
for i in range(1,1000):
for j in range(2,40):
if 1000-i**j < min and 0 < 1000-i**j:
min = 1000-i**j
print(1000-min)
|
s170111937
|
Accepted
| 62 | 9,084 | 139 |
x = int(input())
min = 10000
for i in range(1,1000):
for j in range(2,40):
if x-i**j < min and 0 <= x-i**j:
min = x-i**j
print(x-min)
|
s734721496
|
p02255
|
u546285759
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,524 | 235 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
N = int(input())
seq = list(map(int, input().split()))
for i in range(1, N):
key = seq[i]
j = i-1
while j >= 0 and seq[j] > key:
seq[j+1] = seq[j]
j -= 1
seq[j+1] = key
print(" ".join(map(str, seq)))
|
s050615102
|
Accepted
| 20 | 7,652 | 266 |
N = int(input())
seq = list(map(int, input().split()))
print(" ".join(map(str, seq)))
for i in range(1, N):
key = seq[i]
j = i-1
while j >= 0 and seq[j] > key:
seq[j+1] = seq[j]
j -= 1
seq[j+1] = key
print(" ".join(map(str, seq)))
|
s142784696
|
p03573
|
u098968285
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 90 |
You are given three integers, A, B and C. Among them, two are the same, but the remaining one is different from the rest. For example, when A=5,B=7,C=5, A and C are the same, but B is different. Find the one that is different from the rest among the given three integers.
|
a, b, c = map(int, input().split())
if a == b or a == c:
print(a)
else:
print(b)
|
s877394467
|
Accepted
| 17 | 2,940 | 105 |
a, b, c = map(int, input().split())
if a == b:
print(c)
elif a == c:
print(b)
else:
print(a)
|
s603785334
|
p03545
|
u694467277
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 307 |
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
x=[int(k) for k in input()]
for i in range(2*3):
total=x[0]
bag=[""]*3
for j in range(3):
if (i>>j)&1:
bag[j]="+"
total+=x[j+1]
else:
bag[j]="-"
total-=x[j+1]
if total==7:
print("{}{}{}{}{}{}{}=7".format(x[0],bag[0],x[1],bag[1],x[2],bag[2],x[3]))
break
|
s701560032
|
Accepted
| 18 | 3,064 | 315 |
x=[int(k) for k in input()]
for i in range(2**3):
total=x[0]
bag=[""]*3
for j in range(3):
if (i>>j)&1:
bag[j]="+"
total+=x[j+1]
else:
bag[j]="-"
total-=x[j+1]
if total==7:
print("{}{}{}{}{}{}{}=7".format(x[0],bag[0],x[1],bag[1],x[2],bag[2],x[3]))
break
|
s891215155
|
p03251
|
u367130284
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 2,940 | 85 |
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
_,b,c,d,*e=map(int,open(0).read().split());print((max(e[:b])<min(e[b:]))*"No "+"War")
|
s673093048
|
Accepted
| 18 | 3,060 | 173 |
n,m,x,y=map(int,input().split())
zmin=max(map(int,input().split()))
zmax=min(map(int,input().split()))
if max(zmin,x)<min(zmax,y):
print("No War")
else:
print("War")
|
s423150601
|
p03110
|
u923270446
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 2,940 | 188 |
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
n = int(input())
xu = [input().split() for i in range(n)]
ans = 0
for i in xu:
if i[1] == "JPY":
ans += int(i[0])
else:
ans += int(float(i[0]) * 380000)
print(ans)
|
s421172880
|
Accepted
| 18 | 3,060 | 198 |
n = int(input())
xu = [input().split() for i in range(n)]
ans = 0
for i in xu:
if i[1] == "JPY":
ans += int(i[0])
elif i[1] == "BTC":
ans += float(i[0]) * 380000.0
print(ans)
|
s765096176
|
p03369
|
u361826811
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 245 |
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
# import numpy as np
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
S = readline().decode('utf8')
print(1000-S.count('x'))
|
s227723748
|
Accepted
| 17 | 2,940 | 250 |
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(1000-S.count('x')*100)
|
s568014279
|
p03997
|
u483645888
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 69 |
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*0.5)
|
s218961220
|
Accepted
| 18 | 2,940 | 76 |
a = int(input())
b = int(input())
h = int(input())
print(round((a+b)*h*0.5))
|
s698352521
|
p03524
|
u131881594
| 2,000 | 262,144 |
Wrong Answer
| 26 | 3,444 | 63 |
Snuke has a string S consisting of three kinds of letters: `a`, `b` and `c`. He has a phobia for palindromes, and wants to permute the characters in S so that S will not contain a palindrome of length 2 or more as a substring. Determine whether this is possible.
|
from collections import Counter
s=Counter(input())
print("NO")
|
s178982001
|
Accepted
| 26 | 3,444 | 295 |
from collections import Counter
s=Counter(input())
b,m=max(s.values()),min(s.values())
kind = len(s.keys())
if kind==3:
if b-m<=1: print("YES")
else: print("NO")
elif kind==2:
if b==1 and m==1: print("YES")
else: print("NO")
else:
if b==1: print("YES")
else: print("NO")
|
s480557681
|
p02398
|
u015712946
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 101 |
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
|
a,b,c = map(int, input().split())
n = 0
for i in range(a, b+1):
if i % c == 0:
n += 1
print(n)
|
s336911140
|
Accepted
| 20 | 5,604 | 87 |
a,b,c = map(int, input().split())
print(len([1 for i in range(a, b+1) if c % i == 0]))
|
s169496566
|
p03486
|
u077852398
| 2,000 | 262,144 |
Wrong Answer
| 30 | 8,916 | 86 |
You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s' < t' for the lexicographic order.
|
n = sorted(input())
m = sorted(input())
if n<m:
print('yes')
else:
print('no')
|
s969941176
|
Accepted
| 30 | 9,100 | 111 |
s = sorted(list(input()))
t = sorted(list(input()),reverse=True)
if s<t:
print('Yes')
else:
print('No')
|
s462314433
|
p04012
|
u690781906
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,316 | 148 |
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
import collections
w = input()
c = collections.Counter(w)
for k, v in c.items():
if v % 2 == 1:
print('NO')
exit()
print('YES')
|
s610300323
|
Accepted
| 21 | 3,316 | 149 |
import collections
w = input()
c = collections.Counter(w)
for k, v in c.items():
if v % 2 == 1:
print('No')
exit()
print('Yes')
|
s031763901
|
p02402
|
u539753516
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,572 | 64 |
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
a = list(map(int, input().split()))
print(min(a),max(a),sum(a))
|
s001459722
|
Accepted
| 20 | 6,540 | 72 |
input()
a = list(map(int, input().split()))
print(min(a),max(a),sum(a))
|
s391311244
|
p03992
|
u727787724
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 83 |
This contest is `CODE FESTIVAL`. However, Mr. Takahashi always writes it `CODEFESTIVAL`, omitting the single space between `CODE` and `FESTIVAL`. So he has decided to make a program that puts the single space he omitted. You are given a string s with 12 letters. Output the string putting a single space between the first 4 letters and last 8 letters in the string s.
|
s=list(input())
a=s[0]+s[1]+s[2]+s[3]
b=''
for i in range(8):
b=s[4+i]
print(a,b)
|
s931961419
|
Accepted
| 17 | 2,940 | 85 |
s=list(input())
a=s[0]+s[1]+s[2]+s[3]
b=''
for i in range(8):
b+=s[4+i]
print(a,b)
|
s285703095
|
p03456
|
u313111801
| 2,000 | 262,144 |
Wrong Answer
| 25 | 9,188 | 113 |
AtCoDeer the deer has found two positive integers, a and b. Determine whether the concatenation of a and b in this order is a square number.
|
a,b=map(int,input().split())
x=a+b
q=[i*i for i in range(1000)]
if x in q:
print("Yes")
else:
print("No")
|
s653971232
|
Accepted
| 27 | 9,016 | 110 |
x=int("".join(input().split()))
q=[i*i for i in range(1000)]
if x in q:
print("Yes")
else:
print("No")
|
s124969453
|
p03731
|
u712187387
| 2,000 | 262,144 |
Wrong Answer
| 68 | 26,708 | 247 |
In a public bath, there is a shower which emits water for T seconds when the switch is pushed. If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds. N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it. How long will the shower emit water in total?
|
N,T = map(int,input().split())
t=list(map(int,input().split()))
def count_time(N,T,t):
cnt = T
for i in range(N-1):
tt =t[i+1]-t[i]
if tt <=T:
cnt += tt
else:
cnt += T
return cnt
|
s246549826
|
Accepted
| 98 | 25,200 | 273 |
N,T = map(int,input().split())
t=list(map(int,input().split()))
def count_time(N,T,t):
cnt = T
for i in range(N-1):
tt =t[i+1]-t[i]
if tt <=T:
cnt += tt
else:
cnt += T
return cnt
print(count_time(N,T,t))
|
s468536888
|
p03361
|
u202634017
| 2,000 | 262,144 |
Wrong Answer
| 37 | 9,228 | 660 |
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j). Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`. However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black. Determine if square1001 can achieve his objective.
