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
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s100984830
|
p04012
|
u737840172
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 350
|
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.
|
# Vicfred
# implementation
w = input()
english = {}
for ch in w:
if ch in english:
english[ch] += 1
else:
english[ch] = 1
beautiful = True
for value in english.values():
if value%2 == 1:
beautiful = False
if beautiful:
print("YES")
else:
print("NO")
|
s072770482
|
Accepted
| 17
| 2,940
| 350
|
# Vicfred
# implementation
w = input()
english = {}
for ch in w:
if ch in english:
english[ch] += 1
else:
english[ch] = 1
beautiful = True
for value in english.values():
if value%2 == 1:
beautiful = False
if beautiful:
print("Yes")
else:
print("No")
|
s428252452
|
p03836
|
u798129018
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,060
| 166
|
Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up. Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1. Here, both the x\- and y-coordinates before and after each movement must be integers. He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy). Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty). Under this condition, find a shortest path for him.
|
sx,sy,tx,ty = map(int,input().split())
print("U"*(ty-sy)+"R"*(tx-sx)+"D"*(ty-sy)+"L"*(ty-sy)+"L"+"U"*(ty-sy+1)+"R"*(tx-sx+1)+"D"+"R"+"D"*(ty-sy+1)+"L"*(tx-sy+1)+"U")
|
s848363094
|
Accepted
| 17
| 3,060
| 166
|
sx,sy,tx,ty = map(int,input().split())
print("U"*(ty-sy)+"R"*(tx-sx)+"D"*(ty-sy)+"L"*(tx-sx)+"L"+"U"*(ty-sy+1)+"R"*(tx-sx+1)+"D"+"R"+"D"*(ty-sy+1)+"L"*(tx-sx+1)+"U")
|
s612885880
|
p02742
|
u018679195
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 3,060
| 151
|
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 math
r,c = input().split()
r = int(r)
c = int(c)
ans = r*c/2
if r%2==0 and c%2==0:
print(math.floor(ans))
else:
print(math.floor(ans)+1)
|
s237665894
|
Accepted
| 17
| 3,064
| 471
|
from math import floor, ceil
def solve():
h, w = map(int, input().split())
h1 = float(h/2)
w1 = float(w/2)
if(h == 1 or w == 1):
print(1)
exit()
if (h % 2 and w % 2 is 0) or (h % 2 and w % 2 is not 0):
tot = ceil(h1) * ceil(w1) + floor(h1) * floor(w1)
else:
if h % 2 is 0:
tot = h1 * w
elif w % 2 is 0:
tot = w1 * h
print(int(tot))
if __name__ == "__main__":
solve()
|
s366181761
|
p02612
|
u780698286
| 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(1000 - n)
|
s371923846
|
Accepted
| 26
| 9,092
| 80
|
n = int(input())
if n % 1000 == 0:
print(0)
else:
print(1000 - n % 1000)
|
s053240015
|
p03434
|
u893048163
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 234
|
We have N cards. A number a_i is written on the i-th card. Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first. The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.
|
n = int(input())
a_list = [int(x) for x in input().split()]
alice = bob = []
for i, a in enumerate(sorted(a_list, reverse=True)):
if i % 2 == 0:
alice.append(a)
else:
bob.append(a)
print(sum(alice) - sum(bob))
|
s184762568
|
Accepted
| 17
| 2,940
| 213
|
n = int(input())
a_list = [int(x) for x in input().split()]
alice = bob = 0
for i, a in enumerate(sorted(a_list, reverse=True)):
if i % 2 == 0:
alice += a
else:
bob += a
print(alice - bob)
|
s237676393
|
p03369
|
u382303205
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 25
|
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.
|
print(input().count("o"))
|
s690032544
|
Accepted
| 17
| 2,940
| 35
|
print((input().count("o"))*100+700)
|
s681608140
|
p02694
|
u745861782
| 2,000
| 1,048,576
|
Wrong Answer
| 21
| 9,140
| 136
|
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
x = int(input())
y = 100
count = 0
while (True):
if y <= x:
y *= 1.01
y = int(y)
count += 1
else:
break
print(count)
|
s411208075
|
Accepted
| 23
| 9,156
| 135
|
x = int(input())
y = 100
count = 0
while (True):
if y < x:
y *= 1.01
y = int(y)
count += 1
else:
break
print(count)
|
s953391038
|
p03485
|
u466478199
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 50
|
You are given two positive integers a and b. Let x be the average of a and b. Print x rounded up to the nearest integer.
|
a,b=map(int,input().split())
x=(a+b)/2
print(x//1)
|
s712444220
|
Accepted
| 17
| 2,940
| 95
|
a,b=map(int,input().split())
x=(a+b)/2
if int(x)==x:
print(int(x))
else :
print(int(x)+1)
|
s224140667
|
p02285
|
u007270338
| 2,000
| 131,072
|
Wrong Answer
| 20
| 5,456
| 1
|
Write a program which performs the following operations to a binary search tree $T$ by adding delete operation to B: Binary Search Tree II. * insert $k$: Insert a node containing $k$ as key into $T$. * find $k$: Report whether $T$ has a node containing $k$. * delete $k$: Delete a node containing $k$. * print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively. The operation delete $k$ for deleting a given node $z$ containing key $k$ from $T$ can be implemented by an algorithm which considers the following cases: 1. If $z$ has no children, we modify its parent $z.p$ to replace $z$ with NIL as its child (delete $z$). 2. If $z$ has only a single child, we "splice out" $z$ by making a new link between its child and its parent. 3. If $z$ has two children, we splice out $z$'s successor $y$ and replace $z$'s key with $y$'s key.
|
s279452581
|
Accepted
| 7,360
| 121,640
| 2,261
|
#coding:utf-8
class MakeTree():
def __init__(self, key, p=None, l=None, r=None):
self.key = key
self.p = p
self.l = l
self.r = r
def Insert(root,value):
y = None
x = root
z = MakeTree(value)
while x != None:
y = x
if x.key > z.key:
x = x.l
else:
x = x.r
z.p = y
if y == None:
root = z
elif z.key < y.key:
y.l = z
else:
y.r = z
return root
def Find(u, target):
y = None
x = u
while x != None and target != x.key:
y = x
if x.key > target:
x = x.l
else:
x = x.r
return x
def Delete(u, z):
if z.l == None or z.r == None:
y = z
else:
y = getSuccessor(z)
if y.l != None:
x = y.l
else:
x = y.r
if x != None:
x.p = y.p
if y.p == None:
root = x
elif y == y.p.l:
y.p.l = x
else:
y.p.r = x
if y != z:
z.key = y.key
def getSuccessor(x):
if x.r != None:
return getMinimum(x.r)
y = x.p
while y != None and x == y.r:
x = y
y = y.p
return y
def getMinimum(x):
while x.l != None:
x = x.l
return x
def inParse(u):
if u == None:
return
inParse(u.l)
global inParseList
inParseList.append(u.key)
inParse(u.r)
def preParse(u):
if u == None:
return
global preParseList
preParseList.append(u.key)
preParse(u.l)
preParse(u.r)
root = None
n = int(input())
inParseList = []
preParseList = []
for i in range(n):
order = list(input().split())
if order[0] == "insert":
root = Insert(root, int(order[1]))
elif order[0] == "print":
inParse(root)
preParse(root)
print(" " + " ".join([str(i) for i in inParseList]))
print(" " + " ".join([str(i) for i in preParseList]))
preParseList = []
inParseList = []
elif order[0] == "find":
x = Find(root, int(order[1]))
if x == None:
print("no")
else:
print("yes")
else:
x = Find(root, int(order[1]))
Delete(root, x)
|
|
s059835745
|
p03378
|
u508141157
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 89
|
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 cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
n,m,x=map(int,input().split())
s=sum(int(i)<x for i in input().split())
print(min(s,s-m))
|
s587790830
|
Accepted
| 18
| 2,940
| 89
|
n,m,x=map(int,input().split())
s=sum(int(i)<x for i in input().split())
print(min(s,m-s))
|
s967056432
|
p02854
|
u348868667
| 2,000
| 1,048,576
|
Wrong Answer
| 105
| 26,764
| 156
|
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters. Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times **before** he does the cut: * Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan). * Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen. Find the minimum amount of money needed before cutting the bar into two parts with the same length.