|
h, w = map(int, input().split())
s = [list(input()) for _ in range(h)]
f = True
for i in range(h):
for j in range(w):
if s[i][j] == "#":
dx = [0]
dy = [0]
if i != 0:
dx.append(-1)
if i != h - 1:
dx.append(1)
if j != 0:
dy.append(-1)
if j != w - 1:
dy.append(1)
ff = False
for x in dx:
for y in dy:
if abs(x) == abs(y):
continue
ff |= (s[i + x][j + y] == "#")
f &= ff
print("YES" if f else "NO")
|
s354878825
|
Accepted
| 34 | 9,212 | 660 |
h, w = map(int, input().split())
s = [list(input()) for _ in range(h)]
f = True
for i in range(h):
for j in range(w):
if s[i][j] == "#":
dx = [0]
dy = [0]
if i != 0:
dx.append(-1)
if i != h - 1:
dx.append(1)
if j != 0:
dy.append(-1)
if j != w - 1:
dy.append(1)
ff = False
for x in dx:
for y in dy:
if abs(x) == abs(y):
continue
ff |= (s[i + x][j + y] == "#")
f &= ff
print("Yes" if f else "No")
|
s259353880
|
p02613
|
u015647294
| 2,000 | 1,048,576 |
Wrong Answer
| 144 | 16,256 | 248 |
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 i in range(N):
S.append(input())
AC = S.count("AC")
WA = S.count("WA")
TLE = S.count("TLE")
RE = S.count("RE")
print("AC × " + str(AC))
print("WA × " + str(WA))
print("TLE × " + str(TLE))
print("RE × " + str(RE))
|
s674356944
|
Accepted
| 147 | 16,292 | 244 |
N = int(input())
S = []
for i in range(N):
S.append(input())
AC = S.count("AC")
WA = S.count("WA")
TLE = S.count("TLE")
RE = S.count("RE")
print("AC x " + str(AC))
print("WA x " + str(WA))
print("TLE x " + str(TLE))
print("RE x " + str(RE))
|
s986570718
|
p03050
|
u670180528
| 2,000 | 1,048,576 |
Wrong Answer
| 147 | 2,940 | 100 |
Snuke received a positive integer N from Takahashi. A positive integer m is called a _favorite number_ when the following condition is satisfied: * The quotient and remainder of N divided by m are equal, that is, \lfloor \frac{N}{m} \rfloor = N \bmod m holds. Find all favorite numbers and print the sum of those.
|
n=int(input())
c=0
for i in range(1,int(n**.5)+1):
if n%i==0:
if n//i-i>1:
c+=1
print(c)
|
s118080764
|
Accepted
| 152 | 3,060 | 105 |
n=int(input())
c=0
for i in range(1,int(n**.5)+1):
if n%i==0:
if n//i-i>1:
c+=n//i-1
print(c)
|
s799456658
|
p03609
|
u143318682
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 82 |
We have a sandglass that runs for X seconds. The sand drops from the upper bulb at a rate of 1 gram per second. That is, the upper bulb initially contains X grams of sand. How many grams of sand will the upper bulb contains after t seconds?
|
X, t = map(int, input().split())
if X < t:
print('X - t')
else:
print('0')
|
s012347182
|
Accepted
| 17 | 2,940 | 80 |
X, t = map(int, input().split())
if X > t:
print(X - t)
else:
print('0')
|
s071687768
|
p03814
|
u075303794
| 2,000 | 262,144 |
Wrong Answer
| 52 | 12,044 | 192 |
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 with `Z`.
|
S=[x for x in input()]
T=S[::-1]
start=0
end=0
for i in range(len(S)):
if S[i]=='A':
start=i
break
for i in range(len(T)):
if T[i]=='Z':
end=i
break
print(end-start+1)
|
s363424721
|
Accepted
| 51 | 12,100 | 197 |
S=[x for x in input()]
T=S[::-1]
start=0
end=0
for i in range(len(S)):
if S[i]=='A':
start=i
break
for i in range(len(T)):
if T[i]=='Z':
end=i
break
print(len(S)-end-start)
|
s550352506
|
p03777
|
u904804404
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 59 |
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.
|
if len(input().split())==2:
print("D")
else:
print("H")
|
s160220635
|
Accepted
| 17 | 2,940 | 64 |
if len(set(input().split()))==2:
print("D")
else:
print("H")
|
s958755483
|
p02607
|
u197237612
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 8,996 | 126 |
We have N squares assigned the numbers 1,2,3,\ldots,N. Each square has an integer written on it, and the integer written on Square i is a_i. How many squares i satisfy both of the following conditions? * The assigned number, i, is odd. * The written integer is odd.
|
n = int(input())
a = list(map(int, input().split()))
cnt = 0
for i in range(n, 1):
if i%2 != 0:
cnt += 1
print(cnt)
|
s641570598
|
Accepted
| 23 | 9,108 | 154 |
n = int(input())
a = list(map(int, input().split()))
ans = a[0::2]
cnt = 0
for i in range(len(ans)):
if (ans[i] % 2) != 0:
cnt += 1
print(cnt)
|
s215644787
|
p03157
|
u077291787
| 2,000 | 1,048,576 |
Wrong Answer
| 662 | 20,572 | 870 |
There is a grid with H rows and W columns, where each square is painted black or white. You are given H strings S_1, S_2, ..., S_H, each of length W. If the square at the i-th row from the top and the j-th column from the left is painted black, the j-th character in the string S_i is `#`; if that square is painted white, the j-th character in the string S_i is `.`. Find the number of pairs of a black square c_1 and a white square c_2 that satisfy the following condition: * There is a path from the square c_1 to the square c_2 where we repeatedly move to a vertically or horizontally adjacent square through an alternating sequence of black and white squares: black, white, black, white...
|
def dfs(sx: int, sy: int) -> None:
color, stack = [0] * 2, [(sx, sy, S[sx][sy])]
while stack:
x, y, col = stack.pop()
color[col] += 1
S[x][y] = 2 # make the grid visited
for dx, dy in dxy:
nx, ny = x + dx, y + dy
if 0 <= nx < H and 0 <= ny < W and S[nx][ny] == col ^ 1:
stack.append((nx, ny, S[nx][ny]))
return color[0] * color[1]
def main():
# separate S to connected components
global H, W, S, dxy
H, W, *S = open(0).read().split()
H, W = int(H), int(W)
S = [[1 if i == "#" else 0 for i in s] for s in S]
dxy = [(1, 0), (-1, 0), (0, 1), (0, -1)]
ans = 0
for i in range(H):
for j in range(W):
if S[i][j] != 2:
ans += dfs(i, j)
print(ans)
if __name__ == "__main__":
main()
|
s558993733
|
Accepted
| 392 | 12,556 | 923 |
def dfs(sx: int, sy: int) -> None:
color, stack = [0] * 2, [(sx, sy, S[sx][sy])]
S[sx][sy] = 2 # make the grid visited
while stack:
x, y, col = stack.pop()
color[col] += 1
for dx, dy in dxy:
nx, ny = x + dx, y + dy
if 0 <= nx < H and 0 <= ny < W and S[nx][ny] == col ^ 1:
stack.append((nx, ny, S[nx][ny]))
S[nx][ny] = 2 # make the grid visited
return color[0] * color[1]
def main():
# separate S to connected components
global H, W, S, dxy
H, W, *S = open(0).read().split()
H, W = int(H), int(W)
S = [[1 if i == "#" else 0 for i in s] for s in S]
dxy = [(1, 0), (-1, 0), (0, 1), (0, -1)]
ans = 0
for i in range(H):
for j in range(W):
if S[i][j] != 2:
ans += dfs(i, j)
print(ans)
if __name__ == "__main__":
main()
|
s188072736
|
p03386
|
u617203831
| 2,000 | 262,144 |
Wrong Answer
| 2,141 | 540,824 | 119 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a,b,k=map(int,input().split())
s=[]
for i in range(b-a+1):
s.append(a+i)
A=s[0:k]
B=s[-k:]
ans=set(A+B)
print(*ans)
|
s236186261
|
Accepted
| 17 | 3,064 | 94 |
a,b,k=map(int,input().split())
r=range(a,b+1)
for i in sorted(set(r[:k])|set(r[-k:])):print(i)
|
s686323533
|
p04029
|
u779073299
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 61 |
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())
if n!=1:
print(n*(n+1)/2)
else:
print(1)
|
s027948445
|
Accepted
| 17 | 2,940 | 66 |
n = int(input())
if n!=1:
print(int(n*(n+1)/2))
else:
print(1)
|
s564947206
|
p03448
|
u131625544
| 2,000 | 262,144 |
Wrong Answer
| 42 | 3,060 | 395 |
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.
|
import sys
input = sys.stdin.readline
class AtCoder:
def main(self):
A, B, C, X = [int(input()) for _ in range(4)]
ans = 0
for a in range(A):
for b in range(B):
for c in range(C):
if 500*a+100*b+50*c == X:
ans +=1
print(ans)
if __name__ == '__main__':
AtCoder().main()
|
s432425923
|
Accepted
| 46 | 3,060 | 420 |
import sys
input = sys.stdin.readline
class AtCoder:
def main(self):
A, B, C, X = [int(input()) for _ in range(4)]
ans = 0
for a in range(A + 1):
for b in range(B + 1):
for c in range(C + 1):
if 500 * a + 100 * b + 50 * c == X:
ans += 1
print(ans)
if __name__ == '__main__':
AtCoder().main()
|
s866301147
|
p03401
|
u325282913
| 2,000 | 262,144 |
Wrong Answer
| 188 | 13,920 | 810 |
There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.
|
N = int(input())
array = list(map(int, input().split()))
root = [abs(array[0])]
first = array[0]
for i in range(1,N):
root.append(abs(first-array[i]))
first = array[i]
root.append(abs(array[N-1]))
array.append(0)
sum = sum(root)
print(root)
for i in range(N-1):
if array[i] >= 0:
if array[i] <= array[i+1]:
print(sum)
else:
print(sum-root[i]*2)
else:
if array[i] >= array[i+1]:
print(sum)
else:
print(sum-root[i]*2)
if array[N-2] > 0:
if array[N-2] >= array[N-1]:
if array[N-1] < 0:
print(sum-abs(array[N-1])*2)
else:
print(sum)
else:
print(sum-root[N-1]*2)
else:
if array[N-2] <= array[N-1]:
print(sum)
else:
print(sum-root[N-1]*2)
|
s676562738
|
Accepted
| 222 | 14,048 | 244 |
N = int(input())
arr = [0] + list(map(int, input().split())) + [0]
ans = 0
for i in range(1,len(arr)):
ans += abs(arr[i]-arr[i-1])
for i in range(1,len(arr)-1):
print(ans-abs(arr[i-1]-arr[i])-abs(arr[i+1]-arr[i])+abs(arr[i-1]-arr[i+1]))
|
s080981423
|
p03049
|
u225388820
| 2,000 | 1,048,576 |
Wrong Answer
| 45 | 3,064 | 337 |
Snuke has N strings. The i-th string is s_i. Let us concatenate these strings into one string after arranging them in some order. Find the maximum possible number of occurrences of `AB` in the resulting string.
|
n=int(input())
x,y,z,ans=0,0,0,0
for i in range(n):
s=input()
if s[0]=='B' and s[-1]=='A':
x+=1
elif s[0]=='B':
y+=1
elif s[-1]=='A':
z+=1
for i in range(len(s)-1):
if s[i]=='A' and s[i+1]=='B':
ans+=1
w=1
x1=x-1
if y==z:
w=0
if x==0:
x1==0
print(min(y,z)+x1+w+ans)
|
s759509864
|
Accepted
| 44 | 3,064 | 323 |
n=int(input())
ba,a,b,ab=0,0,0,0
for i in range(n):
s=input()
if s[0]=='B' and s[-1]=='A':
ba+=1
elif s[0]=='B':
b+=1
elif s[-1]=='A':
a+=1
for i in range(len(s)-1):
if s[i]=='A' and s[i+1]=='B':
ab+=1
if a==b and a==0 and ba!=0:
ba-=1
print(min(a,b)+ba+ab)
|
s517944579
|
p03407
|
u582104670
| 2,000 | 262,144 |
Wrong Answer
| 31 | 9,116 | 80 |
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 = list(map(int,input().split()))
if (A+B)>=C:
print('Yes')
print('No')
|
s255499839
|
Accepted
| 27 | 9,152 | 90 |
A,B,C = list(map(int,input().split()))
if (A+B)>=C:
print('Yes')
else:
print('No')
|
s250381725
|
p03196
|
u941753895
| 2,000 | 1,048,576 |
Wrong Answer
| 121 | 6,736 | 781 |
There are N integers a_1, a_2, ..., a_N not less than 1. The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P. Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N.
|
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
def LI(): return list(map(int,input().split()))
def I(): return int(input())
def LS(): return input().split()
def S(): return input()
# Factoring by trial split
def getPrimeList(n):
l=[]
t=int(math.sqrt(n))+1
for a in range(2,t):
while n%a==0:
n//=a
l.append(a)
if n!=1:
l.append(n)
return l
def main():
n,p=LI()
if n==3 and p==24:
exit()
if n==1:
print(p)
exit()
l=getPrimeList(p)
if len(l)==0:
print(1)
exit()
a=l[0]
c=1
sm=1
for x in l[1:]:
if a==x:
c+=1
else:
c=1
a=x
if c==n:
sm*=a
return sm
print(main())
|
s392617290
|
Accepted
| 121 | 5,336 | 758 |
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
def LI(): return list(map(int,input().split()))
def I(): return int(input())
def LS(): return input().split()
def S(): return input()
# Factoring by trial split
def getPrimeList(n):
l=[]
t=int(math.sqrt(n))+1
for a in range(2,t):
while n%a==0:
n//=a
l.append(a)
if n!=1:
l.append(n)
return l
def main():
n,p=LI()
if n==1:
print(p)
exit()
l=getPrimeList(p)
if len(l)==0:
print(1)
exit()
a=l[0]
c=1
sm=1
for x in l[1:]:
if a==x:
c+=1
else:
c=1
a=x
if c==n:
sm*=a
c=0
return sm
print(main())
|
s472229462
|
p03643
|
u825528847
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 242 |
This contest, _AtCoder Beginner Contest_ , is abbreviated as _ABC_. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example, ABC680 is the 680th round of ABC. What is the abbreviation for the N-th round of ABC? Write a program to output the answer.
|
N = int(input())
maxv = 0
ans = 1
for i in range(N, 0, -1):
if i % 2:
continue
tmp = 0
tmpv = i
while i % 2 == 0:
tmp += 1
i = i // 2
if tmp > maxv:
maxv = tmp
ans = tmpv
print(ans)
|
s178368700
|
Accepted
| 17 | 2,940 | 23 |
print("ABC" + input())
|
s641062018
|
p04029
|
u327532412
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 65 |
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 x in range(n):
ans += x
print(ans)
|
s144384280
|
Accepted
| 17 | 2,940 | 67 |
n = int(input())
ans = 0
for x in range(n+1):
ans += x
print(ans)
|
s399183938
|
p03719
|
u402629484
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 73 |
You are given three integers A, B and C. Determine whether C is not less than A and not greater than B.
|
A, B, C = map(int, input().split())
print('YES' if A <= C <= B else 'NO')
|
s871847176
|
Accepted
| 17 | 2,940 | 73 |
A, B, C = map(int, input().split())
print('Yes' if A <= C <= B else 'No')
|
s301727422
|
p02412
|
u203222829
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 357 |
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
# coding: utf-8
while True:
n, x = map(int, input().split())
comb = 0
if n == 0 and x == 0:
exit()
for i in range(1, n+1):
for j in range(i+1, n+1):
for k in range(j+1, n+1):
print('{} {} {}'.format(i, j, k))
if i + j + k == x:
comb += 1
print(comb)
|
s640290451
|
Accepted
| 490 | 5,600 | 359 |
# coding: utf-8
while True:
n, x = map(int, input().split())
comb = 0
if n == 0 and x == 0:
exit()
for i in range(1, n+1):
for j in range(i+1, n+1):
for k in range(j+1, n+1):
# print('{} {} {}'.format(i, j, k))
if i + j + k == x:
comb += 1
print(comb)
|
s217009362
|
p03351
|
u808569469
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,124 | 179 |
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)
AC = abs(a - c)
if AB < d and BC < d:
print("Yes")
elif AC < d:
print("YES")
else:
print("No")
|
s798977119
|
Accepted
| 28 | 9,100 | 183 |
a, b, c, d = map(int, input().split())
AB = abs(a - b)
BC = abs(b - c)
AC = abs(a - c)
if AC <= d:
print("Yes")
elif AB <= d and BC <= d:
print("Yes")
else:
print("No")
|
s085980991
|
p03353
|
u191874006
| 2,000 | 1,048,576 |
Wrong Answer
| 36 | 4,880 | 192 |
You are given a string s. Among the **different** substrings of s, print the K-th lexicographically smallest one. A substring of s is a string obtained by taking out a non-empty contiguous part in s. For example, if s = `ababc`, `a`, `bab` and `ababc` are substrings of s, while `ac`, `z` and an empty string are not. Also, we say that substrings are different when they are different as strings. Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.
|
#!usr/bin/env python3
s = input()
K = int(input())
l = {s}
for n in range(1, K+1):
for i in range(0, len(s)-n+1):
l.add(s[i:i+n])
print(l)
l = list(l)
l.sort()
print(l[K-1])
|
s486640286
|
Accepted
| 45 | 6,128 | 673 |
#!/usr/bin/env python3
#ABC97 C
import sys
import math
import bisect
import time
sys.setrecursionlimit(1000000000)
from heapq import heappush, heappop,heappushpop
from collections import defaultdict
from itertools import accumulate
from collections import Counter
from collections import deque
from operator import itemgetter
from itertools import permutations
mod = 10**9 + 7
inf = float('inf')
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
f = defaultdict(lambda : 0)
s = input()
k = I()
n = len(s)
for i in range(1,k+1):
for j in range(n-i+1):
f[s[j:j+i]] = 1
f = sorted(f.items())
print(f[k-1][0])
|
s297786147
|
p03605
|
u419963262
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,188 | 44 |
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
print("YES")if "9"in input()else print("No")
|
s961567205
|
Accepted
| 18 | 2,940 | 45 |
print("Yes")if "9"in input()else print("No")
|
s909938734
|
p03962
|
u268516119
| 2,000 | 262,144 |
Wrong Answer
| 23 | 3,444 | 440 |
AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him.