|
N = int(input())
A = list(map(int,input().split()))
tmp = 0
std = sum(A)/2
index = 0
while tmp < std:
tmp += A[index]
index += 1
print(int(tmp-std))
|
s364602427
|
Accepted
| 108
| 26,220
| 261
|
N = int(input())
A = list(map(int,input().split()))
mid = sum(A)/2
length = 0
i = 0
while True:
if length >= mid:
length = min(abs(length-mid),abs(length-mid-A[i-1]))+mid
break
length += A[i]
i += 1
print(int(length-(sum(A)-length)))
|
s418889080
|
p03575
|
u629350026
| 2,000
| 262,144
|
Wrong Answer
| 2,103
| 3,064
| 441
|
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges.
|
n,m=map(int,input().split())
temp=[[] for i in range(n)]
for i in range(m):
a,b=map(int,input().split())
a=a-1
b=b-1
temp[a].append(b)
temp[b].append(a)
count=0
print(temp)
for i in range(n):
if len(temp[i])==1:
count=count+1
l=temp[i][0]
while True:
if len(temp[l])==2:
count=count+1
for j in range(2):
if temp[l][j]!=i:
l=temp[l][j]
else:
break
print(count)
|
s490900373
|
Accepted
| 22
| 3,444
| 370
|
import copy
n,m=map(int,input().split())
temp=[[] for i in range(n)]
for i in range(m):
a,b=map(int,input().split())
a=a-1
b=b-1
temp[a].append(b)
temp[b].append(a)
count=0
a=[]
c=True
while c:
c=False
for i in range(n):
if len(temp[i])==1 and i not in a:
temp[temp[i][0]].remove(i)
a.append(i)
count=count+1
c=True
print(count)
|
s274674354
|
p03636
|
u379424722
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 110
|
The word `internationalization` is sometimes abbreviated to `i18n`. This comes from the fact that there are 18 letters between the first `i` and the last `n`. You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.
|
N = list(input())
M = []
M.append(N[0])
M.append(str(len(N)))
M.append(N[-1])
date = ''.join(M)
print(date)
|
s765029832
|
Accepted
| 17
| 3,060
| 114
|
N = list(input())
M = []
M.append(N[0])
M.append(str(len(N) - 2))
M.append(N[-1])
date = ''.join(M)
print(date)
|
s406741225
|
p02603
|
u911619829
| 2,000
| 1,048,576
|
Wrong Answer
| 30
| 9,192
| 215
|
To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives. He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows: * A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day. In the i-th day, M-kun can make the following trade **any number of times** (possibly zero), **within the amount of money and stocks that he has at the time**. * Buy stock: Pay A_i yen and receive one stock. * Sell stock: Sell one stock for A_i yen. What is the maximum possible amount of money that M-kun can have in the end by trading optimally?
|
N=int(input())
A=list(map(int,input().split()))
G=1000
K=0
for i in range(N-1):
if A[i+1]>A[i]:
K=int(G/A[i])
G=G % A[i]
if A[i+1]<A[i]:
G=G+A[i]*K
K=0
print(G)
print(K)
G=G+A[N-1]*K
print(G)
|
s577587922
|
Accepted
| 26
| 9,084
| 191
|
N=int(input())
A=list(map(int,input().split()))
G=1000
K=0
for i in range(N-1):
if A[i+1]>A[i]:
K=K+G//A[i]
G=G % A[i]
if A[i+1]<A[i]:
G=G+A[i]*K
K=0
G=G+A[N-1]*K
print(G)
|
s352693468
|
p03860
|
u314050667
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 39
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
s = input().split()
print("A"+s[0]+"C")
|
s358423401
|
Accepted
| 17
| 2,940
| 42
|
s = input().split()
print("A"+s[1][0]+"C")
|
s000836918
|
p03795
|
u451017206
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 56
|
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())
x = 800 * n
y = n //15 * 100
print(x-y)
|
s466376978
|
Accepted
| 18
| 2,940
| 59
|
n = int(input())
x = 800 * n
y = (n //15) * 200
print(x-y)
|
s039731774
|
p02608
|
u802234211
| 2,000
| 1,048,576
|
Wrong Answer
| 900
| 16,944
| 207
|
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())
list_ = [0]*1000000
for i in range(100):
for j in range(100):
for k in range(100):
list_[i**2 + j**2 + k**2+i*j+j*k+k*i - 1] += 1
for i in range(n):
print(list_[i])
|
s498333297
|
Accepted
| 1,190
| 16,804
| 213
|
n = int(input())
list_ = [0]*1000000
for i in range(1,100):
for j in range(1,100):
for k in range(1,100):
list_[i**2 + j**2 + k**2+i*j+j*k+k*i - 1] += 1
for i in range(n):
print(list_[i])
|
s966432151
|
p03759
|
u974935538
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 86
|
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")
|
s057382279
|
Accepted
| 18
| 2,940
| 86
|
a,b,c = map(int,input().split())
if b-a == c-b:
print("YES")
else:
print("NO")
|
s515893705
|
p03378
|
u003505857
| 2,000
| 262,144
|
Wrong Answer
| 28
| 9,172
| 207
|
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 cost of 1. It is guaranteed that there is no toll gate in Square 0, Square X and Square N. Find the minimum cost incurred before reaching the goal.
|
N, M, X = map(int, input().split())
A = list(map(int, input().split()))
for i , a in enumerate(A):
if X < a:
break
if i > len(A) / 2:
ans = len(A[:i])
else:
ans = len(A[:i])
print(ans)
|
s961205436
|
Accepted
| 29
| 8,972
| 295
|
n, m, x = map(int, input().split())
a = list(map(int, input().split()))
cnt1 = 0
cnt2 = 0
x1 = x
x2 = x
while x1 < n+1:
x1 += 1
if x1 in a:
cnt1 += 1
while x2 > 0:
x2 -= 1
if x2 in a:
cnt2 += 1
if cnt1 < cnt2:
ans = cnt1
else:
ans = cnt2
print(ans)
|
s013584710
|
p03761
|
u982594421
| 2,000
| 262,144
|
Wrong Answer
| 19
| 3,060
| 189
|
Snuke loves "paper cutting": he cuts out characters from a newspaper headline and rearranges them to form another string. He will receive a headline which contains one of the strings S_1,...,S_n tomorrow. He is excited and already thinking of what string he will create. Since he does not know the string on the headline yet, he is interested in strings that can be created regardless of which string the headline contains. Find the longest string that can be created regardless of which string among S_1,...,S_n the headline contains. If there are multiple such strings, find the lexicographically smallest one among them.
|
n = int(input())
ans = set()
for _ in range(n):
s = input().rstrip()
if len(s):
ans |= set(s)
else:
ans = ans.intersection(set(s))
ans = ''.join(sorted(list(ans)))
print(ans)
|
s953084541
|
Accepted
| 22
| 3,316
| 365
|
from collections import Counter
n = int(input())
counter = Counter(input())
for _ in range(n - 1):
counter2 = Counter(input())
for k in counter.keys():
if counter2[k] > 0:
counter[k] = min(counter[k], counter2[k])
else:
counter[k] = 0
ans = ''
for k in sorted(counter.keys()):
ans += k * counter[k]
print(ans)
|
s100579409
|
p03545
|
u520276780
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 589
|
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.
|
import sys
s = input()
a = int(s[0])
b=[]
for i in range(3):
b.append(int(s[i+1]))
print(a,*b)
for i in range(8):
tmp = a
for j in range(3):
if (i>>j) & 1 == 0:
tmp += b[j]
else:
tmp -= b[j]
if tmp ==7:
ans = [str(a)]
for k in range(3):
if (i>>k) & 1 == 0:
ans.append("+")
else:
ans.append("-")
ans.append(str(b[k]))
print(''.join(ans),end="")
print("=7")
sys.exit()
|
s750972546
|
Accepted
| 18
| 3,064
| 535
|
import sys
s = input()
a = int(s[0])
b=[]
for i in range(3):
b.append(int(s[i+1]))
for i in range(8):
tmp = a
for j in range(3):
if (i>>j) & 1 == 0:
tmp += b[j]
else:
tmp -= b[j]
if tmp == 7:
ans = [str(a)]
for k in range(3):
if (i>>k) & 1 == 0:
ans.append("+")
else:
ans.append("-")
ans.append(str(b[k]))
print(''.join(ans),end="")
print("=7")
sys.exit()
|
s946134935
|
p03167
|
u758815106
| 2,000
| 1,048,576
|
Wrong Answer
| 614
| 48,424
| 659
|
There is a grid with H horizontal rows and W vertical columns. Let (i, j) denote the square at the i-th row from the top and the j-th column from the left. For each i and j (1 \leq i \leq H, 1 \leq j \leq W), Square (i, j) is described by a character a_{i, j}. If a_{i, j} is `.`, Square (i, j) is an empty square; if a_{i, j} is `#`, Square (i, j) is a wall square. It is guaranteed that Squares (1, 1) and (H, W) are empty squares. Taro will start from Square (1, 1) and reach (H, W) by repeatedly moving right or down to an adjacent empty square. Find the number of Taro's paths from Square (1, 1) to (H, W). As the answer can be extremely large, find the count modulo 10^9 + 7.
|
def mod(num):
return num % (10 ** 9 + 7)
H, W = map(int, input().split())
dp = [[0] * W for _ in range(H)]
for i in range(H):
a = input()
for j in range(W):
if a[j] == "#":
dp[i][j] = -1
dp[0][0] = 1
for col in range(1, W):
if dp[0][col] != 0:
dp[0][col] = dp[0][col - 1]
else:
dp[0][col] = 0
for row in range(1, H):
for col in range(W):
if dp[row][col] != -1:
if col == 0:
dp[row][col] = dp[row - 1][col]
else:
dp[row][col] = mod(dp[row - 1][col] + dp[row][col - 1])
else:
dp[row][col] = 0
print(dp[-1][-1])
|
s782640433
|
Accepted
| 637
| 48,144
| 660
|
def mod(num):
return num % (10 ** 9 + 7)
H, W = map(int, input().split())
dp = [[0] * W for _ in range(H)]
for i in range(H):
a = input()
for j in range(W):
if a[j] == "#":
dp[i][j] = -1
dp[0][0] = 1
for col in range(1, W):
if dp[0][col] != -1:
dp[0][col] = dp[0][col - 1]
else:
dp[0][col] = 0
for row in range(1, H):
for col in range(W):
if dp[row][col] != -1:
if col == 0:
dp[row][col] = dp[row - 1][col]
else:
dp[row][col] = mod(dp[row - 1][col] + dp[row][col - 1])
else:
dp[row][col] = 0
print(dp[-1][-1])
|
s020194420
|
p02612
|
u218757284
| 2,000
| 1,048,576
|
Wrong Answer
| 35
| 9,140
| 32
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
print(n % 1000)
|
s849268300
|
Accepted
| 24
| 9,152
| 76
|
n = int(input())
if n % 1000 == 0:
print(0)
else:
print(1000 - n % 1000)
|
s876646075
|
p04046
|
u497625442
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 706
|
We have a large square grid with H rows and W columns. Iroha is now standing in the top-left cell. She will repeat going right or down to the adjacent cell, until she reaches the bottom-right cell. However, she cannot enter the cells in the intersection of the bottom A rows and the leftmost B columns. (That is, there are A×B forbidden cells.) There is no restriction on entering the other cells. Find the number of ways she can travel to the bottom-right cell. Since this number can be extremely large, print the number modulo 10^9+7.
|
# E.py#
#K, N = map(int,input().split())
P = 10**9+7
H, W, A, B = map(int,"10 7 3 4".split())
def egcd(a, b):
if a == 0:
return (b, 0, 1)
else:
g, y, x = egcd(b % a, a)
return (g, x - (b // a) * y, y)
def modinv(a, m):
g, x, y = egcd(a, m)
if g != 1:
raise Exception('modular inverse does not exist')
else:
return x % m
def fact(i):
t = 1
while i > 0:
t = (t * i) % P
i -= 1
return t
def comb(i,j):
return (fact(i+j) * modinv(fact(i),P) * modinv(fact(j),P)) % P
s = 0
i = 0
while H-A-i > 0 and B+i+1 <= W and A+i <= H and W-B-i >= 0:
# print((H-A-i,B+i+1))
s = (s + comb(H-A-i-1,B+i+1-1) * comb(A+i,W-B-i-1)) % P
i += 1
print(s)
|
s679606905
|
Accepted
| 324
| 18,804
| 576
|
H, W, A, B = map(int,input().split())
P = 10**9+7
# H, W, A, B = map(int,"2 3 1 1".split())
M = H+W-1
factlist = [1] * (H+W)
factinvlist = [1] * (H+W)
t = 1
for i in range(M):
t = (t * (i+1)) % P
factlist[i+1] = t
t = pow(factlist[M],P-2,P)
factinvlist[M] = t
for i in range(M):
t = (t * (M-i)) % P
factinvlist[M-i-1] = t
def comb(i,j):
return (factlist[i+j] * factinvlist[i] * factinvlist[j]) % P
s = 0
i = 0
while H-A-i-1 >= 0 and B+i <= W and A+i <= H and W-B-i-1 >= 0:
# print((H-A-i,B+i+1))
s = (s + comb(H-A-i-1,B+i) * comb(A+i,W-B-i-1)) % P
i += 1
print(s)
|
s628252175
|
p03971
|
u013202780
| 2,000
| 262,144
|
Wrong Answer
| 100
| 4,016
| 228
|
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these. Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests. * A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B. * An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students. A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these. Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
|
n,a,b=map(int,input().split())
c=0
l=a+b
for i in input():
if i=="a" and l>0:
l-=1
print ("YES")
elif i=="b" and l>0 and b>0:
print ("YES")
l-=1
b-=1
else:
print ("NO")
|
s941171939
|
Accepted
| 92
| 4,016
| 238
|
n,a,b=map(int,input().split())
l=a+b
for i in input():
if i=="a" and l>0:
l-=1
print ("Yes")
elif i=="b" and l>0 and b>0:
print ("Yes")
l-=1
b-=1
else:
print ("No")
|
s081657587
|
p03379
|
u171065106
| 2,000
| 262,144
|
Wrong Answer
| 168
| 30,856
| 182
|
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
n = int(input())
x = sorted(list(map(int, input().split())))
a = x[len(x)//2]
b = x[(len(x)//2) - 1]
for _ in range(len(x)//2):
print(a)
for _ in range(len(x)//2):
print(b)
|
s574114792
|
Accepted
| 179
| 30,832
| 190
|
n = int(input())
x = list(map(int, input().split()))
y = sorted(x)
a = y[len(x)//2]
b = y[(len(x)//2) - 1]
for i in range(n):
if x[i] <= b:
print(a)
else:
print(b)
|
s447367354
|
p02407
|
u025362139
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,572
| 58
|
Write a program which reads a sequence and prints it in the reverse order.
|
i = list(map(int, input().split()))
i.reverse()
print(i)
|
s557969750
|
Accepted
| 20
| 5,624
| 187
|
N = int(input())
i = list(map(int, input().split()))
i.reverse()
for idx, val in enumerate(i):
if idx == N-1:
print(val)
else:
print(val, ' ', sep = '', end='')
|
s140551070
|
p02678
|
u453815934
| 2,000
| 1,048,576
|
Wrong Answer
| 2,207
| 48,016
| 511
|
There is a cave. The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside. It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage. Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1. * If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible. Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.