|
# coding: utf-8
# Your code here!
import copy
S=list(input())
N=len(S)
point=0
hand=copy.deepcopy(S)
panum=sum([j=="p" for j in hand])
pamax=N//2
for i in range(N):
if S[i]=="g":
pa=sum([hand[j]=="p" for j in range(i+1)])
if pa<(i+1)//2 and point<pamax-panum:
hand[i]="p"
point+=1
if point>=pamax-panum:
break
print(point)
|
s149802517
|
Accepted
| 17 | 2,940 | 106 |
abc=list(map(int,input().split()))
d=[]
for i in abc:
if i not in d:
d.append(i)
print(len(d))
|
s353541041
|
p02608
|
u444398697
| 2,000 | 1,048,576 |
Wrong Answer
| 33 | 9,116 | 299 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
n = int(input())
ans = [0]*(n+5)
for x in range(1,10):
for y in range(1,10):
for z in range(1,10):
temp = x**2 + y**2 + z**2 + x*y + y*z + z*x
if n>=temp:
print(x,y,z, temp)
ans[temp]+=1
for i in range(1,n+1):
print(ans[i])
|
s483106838
|
Accepted
| 859 | 9,120 | 267 |
n = int(input())
ans = [0]*(n+5)
for x in range(1,100):
for y in range(1,100):
for z in range(1,100):
temp = x**2 + y**2 + z**2 + x*y + y*z + z*x
if n>=temp:
ans[temp]+=1
for i in range(1,n+1):
print(ans[i])
|
s702726817
|
p02612
|
u968649733
| 2,000 | 1,048,576 |
Wrong Answer
| 31 | 9,144 | 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)
|
s633283842
|
Accepted
| 32 | 9,152 | 75 |
N = int(input())
a = N % 1000
if a == 0:
print(0)
else:
print(1000 - a)
|
s595564477
|
p03605
|
u261696185
| 2,000 | 262,144 |
Wrong Answer
| 30 | 8,980 | 70 |
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
X = str(input())
if '9' in X:
print('YES')
else:
print('NO')
|
s647260030
|
Accepted
| 28 | 8,872 | 70 |
X = str(input())
if '9' in X:
print('Yes')
else:
print('No')
|
s513917148
|
p03714
|
u826263061
| 2,000 | 262,144 |
Wrong Answer
| 370 | 38,292 | 539 |
Let N be a positive integer. There is a numerical sequence of length 3N, a = (a_1, a_2, ..., a_{3N}). Snuke is constructing a new sequence of length 2N, a', by removing exactly N elements from a without changing the order of the remaining elements. Here, the score of a' is defined as follows: (the sum of the elements in the first half of a') - (the sum of the elements in the second half of a'). Find the maximum possible score of a'.
|
import heapq
n = int(input())
a = list(map(int, input().split()))
a1 = a[:n]
ax = a[n:2*n]
a2 = list(map(lambda x: -x, a[2*n:]))
print('a2', a2)
sum_a1 = sum(a1)
fdif1 = [sum_a1]
heapq.heapify(a1)
for i in range(n):
p = heapq.heappushpop(a1,ax[i])
fdif1.append(fdif1[-1]+ax[i]-p)
sum_a2 = sum(a2)
fdif2 = [sum_a2]
heapq.heapify(a2)
print('a2', a2)
for i in range(n):
p = heapq.heappushpop(a2,-ax[-1-i])
fdif2.append(fdif2[-1]-ax[-1-i]-p)
fdif2.reverse()
fdif = [fdif1[i]+fdif2[i] for i in range(n)]
print(max(fdif))
|
s683760020
|
Accepted
| 347 | 38,276 | 543 |
import heapq
n = int(input())
a = list(map(int, input().split()))
a1 = a[:n]
ax = a[n:2*n]
a2 = list(map(lambda x: -x, a[2*n:]))
#print('a2', a2)
sum_a1 = sum(a1)
fdif1 = [sum_a1]
heapq.heapify(a1)
for i in range(n):
p = heapq.heappushpop(a1,ax[i])
fdif1.append(fdif1[-1]+ax[i]-p)
sum_a2 = sum(a2)
fdif2 = [sum_a2]
heapq.heapify(a2)
#print('a2', a2)
for i in range(n):
p = heapq.heappushpop(a2,-ax[-1-i])
fdif2.append(fdif2[-1]-ax[-1-i]-p)
fdif2.reverse()
fdif = [fdif1[i]+fdif2[i] for i in range(n+1)]
print(max(fdif))
|
s170174378
|
p03523
|
u127499732
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 294 |
You are given a string S. Takahashi can insert the character `A` at any position in this string any number of times. Can he change S into `AKIHABARA`?
|
def main():
s = str(input())
t = 'AKIHABARA'
i = 0
j = 0
while i < len(t) and j < len(s):
if t[i] == s[j]:
i += 1
j += 1
else:
i += 1
print('YES' if j == len(s) - 1 else 'NO')
if __name__ == '__main__':
main()
|
s280744996
|
Accepted
| 19 | 3,188 | 141 |
def main():
import re
f = re.match('A?KIHA?BA?RA?$',input())
print('YES'if f else 'NO')
if __name__ == '__main__':
main()
|
s778511603
|
p03548
|
u847867174
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,316 | 229 |
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 people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
def getNum():
x,y,z = map(int,input().split())
return x,y,z
def cal(x,y,z):
tmp = x / y / z
if tmp < x:
print(tmp)
else:
print(tmp - 1)
def main():
x,y,z = getNum()
cal(x,y,z)
main()
|
s070279754
|
Accepted
| 17 | 2,940 | 186 |
def getNum():
x,y,z = map(int,input().split())
return x,y,z
def cal(x,y,z):
num = int((x-z) / (y+z))
print(num)
def main():
x,y,z = getNum()
cal(x,y,z)
main()
|
s126715648
|
p03407
|
u569272329
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 88 |
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
A, B, C = map(int, input().split())
if A+B <= C:
print("Yes")
else:
print("No")
|
s087879317
|
Accepted
| 17 | 2,940 | 90 |
A, B, C = map(int, input().split())
if A + B >= C:
print("Yes")
else:
print("No")
|
s233472008
|
p02743
|
u045939752
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 129 |
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
|
a,b,c = map(int, input().split())
ans = False
if c-a-b >= 0 and a*b < (c-a-b)**2:
ans = True
print('Yes' if ans else 'No')
|
s291328886
|
Accepted
| 17 | 2,940 | 134 |
a,b,c = map(int, input().split())
ans = False
if c-a-b >= 0 and 4*a*b < (c-a-b)**2:
ans = True
print('Yes' if ans else 'No')
|
s263412477
|
p02846
|
u693953100
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 393 |
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 the subsequent T_2 minutes, and alternates between these two modes forever. * Aoki runs B_1 meters per minute for the first T_1 minutes, then runs at B_2 meters per minute for the subsequent T_2 minutes, and alternates between these two modes forever. How many times will Takahashi and Aoki meet each other, that is, come to the same point? We do not count the start of the run. If they meet infinitely many times, report that fact.
|
t1,t2 = map(int,input().split())
a1,a2 = map(int,input().split())
b1,b2 = map(int,input().split())
if a1*t1+a2*t2>=b1*t1+b2*t2:
a1,a2,b1,b2 = b1,b2,a1,a2
d = b1*t1+b2*t2 - (a1*t1+a2*t2)
print(d)
if d == 0:
print('inifinity')
elif a1*t1<b1*t1:
print(0)
else:
if (a1*t1 - b1*t1)*2%d ==0:
print(((a1*t1 - b1*t1)*2+d-1)//d)
else:
print((a1*t1 - b1*t1)//d*2+1)
|
s812936939
|
Accepted
| 23 | 3,064 | 383 |
t1,t2 = map(int,input().split())
a1,a2 = map(int,input().split())
b1,b2 = map(int,input().split())
if a1*t1+a2*t2>=b1*t1+b2*t2:
a1,a2,b1,b2 = b1,b2,a1,a2
d = b1*t1+b2*t2 - (a1*t1+a2*t2)
if d == 0:
print('infinity')
elif a1*t1<b1*t1:
print(0)
else:
if (a1*t1 - b1*t1)*2%d ==0:
print(((a1*t1 - b1*t1)*2+d-1)//d)
else:
print((a1*t1 - b1*t1)//d*2+1)
|
s662024131
|
p03377
|
u941047297
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 106 |
There are a total of A + B cats and dogs. Among them, A are known to be cats, but the remaining B are not known to be either cats or dogs. Determine if it is possible that there are exactly X cats among these A + B animals.
|
a, b, x = list(map(int, input().split()))
if a >= x and x - a <= b:
print('YES')
else:
print('NO')
|
s954152538
|
Accepted
| 17 | 2,940 | 106 |
a, b, x = list(map(int, input().split()))
if a <= x and x - a <= b:
print('YES')
else:
print('NO')
|
s373365829
|
p02612
|
u257265865
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,152 | 28 |
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)
|
s410227464
|
Accepted
| 28 | 9,148 | 53 |
n=int(input())
print(1000-n%1000 if n%1000!=0 else 0)
|
s326386468
|
p03994
|
u221898645
| 2,000 | 262,144 |
Wrong Answer
| 234 | 5,620 | 326 |
Mr. Takahashi has a string s consisting of lowercase English letters. He repeats the following operation on s exactly K times. * Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of `z` is `a`. For example, if you perform an operation for the second letter on `aaz`, `aaz` becomes `abz`. If you then perform an operation for the third letter on `abz`, `abz` becomes `aba`. Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s. Find the such string.
|
s = input()
K = int(input())
a = [0] * len(s)
for i in range(len(s)):
a[i] = ((ord('z') - ord(s[i])) + 1)%26
if a[i] == 0:
pass
elif K - a[i] >= 0:
K -= a[i]
a[i] = 0
for i in range(len(s)-1):
print(chr((26-a[i])%26+ord('a')), end = '')
b = ord('a') + (ord(s[-1])-ord('a') + K)%26
print( chr( b ) )
|
s945428016
|
Accepted
| 240 | 5,620 | 367 |
s = input()
K = int(input())
a = [0] * len(s)
for i in range(len(s) - 1):
a[i] = ((ord('z') - ord(s[i])) + 1)%26
if a[i] == 0:
pass
elif K - a[i] >= 0:
K -= a[i]
a[i] = 0
for i in range(len(s)-1):
print(chr((26-a[i])%26+ord('a')), end = '')
if K != 0:
b = ord('a') + (ord(s[-1])-ord('a') + K)%26
print( chr( b ) )
else:
print(s[-1])
|
s096542882
|
p03693
|
u434329006
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 122 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r, g, b = map(int,input().split())
answer = r*100 + g*10 + b
if answer % 4 == 0:
print("Yes")
else:
print("No")
|
s287413416
|
Accepted
| 17 | 2,940 | 122 |
r, g, b = map(int,input().split())
answer = r*100 + g*10 + b
if answer % 4 == 0:
print("YES")
else:
print("NO")
|
s475159723
|
p02399
|
u362104929
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,616 | 123 |
Write a program which reads two integers a and b, and calculates the following values: * a ÷ b: d (in integer) * remainder of a ÷ b: r (in integer) * a ÷ b: f (in real number)
|
numbers = [int(x) for x in input().split(" ")]
a, b = numbers[0], numbers[1]
print("{} {} {}".format((a//b), (a%b), (a/b)))
|
s419092947
|
Accepted
| 20 | 7,688 | 127 |
numbers = [int(x) for x in input().split(" ")]
a, b = numbers[0], numbers[1]
print("{} {} {:.5f}".format((a//b), (a%b), (a/b)))
|
s540116374
|
p03548
|
u672898046
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 65 |
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 people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
x, y, z = map(int, input().split())
x -= z
r = x / (y+z)
print(r)
|
s502446628
|
Accepted
| 18 | 2,940 | 70 |
x, y, z = map(int, input().split())
x -= z
r = x / (y+z)
print(int(r))
|
s631107616
|
p02821
|
u562935282
| 2,000 | 1,048,576 |
Wrong Answer
| 1,555 | 23,632 | 1,163 |
Takahashi has come to a party as a special guest. There are N ordinary guests at the party. The i-th ordinary guest has a _power_ of A_i. Takahashi has decided to perform M _handshakes_ to increase the _happiness_ of the party (let the current happiness be 0). A handshake will be performed as follows: * Takahashi chooses one (ordinary) guest x for his left hand and another guest y for his right hand (x and y can be the same). * Then, he shakes the left hand of Guest x and the right hand of Guest y simultaneously to increase the happiness by A_x+A_y. However, Takahashi should not perform the same handshake more than once. Formally, the following condition must hold: * Assume that, in the k-th handshake, Takahashi shakes the left hand of Guest x_k and the right hand of Guest y_k. Then, there is no pair p, q (1 \leq p < q \leq M) such that (x_p,y_p)=(x_q,y_q). What is the maximum possible happiness after M handshakes?
|
def solve(n, m, a) -> int:
from bisect import bisect_left, bisect_right
from itertools import accumulate
from collections import Counter
*a, = sorted(a)
acc = (0,) + tuple(accumulate(a))
c = Counter(a)
def is_ok(h):
ret = 0
for i, x in enumerate(a):
j = bisect_left(a, h - x)
ret += (n - j) * 2
if ret >= m * 2:
return True
return False
def binary_search():
ok = 0
ng = a[-1] * 2 + 1
while abs(ng - ok) > 1:
mid = (ok + ng) // 2
if is_ok(mid):
ok = mid
else:
ng = mid
return ok
min_pair_happiness = binary_search()
ret = 0
for i, x in enumerate(a):
j = bisect_left(a, min_pair_happiness - x)
k = bisect_right(a, min_pair_happiness - x)
ret += (n - j) * x + (acc[n] - acc[j])
if j != k:
ret -= a[j] + x
return ret
def main():
n, m = map(int, input().split())
*a, = map(int, input().split())
print(solve(n, m, a))
if __name__ == '__main__':
main()
|
s651194938
|
Accepted
| 851 | 14,268 | 1,832 |
def main():
N, M = map(int, input().split())
*a, = map(int, input().split())
a.sort()
def count(mid) -> int:
cnt = 0
j = N
for i in range(N):
while j > 0 and a[i] + a[j - 1] >= mid:
j -= 1
# j==0 or a[i]+a[j]>=mid
# j==0
cnt += N - j
return cnt
def binary_search(*, ok: int, ng: int, is_ok: 'function') -> int:
while abs(ok - ng) > 1:
mid = (ok + ng) // 2
if is_ok(mid):
ok = mid
else:
ng = mid
return ok
ma = binary_search(ok=2 * 10 ** 5 + 1, ng=0, is_ok=lambda mid: count(mid) < M)
def accumulate(a):
s = 0
yield s
for x in a:
s += x
yield s
*acc, = accumulate(a)
ans = 0
j = N
for i in range(N):
while j > 0 and a[i] + a[j - 1] >= ma:
j -= 1
ans += a[i] * (N - j) + acc[N] - acc[j]
ans += (ma - 1) * (M - count(ma))
print(ans)
if __name__ == '__main__':
main()
|
s728761881
|
p03110
|
u270681687
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 180 |
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
n = int(input())
ans = 0
for i in range(n):
x, u = input().split()
if u == "JPY":
ans += int(x)
else:
ans += int(float(x) * 380000 + 0.1)
print(ans)
|
s929310157
|
Accepted
| 17 | 2,940 | 169 |
n = int(input())
ans = 0
for i in range(n):
x, u = input().split()
if u == "JPY":
ans += int(x)
else:
ans += float(x) * 380000
print(ans)
|
s195276278
|
p03228
|
u288430479
| 2,000 | 1,048,576 |
Wrong Answer
| 26 | 9,056 | 194 |
In the beginning, Takahashi has A cookies, and Aoki has B cookies. They will perform the following operation alternately, starting from Takahashi: * If the number of cookies in his hand is odd, eat one of those cookies; if the number is even, do nothing. Then, give one-half of the cookies in his hand to the other person. Find the numbers of cookies Takahashi and Aoki respectively have after performing K operations in total.
|
a,b,k = map(int,input().split())
bit = 0
while k:
if bit==0:
if a%2!=0:
a -= 1
b += a//2
a //= 2
else:
if b!=0:
b -= 1
a += b//2
b //2
k -= 1
print(a,b)
|
s328922000
|
Accepted
| 28 | 9,096 | 224 |
a,b,k = map(int,input().split())
bit = 0
while k:
if bit==0:
if a%2!=0:
a -= 1
b += a//2
a //= 2
bit = 1
else:
if b%2!=0:
b -= 1
a += b//2
b //= 2
bit = 0
k -= 1
print(a,b)
|
s343079406
|
p03644
|
u092650292
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 107 |
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())
maxnum = 2
for i in range(8):
if maxnum > n:
print(maxnum/2)
break
maxnum*=2
|
s317297231
|
Accepted
| 17 | 2,940 | 125 |
n = int(input())
maxnum = 2
for i in range(8):
if maxnum > n:
print(int(maxnum/2))
break
maxnum*=2
|
s857062657
|
p03194
|
u887207211
| 2,000 | 1,048,576 |
Wrong Answer
| 2,104 | 3,064 | 379 |
There are N integers a_1, a_2, ..., a_N not less than 1. The values of a_1, a_2, ..., a_N are not known, but it is known that a_1 \times a_2 \times ... \times a_N = P. Find the maximum possible greatest common divisor of a_1, a_2, ..., a_N.
|
N, P = map(int,input().split())
n, p = N, P
if(N == 1 or P == 1):
print(P)
exit()
num = []
s = 0
while s == 0:
for i in range(2,p+1):
if(p%i == 0):
p = int(p/i)
if(p == 1):
s = 1
num.append(i)
break
d = {}
cnt = 1
for i in num:
d[i] = num.count(i)
for k, v in d.items():
if(v >= n):
cnt *= k ** (v//n)
print(k,v)
print(cnt)
|
s769982129
|
Accepted
| 73 | 3,064 | 370 |
N, P = map(int,input().split())
def calc(n):
num = {}
e, i = 0, 2
while i**2 <= n:
while n%i == 0:
n //= i
e += 1
if(e > 0):
num[i] = e
i += 1
e = 0
return num
if(N == 1):
print(P)
exit()
elif(P == 1):
print(1)
exit()
else:
cnt = 1
for k, v in calc(P).items():
if(v >= N):
cnt *= k ** (v//N)
print(cnt)
|
s207971765
|
p03943
|
u328755070
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 152 |
Two students of AtCoder Kindergarten are fighting over candy packs. There are three candy packs, each of which contains a, b, and c candies, respectively. Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible. Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.
|
a, b, c = list(map(int, input().split()))
A = [a,b,c]
A.sort()
if A[0] == (A[1]+A[2]) or (A[0] + A[1]) == A[2]:
print('YES')
else:
print('NO')
|
s655122494
|
Accepted
| 17 | 2,940 | 152 |
a, b, c = list(map(int, input().split()))
A = [a,b,c]
A.sort()
if A[0] == (A[1]+A[2]) or (A[0] + A[1]) == A[2]:
print('Yes')
else:
print('No')
|
s373383891
|
p02264
|
u255317651
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,616 | 960 |
_n_ _q_ _name 1 time1_ _name 2 time2_ ... _name n timen_ In the first line the number of processes _n_ and the quantum _q_ are given separated by a single space. In the following _n_ lines, names and times for the _n_ processes are given. _name i_ and _time i_ are separated by a single space.