|
from collections import deque
import sys
a,b=map(int,input().split())
List=[list(map(int,input().split())) for i in range(b)]
l=[0]*(a+1)
line=deque([1])
while line:
i=line.popleft()
bye=[]
for j in range(len(List)):
x,y=List[j]
if x==i:
if l[y]==0:
l[y]=i
bye+=[j]
line.append(y)
if y==i:
if l[x]==0:
l[x]=i
bye+=[j]
line.append(x)
for p in range(len(bye)):
List.pop(bye[p]-p)
print("yes")
for h in range(a-1):
print(l[h+2])
|
s373732376
|
Accepted
| 519
| 34,412
| 612
|
import math
import sys
from collections import deque
import heapq
import copy
import itertools
from itertools import permutations
from itertools import combinations
import bisect
def mi() : return map(int,sys.stdin.readline().split())
def ii() : return int(sys.stdin.readline().rstrip())
def i() : return sys.stdin.readline().rstrip()
a,b=mi()
l=[[] for i in range(a)]
for j in range(b):
x,y=mi()
l[x-1].append(y-1)
l[y-1].append(x-1)
lst=[-1]*a
q=deque([0])
while q:
s=q.popleft()
for j in l[s]:
if lst[j]==-1:
lst[j]=s
q.append(j)
print("Yes")
for k in range(a-1):
print(lst[k+1]+1)
|
s200095021
|
p03447
|
u317710033
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 149
|
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?
|
total = int(input())
first_cake = int(input())
second_cake = int(input())
total =- first_cake
while total >= 0:
total =- second_cake
print(total)
|
s136837225
|
Accepted
| 17
| 2,940
| 161
|
total = int(input())
first_cake = int(input())
second_cake = int(input())
total -= first_cake
while total >= second_cake:
total -= second_cake
print(total )
|
s056710278
|
p03671
|
u722535636
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 70
|
Snuke is buying a bicycle. The bicycle of his choice does not come with a bell, so he has to buy one separately. He has very high awareness of safety, and decides to buy two bells, one for each hand. The store sells three kinds of bells for the price of a, b and c yen (the currency of Japan), respectively. Find the minimum total price of two different bells.
|
print(sum(sorted(list(map(int,input().split())), reverse = True)[:2]))
|
s027439714
|
Accepted
| 17
| 2,940
| 54
|
print(sum(sorted(list(map(int,input().split())))[:2]))
|
s656535780
|
p03696
|
u024442309
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 218
|
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 `(`, X and `)` in this order is also a correct bracket sequence. * If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence. * Every correct bracket sequence can be derived from the rules above. Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.
|
N = int(input())
S = input()
nl = 0
nr = 0
for i in S:
if i == ')':
if nr > 0:
nr -= 1
else:
nl += 1
else:
nr += 1
print(nr)
print(nl)
print('('*nl + S + ')'*nr)
|
s493831568
|
Accepted
| 17
| 2,940
| 198
|
N = int(input())
S = input()
nl = 0
nr = 0
for i in S:
if i == ')':
if nr > 0:
nr -= 1
else:
nl += 1
else:
nr += 1
print('('*nl + S + ')'*nr)
|
s477854525
|
p03448
|
u292220197
| 2,000
| 262,144
|
Wrong Answer
| 54
| 3,060
| 283
|
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(), 10)
b = int(input(), 10)
c = int(input(), 10)
x = int(input(), 10)
ptn = 0
for i in range(a+1):
for j in range(b+1):
for t in range(c+1):
total = (a * 500) + (b * 100) + (c * 50)
if x == total:
ptn += 1
print(ptn)
|
s245440571
|
Accepted
| 51
| 3,060
| 221
|
a = int(input())
b = int(input())
c = int(input())
x = int(input())
cnt = 0
for i in range(0,a+1):
for j in range(0,b+1):
for k in range(0,c+1):
if x == 500*i + 100*j + 50*k:
cnt = cnt + 1
print(cnt)
|
s616208576
|
p02612
|
u376812964
| 2,000
| 1,048,576
|
Wrong Answer
| 36
| 9,144
| 33
|
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
print(N % 1000)
|
s919750420
|
Accepted
| 28
| 9,136
| 75
|
N = int(input()) % 1000
if N == 0:
print(0)
else:
print(1000 - N)
|
s387824145
|
p02647
|
u878384274
| 2,000
| 1,048,576
|
Wrong Answer
| 2,206
| 32,196
| 252
|
We have N bulbs arranged on a number line, numbered 1 to N from left to right. Bulb i is at coordinate i. Each bulb has a non-negative integer parameter called intensity. When there is a bulb of intensity d at coordinate x, the bulb illuminates the segment from coordinate x-d-0.5 to x+d+0.5. Initially, the intensity of Bulb i is A_i. We will now do the following operation K times in a row: * For each integer i between 1 and N (inclusive), let B_i be the number of bulbs illuminating coordinate i. Then, change the intensity of each bulb i to B_i. Find the intensity of each bulb after the K operations.
|
N , K = map(int,input().split())
Lum = [a for a in map(int,input().split())]
for i1 in range(K):
B = [0 for i in range(N)]
for i in range(N):
for j in range(max(0,i-Lum[i]),min(i+Lum[i]+1,N)):
B[j]+=1
Lum = B
print(Lum)
|
s365698432
|
Accepted
| 895
| 136,004
| 560
|
import numpy as np
from numba import jit
N , K = map(int,input().split())
Lum = np.array(list(map(int,input().split())))
@jit("i8,i8,i8[:]")
def fn(N,K,A):
K1 = min(K,41)
for i1 in range(K1):
B = np.zeros(N,dtype=np.int64)
for i in range(N):
l = max(0,i-A[i])
r = min(N-1,i+A[i])
B[l]+=1
if r+1<N:
B[r+1]-=1
for i in range(1,N):
B[i]+=B[i-1]
A = B
return A
Lum = list(fn(N,K,Lum))
Lum = [str(l) for l in Lum]
Lum=" ".join(Lum)
print(Lum)
|
s021674248
|
p03457
|
u167501921
| 2,000
| 262,144
|
Wrong Answer
| 841
| 3,188
| 458
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N=int(input())
oldn,oldx,oldy=0,0,0
#print(x,y)
no=0
for i in range(N):
nxy=input().split()
n,x,y=int(nxy[0]),int(nxy[1]),int(nxy[2])
# print(n,x,y)
ndiff = n - oldn
distance = abs(x - oldx) + abs(y - oldy)
# print(ndiff,distance)
oldn,oldx,oldy=n,x,y
if distance > ndiff:
print("No")
no=1
break
print((ndiff-distance)%2)
if (ndiff-distance)%2==0:
continue
else:
no=1
print("No")
break
if no==0:
print("Yes")
|
s700230684
|
Accepted
| 352
| 3,064
| 459
|
N=int(input())
oldn,oldx,oldy=0,0,0
#print(x,y)
no=0
for i in range(N):
nxy=input().split()
n,x,y=int(nxy[0]),int(nxy[1]),int(nxy[2])
# print(n,x,y)
ndiff = n - oldn
distance = abs(x - oldx) + abs(y - oldy)
# print(ndiff,distance)
oldn,oldx,oldy=n,x,y
if distance > ndiff:
print("No")
no=1
break
# print((ndiff-distance)%2)
if (ndiff-distance)%2==0:
continue
else:
no=1
print("No")
break
if no==0:
print("Yes")
|
s715205293
|
p02645
|
u191960840
| 2,000
| 1,048,576
|
Wrong Answer
| 27
| 8,956
| 30
|
When you asked some guy in your class his name, he called himself S, where S is a string of length between 3 and 20 (inclusive) consisting of lowercase English letters. You have decided to choose some three consecutive characters from S and make it his nickname. Print a string that is a valid nickname for him.