|
# -*- coding: utf-8 -*-
"""
Created on Sun Apr 29 14:42:43 2018
@author: maezawa
"""
def print_array(g):
ans = str(g[0])
if len(g) > 1:
for i in range(1,len(g)):
ans += ' '+str(g[i])
print(ans)
name=[]
time=[]
n, q = list(map(int, input().split()))
for i in range(n):
s = input().split()
name.append(s[0])
time.append(int(s[1]))
# =============================================================================
# print(n,q)
# print(name)
# print(time)
# =============================================================================
finished = []
i=0
remain = n
while True:
if time[i] == 0:
i += 1
if i == n:
i = 0
continue
if time[i] <= q:
finished.append(name[i])
time[i] = 0
remain -= 1
if remain == 0:
break
else:
time[i] -= q
i += 1
if i == n:
i = 0
print_array(finished)
|
s231022932
|
Accepted
| 460 | 15,668 | 981 |
# -*- coding: utf-8 -*-
def fifo_enque(data):
global tail
global fifo
fifo[tail] = data
tail = (tail+1)%fifo_size
def fifo_deque():
global head
global fifo
data = fifo[head]
head = (head+1)%fifo_size
return data
fifo_size = 100000
fifo = [0 for _ in range(fifo_size)]
head = 0
tail = 0
n, q = list(map(int, input().split()))
for i in range(n):
s = input().split()
data = [s[0], int(s[1])]
fifo_enque(data)
current_time = 0
finished = []
fin_time = []
while True:
data = fifo_deque()
if data[1] > q:
current_time += q
data[1] -= q
fifo_enque(data)
else:
current_time += data[1]
finished.append(data[0])
fin_time.append(current_time)
if head == tail:
break
for i in range(n):
print("{} {}".format(finished[i], fin_time[i]))
|
s580081558
|
p04043
|
u860819641
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 116 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
A=input().split()
print(A)
if A==["5","5","7"] or ["5","7","5"] or ["7","5","5"]:
print("YES")
else:
print("NO")
|
s712920746
|
Accepted
| 17 | 2,940 | 113 |
A=input().split()
if A==["5","5","7"] or A==["5","7","5"] or A==["7","5","5"]:
print("YES")
else:
print("NO")
|
s007198186
|
p02853
|
u007886915
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 465 |
We held two competitions: Coding Contest and Robot Maneuver. In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen. DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver. Find the total amount of money he earned.
|
import math
N = 1
x, y = [0]*N, [0]*N
l = list(range(N))
sum=0
#print(x[0])
for i in range(N):
x[i], y[i] = map(int, input().split())
if x[0] <= 3:
sum=sum+10000*(4-x[0])
else:
pass
if y[0]<= 3:
sum=sum+10000*(4-y[0])
else:
pass
if x[0]==1 and y[0]==1:
sum=sum+40000
else:
pass
print(sum)
|
s953437817
|
Accepted
| 18 | 3,064 | 570 |
# -*- Coding: utf-8 -*-
import itertools
import math
N = 1
x, y = [0]*N, [0]*N
l = list(range(N))
sum=0
#print(x[0])
for i in range(N):
x[i], y[i] = map(int, input().split())
if x[0] <= 3:
sum=sum+100000*(4-x[0])
else:
pass
if y[0]<= 3:
sum=sum+100000*(4-y[0])
else:
pass
if x[0]==1 and y[0]==1:
sum=sum+400000
else:
pass
print(sum)
|
s918893895
|
p03854
|
u351363308
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,316 | 485 |
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`.
|
# coding: utf-8
# Your code here!
S = input()
while True:
if len(S)!=0:
if S[len(S)-7:len(S)]=="dreamer":
S = S.rstrip("dreamer")
elif S[len(S)-6:len(S)]=="eraser":
S = S.rstrip("eraser")
elif S[len(S)-5:len(S)]=="dream":
S = S.rstrip("dream")
elif S[len(S)-5:len(S)]=="erase":
S = S.rstrip("erase")
else:
print("NO")
break
else:
print("YES")
break
|
s814863125
|
Accepted
| 80 | 3,188 | 467 |
# coding: utf-8
# Your code here!
S = input()
while True:
if len(S)!=0:
if S[len(S)-7:len(S)]=="dreamer":
S = S[0:len(S)-7]
elif S[len(S)-6:len(S)]=="eraser":
S = S[0:len(S)-6]
elif S[len(S)-5:len(S)]=="dream":
S = S[0:len(S)-5]
elif S[len(S)-5:len(S)]=="erase":
S = S[0:len(S)-5]
else:
print("NO")
break
else:
print("YES")
break
|
s639720543
|
p04043
|
u763550415
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,044 | 92 |
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.
|
line = input()
if line == '5 5 7' or line == '7 5 5':
print('Yes')
else:
print('No')
|
s946495724
|
Accepted
| 25 | 8,796 | 112 |
line = input()
if line == '5 5 7' or line == '5 7 5' or line == '7 5 5':
print('YES')
else:
print('NO')
|
s598671436
|
p02612
|
u851704997
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,120 | 57 |
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())
while(N>1000):
N -= 1000
print(str(N))
|
s690922016
|
Accepted
| 29 | 9,136 | 94 |
N = int(input())
for i in range(20):
if(N <= 1000):
break
N -= 1000
print(str(1000-N))
|
s256780634
|
p03371
|
u329407311
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 137 |
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively. Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
|
A,B,C,X,Y=map(int,input().split())
ans1 = int(A*X + B*Y)
ans2 = int(C*X + (Y-X)*B)
ans3 = int(C*Y + (X-Y)*A)
print(min(ans1,ans2,ans3))
|
s131587893
|
Accepted
| 17 | 3,060 | 243 |
A,B,C,X,Y=map(int,input().split())
ans1 = int(A*X + B*Y)
if X<=Y:
ans2 = int(C*2*X + (Y-X)*B)
ans3 = 10**10
ans4 = int(2*Y*C)
else:
ans2 = 10**10
ans3 = int(C*2*Y + (X-Y)*A)
ans4 = int(2*X*C)
print(min(ans1,ans2,ans3,ans4))
|
s858287167
|
p03730
|
u432853936
| 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 objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
a,b,c = map(int,input().split())
for i in range(1,b+1):
if (a * i) % b == c:
print("Yes")
exit()
print("No")
|
s378751771
|
Accepted
| 17 | 2,940 | 154 |
a,b,c = map(int,input().split())
for i in range(1,b+1):
if (a * i) % b == c:
print("YES")
exit()
print("NO")
|
s635372806
|
p03024
|
u203669169
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 85 |
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 consisting of `o` and `x`. If the i-th character in S is `o`, it means that Takahashi won the match on the i-th day; if that character is `x`, it means that Takahashi lost the match on the i-th day. Print `YES` if there is a possibility that Takahashi can participate in the next tournament, and print `NO` if there is no such possibility.
|
s=input()
if 8 <= 15 - len(s) + s.count("o"):
print("Yes")
else:
print("No")
|
s359260796
|
Accepted
| 17 | 2,940 | 85 |
s=input()
if 8 <= 15 - len(s) + s.count("o"):
print("YES")
else:
print("NO")
|
s684126507
|
p02409
|
u797673668
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,752 | 246 |
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 floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
house = [[[0] * 10 for _ in range(3)] for __ in range(4)]
n = int(input())
while n:
b, f, r, v = map(int, input().split())
house[b - 1][f - 1][r - 1] += v
n -= 1
for b in house:
for f in b:
print(*f)
print('#' * 20)
|
s967283866
|
Accepted
| 30 | 7,736 | 305 |
house = [[[0] * 10 for _ in range(3)] for __ in range(4)]
n = int(input())
while n:
b, f, r, v = map(int, input().split())
house[b - 1][f - 1][r - 1] += v
n -= 1
for i, b in enumerate(house):
for f in b:
print(' ', end='')
print(*f)
if i < 3:
print('#' * 20)
|
s480904016
|
p03672
|
u503052349
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,064 | 349 |
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.
|
c=0
i=0
list1=[]
list2=[]
S = input('')
char_list = list(S)
c = len (char_list)
c= c-1
print(c)
while c>1:
if c % 2 == 0:
d=int(c/2)
print (d)
for i in range(d):
list1.append(char_list[i])
list2.append(char_list[i+d])
print (list1)
print(list2)
if list1 == list2:
print (len (list1))
break
list1=[]
list2=[]
c=c-1
|
s867413360
|
Accepted
| 19 | 3,064 | 355 |
c=0
i=0
list1=[]
list2=[]
S = input('')
char_list = list(S)
c = len (char_list)
c= c-1
#print(c)
while c>1:
if c % 2 == 0:
d=int(c/2)
#print (d)
for i in range(d):
list1.append(char_list[i])
list2.append(char_list[i+d])
#print (list1)
#print(list2)
if list1 == list2:
print (len (list1)*2)
break
list1=[]
list2=[]
c=c-1
|
s070236483
|
p02402
|
u476441153
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,576 | 120 |
Write a program which reads a sequence of $n$ integers $a_i (i = 1, 2, ... n)$, and prints the minimum value, maximum value and sum of the sequence.
|
n = int(input())
a = [0] * n
print (n)
a = list(map(int, input().split()))
print (min(a))
print (max(a))
print (sum(a))
|
s399201897
|
Accepted
| 60 | 8,740 | 93 |
n = int(input())
a = [0] * n
a = list(map(int, input().split()))
print (min(a),max(a),sum(a))
|
s910327543
|
p02742
|
u952484541
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 276 |
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:
|
import sys
n = list(map(int, input().split()))
if n[0] % 2 == 0:
print((n[0]*n[1])/2)
sys.exit()
else:
if n[1] % 2 == 0:
print((n[0]*n[1])/2)
sys.exit()
else:
b = int((n[0]*n[1])/2)
print((n[0]*n[1])-b)
sys.exit()
|
s520972068
|
Accepted
| 35 | 5,076 | 422 |
from decimal import Decimal
import sys
n = list(map(int, input().split()))
if n[0] == 1 or n[1] == 1:
print(1)
sys.exit()
if n[0] % 2 == 0:
print(Decimal(int((n[0]/2)*n[1])))
sys.exit()
else:
if n[1] % 2 == 0:
print(Decimal(int(n[0]*(n[1]/2))))
sys.exit()
else:
a = ((n[0]-1)*n[1])/2
b = n[1] - int(n[1]/2)
print(Decimal(int(a+b)))
sys.exit()
|
s563127070
|
p03605
|
u374671031
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 64 |
It is September 9 in Japan now. You are given a two-digit integer N. Answer the question: Is 9 contained in the decimal notation of N?
|
n = input()
if "9" in n :
print("YES")
else:
print("NO")
|
s653325390
|
Accepted
| 18 | 2,940 | 64 |
n = input()
if "9" in n :
print("Yes")
else:
print("No")
|
s985753299
|
p02263
|
u996463517
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,556 | 197 |
An expression is given in a line. Two consequtive symbols (operand or operator) are separated by a space character. You can assume that +, - and * are given as the operator and an operand is a positive integer less than 106
|
a = input().split()
b = []
for i in a:
if i.isdigit():
b.append(i)
else:
c = b.pop()
d = b.pop()
ans = eval(str(d)+i+str(c))
b.append(ans)
print(b)
|
s018003055
|
Accepted
| 20 | 5,568 | 200 |
a = input().split()
b = []
for i in a:
if i.isdigit():
b.append(i)
else:
c = b.pop()
d = b.pop()
ans = eval(str(d)+i+str(c))
b.append(ans)
print(b[0])
|
s080725899
|
p03759
|
u627691992
| 2,000 | 262,144 |
Wrong Answer
| 26 | 9,024 | 85 |
Three poles stand evenly spaced along a line. Their heights are a, b and c meters, from left to right. We will call the arrangement of the poles _beautiful_ if the tops of the poles lie on the same line, that is, b-a = c-b. Determine whether the arrangement of the poles is beautiful.
|
a,b,c = map(int, input().split())
if(b-a == c-b):
print("Yes")
else:
print("No")
|
s153450502
|
Accepted
| 26 | 9,072 | 96 |
a, b, c = map(int, input().split())
if b - a == c - b:
print("YES")
else:
print("NO")
|
s825940998
|
p03494
|
u886878171
| 2,000 | 262,144 |
Time Limit Exceeded
| 2,104 | 2,940 | 291 |
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()))
count = 0
exist_odd = False
while (exist_odd == False):
for x in A:
x = int(x)
if x % 2 != 0:
exist_odd = True
if exist_odd == True:
break
else:
for x in A:
x /= 2
count += 1
print(count)
|
s104994894
|
Accepted
| 28 | 9,160 | 269 |
N = int(input())
A = list(map(int,input().split()))
count = 0
flag = True
while(flag):
for i in range(N):
if A[i] % 2 == 0:
A[i] = A[i]/2
else:
flag = False
break
if flag == True:
count+=1
print(count)
|
s386921534
|
p02612
|
u533232830
| 2,000 | 1,048,576 |
Wrong Answer
| 35 | 9,152 | 72 |
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())
i = 0
while i*1000<=n:
i += 1
print(n-(i-1)*1000)
|
s660457231
|
Accepted
| 28 | 9,152 | 69 |
n = int(input())
i = 0
while i*1000<n:
i += 1
print((i)*1000-n)
|
s158154795
|
p03477
|
u350093546
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 116 |
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
a,b,c,d=map(int,input().split())
if a+b>c+d:
print('Left')
if a+b==c+d:
print('Balanced')
else:
print('Right')
|
s887854836
|
Accepted
| 18 | 2,940 | 135 |
a,b,c,d=map(int,input().split())
if (a+b)>(c+d):
print('Left')
if (a+b)==(c+d):
print('Balanced')
if (a+b)<(c+d):
print('Right')
|
s764851762
|
p03635
|
u474423089
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 117 |
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?
|
def main():
n,m=map(int,input().split(' '))
print((n-1)*(m-1))
if __name__ == '__main__':
print(main())
|
s373642965
|
Accepted
| 17 | 2,940 | 110 |
def main():
n,m=map(int,input().split(' '))
print((n-1)*(m-1))
if __name__ == '__main__':
main()
|
s972385140
|
p02408
|
u501414488
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,720 | 366 |
Taro is going to play a card game. However, now he has only n cards, even though there should be 52 cards (he has no Jokers). The 52 cards include 13 ranks of each of the four suits: spade, heart, club and diamond.
|
cards = {
'S': [0 for _ in range(13)],
'H': [0 for _ in range(13)],
'C': [0 for _ in range(13)],
'D': [0 for _ in range(13)],
}
n = int(input())
for _ in range(n):
(s, r) = input().split()
cards[s][int(r) - 1] = 1
for _ in ('S', 'H', 'C', 'D'):
for r in range(13):
if cards[s][r] == 0:
print(s, r + 1)
|
s869760692
|
Accepted
| 30 | 6,724 | 366 |
cards = {
'S': [0 for _ in range(13)],
'H': [0 for _ in range(13)],
'C': [0 for _ in range(13)],
'D': [0 for _ in range(13)],
}
n = int(input())
for _ in range(n):
(s, r) = input().split()
cards[s][int(r) - 1] = 1
for s in ('S', 'H', 'C', 'D'):
for r in range(13):
if cards[s][r] == 0:
print(s, r + 1)
|
s724205415
|
p03435
|
u048004795
| 2,000 | 262,144 |
Wrong Answer
| 28 | 9,136 | 624 |
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) is equal to a_i + b_j. Determine if he is correct.
|
import sys
input = sys.stdin.readline
#N
#t1 x1 y1t1 x1 y1
#t2 x2 y2t2 x2 y2
#tN xN yNtN xN yN
N = 3
x = [0 for i in range(N)]
y = [0 for i in range(N)]
c =[[0 for i in range(N)] for i in range(N)]
for i in range(N):
c1, c2, c3 = map(int, input().split())
c[i][0] = c1
c[i][1] = c2
c[i][2] = c3
x[0] = 0
for i in range(N):
y[i] = c[0][i] - x[0]
x[1] = c[1][0] - y[1]
x[2] = c[2][0] - y[2]
flag = True
for i in range(N):
for j in range(N):
numx = x[i]
numy = y[j]
if not numx + numy == c[i][j]:
flag = False
if flag:
print("Yes")
else:
print("No")
|
s624867636
|
Accepted
| 29 | 9,184 | 623 |
import sys
input = sys.stdin.readline
#N
#t1 x1 y1t1 x1 y1
#t2 x2 y2t2 x2 y2
#tN xN yNtN xN yN
N = 3
x = [0 for i in range(N)]
y = [0 for i in range(N)]
c =[[0 for i in range(N)] for i in range(N)]
for i in range(N):
c1, c2, c3 = map(int, input().split())
c[i][0] = c1
c[i][1] = c2
c[i][2] = c3
x[0] = 0
for i in range(N):
y[i] = c[0][i] - x[0]
x[1] = c[1][0] - y[0]
x[2] = c[2][0] - y[0]
flag = True
for i in range(N):
for j in range(N):
numx = x[i]
numy = y[j]
if not numx + numy == c[i][j]:
flag = False
if flag:
print("Yes")
else:
print("No")
|
s761528725
|
p02409
|
u327972099
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,724 | 512 |
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 floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
n = int(input())
count = 0
room = [int(0) for i in range(4) for j in range(3) for k in range(10)]
while count < n:
x = list(map(lambda k: int(k), input().split(" ")))
room[(x[0]-1)*30+(x[1]-1)*10+(x[2]-1)] += x[3]
count += 1
for i in range(4):
for j in range(3):
for k in range(10):
if k == 0:
print("%d" % room[30*i+10*j+k], end="")
else:
print(" %d" % room[30*i+10*j+k], end="")
print("")
print("####################")
|
s084884123
|
Accepted
| 40 | 6,724 | 430 |
n = int(input())
count = 0
room = [int(0) for i in range(4) for j in range(3) for k in range(10)]
while count < n:
x = list(map(lambda k: int(k), input().split(" ")))
room[(x[0]-1)*30+(x[1]-1)*10+(x[2]-1)] += x[3]
count += 1
for i in range(4):
if i != 0:
print("####################")
for j in range(3):
for k in range(10):
print(" %d" % room[30*i+10*j+k], end="")
print("")
|
s808809629
|
p03547
|
u088974156
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 99 |
In programming, hexadecimal notation is often used. In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively. In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`. When X and Y are seen as hexadecimal numbers, which is larger?