|
S = str(input())
print(S[0:2])
|
s354882271
|
Accepted
| 29
| 8,916
| 30
|
S = str(input())
print(S[0:3])
|
s886731081
|
p02402
|
u316584871
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,592
| 119
|
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())
l = input().split()
l.sort()
sum = 0
for i in range(n):
sum = sum+int(l[i])
print(l[0],l[-1],sum)
|
s789610085
|
Accepted
| 20
| 6,588
| 135
|
n = int(input())
l = list(map(int, input().split()))
l.sort()
sum = 0
for i in range(n):
sum = sum+int(l[i])
print(l[0],l[-1],sum)
|
s193759479
|
p03679
|
u695079172
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 225
|
Takahashi has a strong stomach. He never gets a stomachache from eating something whose "best-by" date is at most X days earlier. He gets a stomachache if the "best-by" date of the food is X+1 or more days earlier, though. Other than that, he finds the food delicious if he eats it not later than the "best-by" date. Otherwise, he does not find it delicious. Takahashi bought some food A days before the "best-by" date, and ate it B days after he bought it. Write a program that outputs `delicious` if he found it delicious, `safe` if he did not found it delicious but did not get a stomachache either, and `dangerous` if he got a stomachache.
|
def main():
x,a,b = map(int,input().split())
now = b - a
if now < 0:
print("delicious")
elif now <= x:
print("save")
else:
print("dangerous")
if __name__ == '__main__':
main()
|
s687239823
|
Accepted
| 17
| 2,940
| 226
|
def main():
x,a,b = map(int,input().split())
now = b - a
if now <= 0:
print("delicious")
elif now <= x:
print("safe")
else:
print("dangerous")
if __name__ == '__main__':
main()
|
s933047832
|
p03963
|
u288430479
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 53
|
There are N balls placed in a row. AtCoDeer the deer is painting each of these in one of the K colors of his paint cans. For aesthetic reasons, any two adjacent balls must be painted in different colors. Find the number of the possible ways to paint the balls.
|
n,k = map(int,input().split())
print(k+(k-1)**(n-1))
|
s507494333
|
Accepted
| 17
| 2,940
| 52
|
n,k = map(int,input().split())
print(k*(k-1)**(n-1))
|
s860517316
|
p03379
|
u896726004
| 2,000
| 262,144
|
Wrong Answer
| 323
| 25,228
| 286
|
When l is an odd number, the median of l numbers a_1, a_2, ..., a_l is the (\frac{l+1}{2})-th largest value among a_1, a_2, ..., a_l. You are given N numbers X_1, X_2, ..., X_N, where N is an even number. For each i = 1, 2, ..., N, let the median of X_1, X_2, ..., X_N excluding X_i, that is, the median of X_1, X_2, ..., X_{i-1}, X_{i+1}, ..., X_N be B_i. Find B_i for each i = 1, 2, ..., N.
|
n = int(input())
x = list(map(int, input().split()))
if n==2:
print(x[1])
print(x[0])
exit()
x_sort = sorted(x)
med1_ix = n//2-1
med2_ix = n//2
med1 = x_sort[med1_ix]
med2 = x_sort[med2_ix]
for i in x:
if i <= med1:
print(med1)
else:
print(med2)
|
s999215213
|
Accepted
| 286
| 26,772
| 233
|
n = int(input())
x = list(map(int, input().split()))
x_sort = sorted(x)
med1_ix = n//2-1
med2_ix = n//2
med1 = x_sort[med1_ix]
med2 = x_sort[med2_ix]
for i in x:
if i <= med1:
print(med2)
else:
print(med1)
|
s974315352
|
p03645
|
u301823349
| 2,000
| 262,144
|
Wrong Answer
| 2,105
| 17,312
| 272
|
In Takahashi Kingdom, there is an archipelago of N islands, called Takahashi Islands. For convenience, we will call them Island 1, Island 2, ..., Island N. There are M kinds of regular boat services between these islands. Each service connects two islands. The i-th service connects Island a_i and Island b_i. Cat Snuke is on Island 1 now, and wants to go to Island N. However, it turned out that there is no boat service from Island 1 to Island N, so he wants to know whether it is possible to go to Island N by using two boat services. Help him.
|
N,M=map(int, input().split())
p=[]
for x in range(M):
p.append(input())
f=0
for x in range(M):
if "1 %s" % x in p:
if "%s N" % x in p:
print("POSSIBLE")
else:
f=f+1
else:
f=f+1
if f==M:
print("IMPOSSIBLE")
|
s039041268
|
Accepted
| 584
| 34,796
| 403
|
N,M=map(int, input().split())
p=[]
for x in range(M):
p.append(input())
A=[]
B=[]
for x in range(M):
a,b=map(int,p[x].split())
A.append(a)
B.append(b)
isl=[]
for x in range(N):
isl.append(0)
for x in range(M):
if A[x]==1:
isl[B[x]]=1
for x in range(M):
if B[x]==N:
isl[A[x]]=isl[A[x]]+1
if 2 in isl:
print("POSSIBLE")
else:
print("IMPOSSIBLE")
|
s922272750
|
p03477
|
u821432765
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 111
|
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=3
B=2
C=1
D=4
L=A+B
R=C+D
if L<R:
print("Right")
if L>R:
print("Left")
if L==R:
print("Balanced")
|
s902206874
|
Accepted
| 18
| 3,060
| 140
|
A,B,C,D=map(int,input().split())
AB=A+B
CD=C+D
if AB==CD:
print("Balanced")
elif AB>CD:
print("Left")
elif AB<CD:
print("Right")
|
s949917742
|
p02257
|
u409571842
| 1,000
| 131,072
|
Wrong Answer
| 30
| 5,660
| 354
|
A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7. Write a program which reads a list of _N_ integers and prints the number of prime numbers in the list.
|
#coding: UTF-8
import sys
import math
class Algo:
@staticmethod
def is_prime(l):
ans = 0
for n in l:
if n == 2:
ans += 1
else:
for i in range(2,int(math.sqrt(n))+2):
if n%i == 0:
ans += 1
print(ans)
N = int(input())
L = []
for i in range(0,N):
L.append(int(input()))
Algo.is_prime(L)
|
s761235905
|
Accepted
| 230
| 5,676
| 387
|
#coding: UTF-8
import sys
import math
class Algo:
@staticmethod
def is_prime(n):
if n == 2:
return True
elif n<2 or n%2 == 0:
return False
else:
for i in range(3,int(math.sqrt(n))+2, 2):
if n%i == 0:
return False
return True
N = int(input())
ans = 0
for i in range(0,N):
if Algo.is_prime(int(input())):
ans += 1
print(ans)
|
s832149203
|
p03557
|
u406138190
| 2,000
| 262,144
|
Wrong Answer
| 577
| 23,360
| 375
|
The season for Snuke Festival has come again this year. First of all, Ringo will perform a ritual to summon Snuke. For the ritual, he needs an altar, which consists of three parts, one in each of the three categories: upper, middle and lower. He has N parts for each of the three categories. The size of the i-th upper part is A_i, the size of the i-th middle part is B_i, and the size of the i-th lower part is C_i. To build an altar, the size of the middle part must be strictly greater than that of the upper part, and the size of the lower part must be strictly greater than that of the middle part. On the other hand, any three parts that satisfy these conditions can be combined to form an altar. How many different altars can Ringo build? Here, two altars are considered different when at least one of the three parts used is different.
|
import bisect
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=list(map(int,input().split()))
a=sorted(a)
b=sorted(b)
c=sorted(c)
sum=0
for i in range(n):
bisect_a=bisect.bisect_left(a,b[i])
print(bisect_a)
bisect_c=bisect.bisect_right(c,b[i])
print(bisect_c)
sum+=bisect_a*(n-bisect_c)
print(sum)
print(sum)
|
s345414669
|
Accepted
| 350
| 22,516
| 326
|
import bisect
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=list(map(int,input().split()))
a=sorted(a)
b=sorted(b)
c=sorted(c)
sum=0
for i in range(n):
bisect_a=bisect.bisect_left(a,b[i])
bisect_c=bisect.bisect_right(c,b[i])
sum+=bisect_a*(n-bisect_c)
print(sum)
|
s460671840
|
p03387
|
u685983477
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 196
|
You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order: * Choose two among A, B and C, then increase both by 1. * Choose one among A, B and C, then increase it by 2. It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.
|
a=[int(i) for i in input().split()]
a.sort()
print(a)
ans=0
while a[1]!=a[2]:
a[1]+=1
a[0]+=1
ans+=1
if a[1]%2==a[0]%2:
ans+=(a[1]-a[0])//2
else:
ans+=(a[1]-a[0])+1
print(ans)
|
s971794954
|
Accepted
| 18
| 3,060
| 211
|
a=[int(i) for i in input().split()]
a.sort()
ans=0
while a[1]!=a[2]:
a[1]+=1
a[0]+=1
ans+=1
if a[1]%2==a[0]%2:
ans+=(a[1]-a[0])//2
else:
ans+=1
a[1]+=1
ans+=(a[1]-a[0])//2
print(ans)
|
s302850758
|
p02388
|
u478810373
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,380
| 10
|
Write a program which calculates the cube of a given integer x.
|
x =2
x**3
|
s870450756
|
Accepted
| 20
| 7,548
| 28
|
x = int(input())
print(x**3)
|
s514604747
|
p02613
|
u208120643
| 2,000
| 1,048,576
|
Wrong Answer
| 144
| 9,204
| 341
|
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.
|
"""B - Judge Status Summary"""
N = int(input())
ac, wa, tle, re = 0, 0, 0, 0
for i in range(N):
S = input()
if S == "AC":
ac = +1
elif S == "WA":
wa = +1
elif S == "TLE":
tle = +1
else:
re = +1
print("AC x "+str(ac))
print("WA x "+str(wa))
print("TLE x "+str(tle))
print("RE x "+str(re))
|
s181685667
|
Accepted
| 144
| 9,044
| 341
|
"""B - Judge Status Summary"""
N = int(input())
ac, wa, tle, re = 0, 0, 0, 0
for i in range(N):
S = input()
if S == "AC":
ac += 1
elif S == "WA":
wa += 1
elif S == "TLE":
tle += 1
else:
re += 1
print("AC x "+str(ac))
print("WA x "+str(wa))
print("TLE x "+str(tle))
print("RE x "+str(re))
|
s153300510
|
p02240
|
u569960318
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,672
| 627
|
Write a program which reads relations in a SNS (Social Network Service), and judges that given pairs of users are reachable each other through the network.
|
def connected(G,s,t):
checked = [False]*len(G)
def main(G,s,t):
if t in G[s]: return True
checked[s] = True
for f in G[s]:
if checked[f]: continue
if main(G,f,t): return True
return False
return main(G,s,t)
if __name__=='__main__':
n,m = list(map(int,input().split()))
G = [[] for _ in range(n)]
for _ in range(m):
s,t = list(map(int,input().split()))
G[s].append(t)
G[t].append(s)
q = int(input())
for _ in range(q):
s,t = list(map(int,input().split()))
print('Yes' if connected(G,s,t) else 'No')
|
s400225314
|
Accepted
| 480
| 33,568
| 840
|
import sys
def make_connected_group(G):
C = [None]*len(G)
group = 0
for i,p in enumerate(C):
if p != None: continue
friends = [i]
while len(friends) > 0:
f = friends.pop()
if C[f] == None:
C[f] = group
friends += G[f]
group += 1
return C
if __name__=='__main__':
n,m = list(map(int,sys.stdin.readline().split()))
G = [[] for _ in range(n)]
for j,line in enumerate(sys.stdin):
if not j < m:
q = int(line)
break
s,t = list(map(int,line.split()))
G[s].append(t)
G[t].append(s)
C = make_connected_group(G)
for j,line in enumerate(sys.stdin):
if not j < q: break
s,t = list(map(int,line.split()))
print('yes' if C[s] == C[t] else 'no')
|
s778492916
|
p03456
|
u083960235
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 200
|
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.
|
l=list(map(str,input().split()))
l=int(l[0]+l[1])
print(l)
temp=False
for i in range(1000):
if i*i==l:
temp=True
break
if temp==True:
print("Yes")
else:
print("No")
|
s438110368
|
Accepted
| 17
| 2,940
| 191
|
l=list(map(str,input().split()))
l=int(l[0]+l[1])
temp=False
for i in range(1000):
if i*i==l:
temp=True
break
if temp==True:
print("Yes")
else:
print("No")
|
s155310751
|
p02400
|
u636711749
| 1,000
| 131,072
|
Wrong Answer
| 30
| 6,804
| 104
|
Write a program which calculates the area and circumference of a circle for given radius r.
|
import math
r = float(input())
S =r * r * math.pi
C =r * math.pi
print('{0:.6f} {1:.6f}'.format(S,C))
|
s498598270
|
Accepted
| 30
| 6,804
| 110
|
import math
r = float(input())
S =r * r * math.pi
C =(r + r) * math.pi
print('{0:.6f} {1:.6f}'.format(S,C))
|
s242579115
|
p03386
|
u167647458
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 197
|
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 = set()
for i in range(a, a+k):
if a <= i <= b:
s.add(i)
for i in range(b-k, b+1):
if a <= i <= b:
s.add(i)
for i in s:
print(i)
|
s722197413
|
Accepted
| 17
| 3,060
| 206
|
a, b, k = map(int, input().split())
s = set()
for i in range(a, a+k):
if a <= i <= b:
s.add(i)
for i in range(b-k+1, b+1):
if a <= i <= b:
s.add(i)
for i in sorted(s):
print(i)
|
s728122646
|
p03435
|
u698479721
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,060
| 324
|
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.
|
c1, c2, c3 = map(int, input().split())
c4, c5, c6 = map(int, input().split())
c7, c8, c9 = map(int, input().split())
if c4-c1==c5-c2 and c5-c2 == c6-c3 and c7-c4 ==c8-c5 and c8-c5 == c9-c6:
if c2-c1 == c5-c4 and c5-c4 == c8-c7 and c3-c2 == c6-c5 and c6-c5 == c9-c8:
print('Yes')
print('No')
else:
print('No')
|
s504262118
|
Accepted
| 17
| 3,060
| 329
|
c1, c2, c3 = map(int, input().split())
c4, c5, c6 = map(int, input().split())
c7, c8, c9 = map(int, input().split())
if c4-c1==c5-c2 and c5-c2 == c6-c3 and c7-c4 ==c8-c5 and c8-c5 == c9-c6:
if c2-c1 == c5-c4 and c5-c4 == c8-c7 and c3-c2 == c6-c5 and c6-c5 == c9-c8:
print('Yes')
else:print('No')
else:
print('No')
|
s276038000
|
p03730
|
u721970149
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 126
|
We ask you to select some number of positive integers, and calculate the sum of them. It is allowed to select as many integers as you like, and as large integers as you wish. You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer. Your objective is to make the sum congruent to C modulo B. Determine whether this is possible. If the objective is achievable, print `YES`. Otherwise, print `NO`.
|
A, B, C = map(int,input().split())
ans = "No"
for i in range(1,B+1) :
if (A*i)%B == C :
ans = "Yes"
print(ans)
|
s756371860
|
Accepted
| 17
| 2,940
| 126
|
A, B, C = map(int,input().split())
ans = "NO"
for i in range(1,B+1) :
if (A*i)%B == C :
ans = "YES"
print(ans)
|
s961435019
|
p03852
|
u089376182
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 55
|
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
print('vowel' if input() in ['aeiou'] else 'consonant')
|
s067232311
|
Accepted
| 17
| 2,940
| 54
|
print('vowel' if input() in 'aeiou' else 'consonant')
|
s466928357
|
p03448
|
u617037231
| 2,000
| 262,144
|
Wrong Answer
| 49
| 3,064
| 193
|
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.