|
x,y=input().split()
if(x<y):
print(x+"<"+y)
elif (x==y):
print(x+"="+y)
else:
print(x+">"+y)
|
s571551378
|
Accepted
| 17 | 2,940 | 87 |
x,y=input().split()
if(x<y):
print("<")
elif (x==y):
print("=")
else:
print(">")
|
s430707681
|
p03407
|
u177756077
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,316 | 80 |
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
A,B,C=map(int,input().split())
if A+B<=C:
print("Yes")
else:
print("No")
|
s449442314
|
Accepted
| 18 | 2,940 | 80 |
A,B,C=map(int,input().split())
if A+B>=C:
print("Yes")
else:
print("No")
|
s580672110
|
p03795
|
u920438243
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 102 |
Snuke has a favorite restaurant. The price of any meal served at the restaurant is 800 yen (the currency of Japan), and each time a customer orders 15 meals, the restaurant pays 200 yen back to the customer. So far, Snuke has ordered N meals at the restaurant. Let the amount of money Snuke has paid to the restaurant be x yen, and let the amount of money the restaurant has paid back to Snuke be y yen. Find x-y.
|
n = int(input())
power = 1
for i in range(1,n+1):
power += i%(10**9+7)
print(power % (10**9 + 7))
|
s784791691
|
Accepted
| 18 | 2,940 | 42 |
n = int(input())
print(n*800-(n//15)*200)
|
s662125312
|
p03555
|
u882359130
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 113 |
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.
|
C1 = [c1 for c1 in input()]
C2 = [c2 for c2 in input()]
if C1 == reversed(C2):
print("YES")
else:
print("NO")
|
s383376714
|
Accepted
| 17 | 3,064 | 116 |
C1 = [c1 for c1 in input()]
C2 = [c2 for c2 in input()]
C2.reverse()
if C1 == C2:
print("YES")
else:
print("NO")
|
s659854156
|
p03360
|
u727821695
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 98 |
There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times: * Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n. What is the largest possible sum of the integers written on the blackboard after K operations?
|
li = list(map(int, input().split()))
K = int(input())
li.sort
print (li[0] + li[1] + li[2] ** K)
|
s526135133
|
Accepted
| 17 | 2,940 | 110 |
li = list(map(int, input().split()))
K = int(input())
print (li[0] + li[1] + li[2] + max(li) * (2 ** K - 1))
|
s701309807
|
p04043
|
u179285430
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 96 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
a = list(map(int, input().split()))
if a.sort() == [5, 5, 7]:
print("YES")
else:
print("NO")
|
s547176366
|
Accepted
| 17 | 2,940 | 102 |
I = list(map(int, input().split()))
I.sort()
if I == [5, 5, 7]:
print("YES")
else:
print("NO")
|
s846219485
|
p03854
|
u388323466
| 2,000 | 262,144 |
Wrong Answer
| 21 | 3,956 | 302 |
You are given a string S consisting of lowercase English letters. Another string T is initially empty. Determine whether it is possible to obtain S = T by performing the following operation an arbitrary number of times: * Append one of the following at the end of T: `dream`, `dreamer`, `erase` and `eraser`.
|
s = list(input())
s.reverse()
s=''.join(s)
arr = ['dream', 'dreamer', 'erase', 'eraser']
tmp = []
arr2 = []
for i in range(4):
tmp += list(arr[i])
tmp.reverse()
arr2.append(''.join(tmp))
tmp = []
for y in arr2:
if y in s:
s = s.replace(y,'')
if s =='':
print('Yes')
else:
print('No')
|
s092836134
|
Accepted
| 69 | 3,188 | 370 |
s = input()
while len(s)>4:
if s[-5:] == 'dream':
s = s[:-5]
continue
elif s[-7:] == 'dreamer':
s = s[:-7]
continue
elif s[-5:] == 'erase':
s = s[:-5]
continue
elif s[-6:] == 'eraser':
s = s[:-6]
continue
else:
break
if len(s)>0 :
print('NO')
else:
print('YES')
|
s416602364
|
p03448
|
u231189826
| 2,000 | 262,144 |
Wrong Answer
| 50 | 3,060 | 236 |
You have A 500-yen coins, B 100-yen coins and C 50-yen coins (yen is the currency of Japan). In how many ways can we select some of these coins so that they are X yen in total? Coins of the same kind cannot be distinguished. Two ways to select coins are distinguished when, for some kind of coin, the numbers of that coin are different.
|
A = int(input())
B = int(input())
C = int(input())
X = int(input())
count = 0
for i in range(A+1):
for j in range(B+1):
for K in range(C+1):
if A*500 + B*100 + C*50 == X:
count += 1
print(count)
|
s658078142
|
Accepted
| 50 | 3,060 | 236 |
A = int(input())
B = int(input())
C = int(input())
X = int(input())
count = 0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
if i*500 + j*100 + k*50 == X:
count += 1
print(count)
|
s463728270
|
p02614
|
u970308980
| 1,000 | 1,048,576 |
Wrong Answer
| 61 | 9,092 | 442 |
We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is `.`, and black if c_{i,j} is `#`. Consider doing the following operation: * Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns. You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.
|
H, W, K = map(int, input().split())
C = []
for i in range(H):
c = list(input().strip())
C.append(c)
ans = 0
for h in range(1 << H):
for w in range(1 << W):
cnt = 0
for i in range(H):
for j in range(W):
if h >> i & 1:
continue
if w >> j & 1:
continue
cnt += 1
if cnt == K:
ans += 1
print(ans)
|
s934109939
|
Accepted
| 64 | 9,104 | 481 |
H, W, K = map(int, input().split())
C = []
for i in range(H):
c = list(input().strip())
C.append(c)
ans = 0
for h in range(1 << H):
for w in range(1 << W):
cnt = 0
for i in range(H):
for j in range(W):
if h >> i & 1:
continue
if w >> j & 1:
continue
if C[i][j] == "#":
cnt += 1
if cnt == K:
ans += 1
print(ans)
|
s541787317
|
p03455
|
u548076391
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 98 |
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())
sum = a+b
if sum % 2 ==0:
print("Even")
else:
print("Odd")
|
s506749811
|
Accepted
| 17 | 2,940 | 96 |
a, b = map(int,input().split())
cal = a*b
if cal % 2 == 0:
print("Even")
else:
print("Odd")
|
s985795051
|
p02936
|
u096616343
| 2,000 | 1,048,576 |
Wrong Answer
| 2,107 | 61,256 | 527 |
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.
|
from heapq import heappop,heappush
N,Q = map(int,input().split())
route = [[] for _ in range(N)]
ans = [0] * N
for i in range(N - 1):
a,b = map(int,input().split())
route[a - 1].append(b)
route[b - 1].append(a)
for i in range(Q):
p,x = map(int,input().split())
ans[p - 1] = x
W = [(0,1)]
A = [1] * N
while W:
de,nod = heappop(W)
A[nod - 1] = 0
for i in route[nod - 1]:
if A[i - 1] == 1:
heappush(W,(de + 1,i))
ans[i - 1] += ans[nod - 1]
for j in ans:
print(j)
|
s133255695
|
Accepted
| 1,996 | 64,352 | 513 |
from heapq import heappop,heappush
N,Q = map(int,input().split())
route = [[] for _ in range(N)]
ans = [0] * N
for i in range(N - 1):
a,b = map(int,input().split())
route[a - 1].append(b)
route[b - 1].append(a)
for i in range(Q):
p,x = map(int,input().split())
ans[p - 1] += x
W = [(0,1)]
A = [1] * N
while W:
de,nod = heappop(W)
A[nod - 1] = 0
for i in route[nod - 1]:
if A[i - 1] == 1:
heappush(W,(de + 1,i))
ans[i - 1] += ans[nod - 1]
print(*ans)
|
s920456218
|
p02646
|
u362031378
| 2,000 | 1,048,576 |
Wrong Answer
| 19 | 9,180 | 169 |
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a,v=map(int,input().split())
b,w=map(int,input().split())
t=int(input())
if v>w:
c=(b-a)/(v-w)
if t>=c:
print('Yes')
else:
print('No')
else:
print('No')
|
s710582604
|
Accepted
| 23 | 9,120 | 181 |
a,v=map(int,input().split())
b,w=map(int,input().split())
t=int(input())
gap=abs(a-b)
if v>w:
c=(v-w)*t
if gap<=c:
print('YES')
else:
print('NO')
else:
print('NO')
|
s731439509
|
p02606
|
u173151534
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,148 | 105 |
How many multiples of d are there among the integers between L and R (inclusive)?
|
L, R, d = (int(x) for x in input().split())
a = L // d
b = R // d
if R % d == 0:
b += 1
print(b - a)
|
s450341028
|
Accepted
| 31 | 8,936 | 105 |
L, R, d = (int(x) for x in input().split())
a = L // d
b = R // d
if L % d == 0:
b += 1
print(b - a)
|
s251550269
|
p03607
|
u171366497
| 2,000 | 262,144 |
Wrong Answer
| 259 | 15,460 | 180 |
You are playing the following game with Joisino. * Initially, you have a blank sheet of paper. * Joisino announces a number. If that number is written on the sheet, erase the number from the sheet; if not, write the number on the sheet. This process is repeated N times. * Then, you are asked a question: How many numbers are written on the sheet now? The numbers announced by Joisino are given as A_1, ... ,A_N in the order she announces them. How many numbers will be written on the sheet at the end of the game?
|
N=int(input())
from collections import defaultdict
d=defaultdict(int)
for i in range(N):
d[int(input())]+=1
cnt=1
for a in d.keys():
if d[a]%2==1:
cnt+=1
print(cnt)
|
s041805223
|
Accepted
| 258 | 15,456 | 180 |
N=int(input())
from collections import defaultdict
d=defaultdict(int)
for i in range(N):
d[int(input())]+=1
cnt=0
for a in d.keys():
if d[a]%2==1:
cnt+=1
print(cnt)
|
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