|
def i():
return int(input())
A = i()
B = i()
C = i()
X = i()
ans = 0
for i in range(A):
for j in range(B):
for k in range(C):
if 500*i+100*j+50*k == X:
ans += 1
print(ans)
|
s665368294
|
Accepted
| 51
| 3,060
| 200
|
def i():
return int(input())
A = i()
B = i()
C = i()
X = i()
ans = 0
for i in range(A+1):
for j in range(B+1):
for k in range(C+1):
if 500*i+100*j+50*k == X:
ans += 1
print(ans)
|
s655648460
|
p03644
|
u502389123
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 2,940
| 207
|
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())
count = 0
counter = []
for i in range(1, N+1):
while i % 2 == 0:
count += 1
counter.append([count, i])
count = 0
counter.sort()
counter.reverse()
print(counter[0][1])
|
s846552788
|
Accepted
| 17
| 3,060
| 238
|
N = int(input())
count = 0
counter = []
for i in range(1, N+1):
num = i
while num % 2 == 0:
count += 1
num /= 2
counter.append([count, i])
count = 0
counter.sort()
counter.reverse()
print(counter[0][1])
|
s647369063
|
p02927
|
u970809473
| 2,000
| 1,048,576
|
Wrong Answer
| 24
| 3,060
| 183
|
Today is August 24, one of the five Product Days in a year. A date m-d (m is the month, d is the date) is called a Product Day when d is a two-digit number, and all of the following conditions are satisfied (here d_{10} is the tens digit of the day and d_1 is the ones digit of the day): * d_1 \geq 2 * d_{10} \geq 2 * d_1 \times d_{10} = m Takahashi wants more Product Days, and he made a new calendar called Takahashi Calendar where a year consists of M month from Month 1 to Month M, and each month consists of D days from Day 1 to Day D. In Takahashi Calendar, how many Product Days does a year have?
|
m,d = map(int, input().split())
res = 0
for i in range(1,m+1):
for j in range(1,d+1):
if j >= 22 and j%10 >= 2 and int(str(j)[0]) *int(str(j)[1]) == m:
res += 1
print(res)
|
s932744564
|
Accepted
| 23
| 2,940
| 187
|
m,d = map(int, input().split())
res = 0
for i in range(1,m+1):
for j in range(1,d+1):
if j >= 22 and j % 10 >= 2 and int(str(j)[0]) * int(str(j)[1]) == i:
res += 1
print(res)
|
s829307282
|
p03050
|
u422711869
| 2,000
| 1,048,576
|
Wrong Answer
| 116
| 3,272
| 438
|
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.
|
from sys import stdin
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
return divisors
n = int(stdin.readline().rstrip())
ds = make_divisors(n)
#print(ds)
sum = 0
for x in ds:
if x==n:
continue
m = int(n/x-1)
# print(x,m)
sum += m
print(sum)
|
s383742698
|
Accepted
| 119
| 3,264
| 504
|
from sys import stdin
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
return divisors
n = int(stdin.readline().rstrip())
ds = make_divisors(n)
#print(ds)
sum = 0
for x in ds:
if x==n:
continue
m = int(n/x-1)
if m==1:
continue
# print(x,m, int(n/m), n%m)
if(int(n/m) == n%m):
sum += m
print(sum)
|
s081115065
|
p03525
|
u368796742
| 2,000
| 262,144
|
Wrong Answer
| 30
| 9,176
| 237
|
In CODE FESTIVAL XXXX, there are N+1 participants from all over the world, including Takahashi. Takahashi checked and found that the _time gap_ (defined below) between the local times in his city and the i-th person's city was D_i hours. The time gap between two cities is defined as follows. For two cities A and B, if the local time in city B is d o'clock at the moment when the local time in city A is 0 o'clock, then the time gap between these two cities is defined to be min(d,24-d) hours. Here, we are using 24-hour notation. That is, the local time in the i-th person's city is either d o'clock or 24-d o'clock at the moment when the local time in Takahashi's city is 0 o'clock, for example. Then, for each pair of two people chosen from the N+1 people, he wrote out the time gap between their cities. Let the smallest time gap among them be s hours. Find the maximum possible value of s.
|
n = int(input())
d = list(map(int,input().split()))
if n >= 12:
count = [0]*24
count[0] += 2
for i in d:
count[i] += 1
count[(24-i)%24] += 1
if max(count) > 2:
print(0)
else:
print(1)
|
s621327431
|
Accepted
| 42
| 9,220
| 539
|
n = int(input())
d = list(map(int,input().split()))
if n >= 12:
count = [0]*24
count[0] += 2
for i in d:
count[i] += 1
count[(24-i)%24] += 1
if max(count) > 2:
print(0)
else:
print(1)
exit()
ans = 0
for i in range(1<<n):
l = [0,24]
for j in range(n):
if i>>j & 1:
l.append(d[j])
else:
l.append(24-d[j])
l.sort()
count = 24
for j in range(len(l)-1):
count = min(count,l[j+1]-l[j])
ans = max(ans,count)
print(ans)
|
s283146449
|
p03386
|
u316233444
| 2,000
| 262,144
|
Wrong Answer
| 2,103
| 3,060
| 171
|
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
A,B,K=map(int,input().split())
ans=[]
for i in range(A,B+1):
if i <= A+K:
ans.append(i)
if i >= B-K:
ans.append(i)
for i in sorted(set(ans)):
print(i)
|
s174072423
|
Accepted
| 17
| 3,060
| 165
|
A,B,K=map(int,input().split())
ans=[]
for i in range(K):
if A+i <= B:
ans.append(A+i)
if B-i >= A:
ans.append(B-i)
for i in sorted(set(ans)):
print(i)
|
s660610992
|
p03730
|
u145600939
| 2,000
| 262,144
|
Wrong Answer
| 18
| 2,940
| 118
|
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())
ans = 'No'
for i in range(1,b+1):
if (a*i)%b == c:
ans = 'Yes'
print(ans)
|
s056197593
|
Accepted
| 17
| 2,940
| 129
|
a,b,c = map(int,input().split())
ans = 'NO'
for i in range(1,b+1):
if (a*i)%b == c:
ans = 'YES'
break
print(ans)
|
s062544841
|
p02928
|
u814986259
| 2,000
| 1,048,576
|
Wrong Answer
| 359
| 3,188
| 206
|
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 ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.
|
N,K=map(int,input().split())
A=list(map(int,input().split()))
ans=0
for i in range(N):
count=0
for j in range(i+1,N):
if A[i]>A[j]:
count+=1
ans+=count
ans+=((N-1)*N // 2)*(K-1)
print(ans)
|
s204508105
|
Accepted
| 611
| 9,320
| 440
|
N, K = map(int, input().split())
A = list(map(int, input().split()))
count = [[0, 0] for i in range(N)]
for i in range(N):
for j in range(i):
if A[j] < A[i]:
count[i][0] += 1
for j in range(i+1, N):
if A[j] < A[i]:
count[i][1] += 1
ans = 0
mod = 10**9 + 7
for i in range(N):
ans += (1+K-1)*(K-1) // 2 * count[i][0]
ans += (1+K)*K // 2 * count[i][1]
ans %= mod
print(ans % mod)
|
s912006159
|
p03386
|
u576434377
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 192
|
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().strip().split())
res = []
for a in range(A,min(A + K,B)):
res.append(a)
for b in range(max(A,B - K + 1),B + 1):
res.append(b)
print(sorted(list(set(res))))
|
s756846836
|
Accepted
| 17
| 3,060
| 209
|
A, B, K = map(int,input().strip().split())
res = []
for a in range(A,min(A + K,B)):
res.append(a)
for b in range(max(A,B - K + 1),B + 1):
res.append(b)
for r in sorted(list(set(res))):
print(r)
|
s837213962
|
p02411
|
u628732336
| 1,000
| 131,072
|
Wrong Answer
| 20
| 7,652
| 394
|
Write a program which reads a list of student test scores and evaluates the performance for each student. The test scores for a student include scores of the midterm examination m (out of 50), the final examination f (out of 50) and the makeup examination r (out of 100). If the student does not take the examination, the score is indicated by -1. The final performance of a student is evaluated by the following procedure: * If the student does not take the midterm or final examination, the student's grade shall be F. * If the total score of the midterm and final examination is greater than or equal to 80, the student's grade shall be A. * If the total score of the midterm and final examination is greater than or equal to 65 and less than 80, the student's grade shall be B. * If the total score of the midterm and final examination is greater than or equal to 50 and less than 65, the student's grade shall be C. * If the total score of the midterm and final examination is greater than or equal to 30 and less than 50, the student's grade shall be D. However, if the score of the makeup examination is greater than or equal to 50, the grade shall be C. * If the total score of the midterm and final examination is less than 30, the student's grade shall be F.
|
while True:
m, f, r = [int(i) for i in input().split()]
if m == f == r == -1:
break
total = m + f
if m == -1 or f == -1 or total < 30:
print("F")
elif total <= 80:
print("A")
elif 65 <= total < 80:
print("B")
elif 50 <= total < 65:
print("C")
elif 30 <= total < 50 and r < 50:
print("D")
print()
|
s037137086
|
Accepted
| 30
| 7,656
| 345
|
while True:
m, f, r = [int(i) for i in input().split()]
if m == f == r == -1:
break
total = m + f
if m == -1 or f == -1 or total < 30:
print("F")
elif total < 50 and r < 50:
print("D")
elif r >= 50 or total < 65:
print("C")
elif total < 80:
print("B")
else:
print("A")
|
s413248696
|
p03377
|
u198274496
| 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())
print("Yes" if A <= X and A + B >= X else "No")
|
s538537871
|
Accepted
| 18
| 2,940
| 84
|
A, B, X = map(int, input().split())
print("YES" if A <= X and A + B >= X else "NO")
|
s212511885
|
p03473
|
u657208344
| 2,000
| 262,144
|
Wrong Answer
| 20
| 3,060
| 30
|
How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?
|
m=int(input())
m=24-m
print(m)
|
s313069745
|
Accepted
| 17
| 2,940
| 30
|
m=int(input())
m=48-m
print(m)
|
s381616907
|
p02972
|
u099450021
| 2,000
| 1,048,576
|
Wrong Answer
| 56
| 7,148
| 211
|
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
N = int(input())
A = [int(s) for s in input().split(' ')]
for i in range(N):
ai = A[i]
if ai == 0:
print(0)
exit(0)
else:
print(1)
print(i)
exit(0)
print(-1)
|
s247654624
|
Accepted
| 791
| 19,816
| 390
|
N = int(input())
A = [0] + [int(s) for s in input().split(' ')]
B = [0] * (N + 1)
count = 0
ret = []
for i in range(N, 0, -1):
ai = A[i]
ii = i
total = 0
while ii <= N:
total += B[ii]
ii += i
if total % 2 == ai:
continue
else:
B[i] = 1
count += 1
ret.append(i)
print(count)
print(' '.join(str(i) for i in ret))
|
s130368850
|
p03494
|
u510630535
| 2,000
| 262,144
|
Wrong Answer
| 29
| 8,832
| 182
|
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 resolve():
import re
N = int(input())
An = input().split()
ret = min(
len((re.findall("0+$", bin(int(a))) + [""])[0])
for a in An)
print(ret)
|
s388354898
|
Accepted
| 36
| 9,884
| 105
|
import re
input()
An=input().split()
print(min(len((re.findall("0+$",bin(int(a)))+[""])[0]) for a in An))
|
s218120971
|
p02613
|
u303711501
| 2,000
| 1,048,576
|
Wrong Answer
| 136
| 16,304
| 200
|
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 = [input() for i in range(N)]
c = [s.count('AC'),s.count('WA'),s.count('TLE'),s.count('RE')]
d = ['AC','WA','TLE','RE']
for i in range(4):
print(d[i]+' '+'×'+' '+str(c[i]) )
|
s442270267
|
Accepted
| 139
| 16,284
| 199
|
N = int(input())
s = [input() for i in range(N)]
c = [s.count('AC'),s.count('WA'),s.count('TLE'),s.count('RE')]
d = ['AC','WA','TLE','RE']
for i in range(4):
print(d[i]+' '+'x'+' '+str(c[i]) )
|
s316544934
|
p03131
|
u578406587
| 2,000
| 1,048,576
|
Time Limit Exceeded
| 2,103
| 3,064
| 409
|
Snuke has one biscuit and zero Japanese yen (the currency) in his pocket. He will perform the following operations exactly K times in total, in the order he likes: * Hit his pocket, which magically increases the number of biscuits by one. * Exchange A biscuits to 1 yen. * Exchange 1 yen to B biscuits. Find the maximum possible number of biscuits in Snuke's pocket after K operations.
|
K, A, B = (int(i) for i in input().split())
num = 1
while True:
if A + 2 - num >= B:
print(K + num)
break
elif num >= A:
if (K // 2) is not 0:
print(num + (K // 2) * (B - A) + (K % 2))
break
else:
if K >= 2 and K >= (A + 2 - num):
K -= (A + 2 - num)
num = B
else:
print(K + num)
break
|
s626332983
|
Accepted
| 17
| 3,060
| 361
|
K, A, B = (int(i) for i in input().split())
num = 1
while True:
if A + 2 - num >= B:
print(K + num)
break
elif num >= A and (K // 2) is not 0:
print(num + (K // 2) * (B - A) + (K % 2))
break
elif K >= 2 and K >= (A + 2 - num):
K -= (A + 2 - num)
num = B
else:
print(K + num)
break
|
s230919198
|
p03719
|
u810356688
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 79
|
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())
if c>=a and c<=b:print("YES")
else:print("NO")
|
s529136963
|
Accepted
| 17
| 2,940
| 79
|
a,b,c = map(int,input().split())
if c>=a and c<=b:print("Yes")
else:print("No")
|
s753413315
|
p03089
|
u121161758
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 135
|
Snuke has an empty sequence a. He will perform N operations on this sequence. In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1). You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.
|
N = int(input())
b = list(map(int, input().split()))
print(b)
for i in range(N):
if b[i] > i:
print("-1")
exit()
|
s290412320
|
Accepted
| 18
| 3,064
| 366
|
N = int(input())
b = list(map(int, input().split()))
#print(b)
ans = []
i = len(b)-1
while(1):
if i == -1:
print("-1")
exit()
if b[i] == i + 1:
#print("check")
ans.append(b.pop(i))
if len(b) == 0:
break
i = len(b)-1
else:
i -= 1
for i in range(len(ans)-1, -1, -1):
print(ans[i])
|
s328073965
|
p03493
|
u233588813
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 26
|
Snuke has a grid consisting of three squares numbered 1, 2 and 3. In each square, either `0` or `1` is written. The number written in Square i is s_i. Snuke will place a marble on each square that says `1`. Find the number of squares on which Snuke will place a marble.
|
a,b,c=input()
print(a+b+c)
|
s758672201
|
Accepted
| 17
| 2,940
| 53
|
a,b,c=input()
a=int(a)
b=int(b)
c=int(c)
print(a+b+c)
|
s142357394
|
p04044
|
u841531687
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 111
|
Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L. She will concatenate all of the strings in some order, to produce a long string. Among all strings that she can produce in this way, find the lexicographically smallest one. Here, a string s=s_1s_2s_3...s_n is _lexicographically smaller_ than another string t=t_1t_2t_3...t_m if and only if one of the following holds: * There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i. * s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.
|
n, l = map(int, input().split())
list = []
for s in range(n - 1):
list.append(input())
print(sorted(list))
|
s349593533
|
Accepted
| 18
| 3,188
| 131
|
n, l = map(int, input().split())
list = []
for s in range(n):
list.append(input())
s_list = sorted(list)
print(*s_list, sep='')
|
s371866323
|
p02842
|
u579015878
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 83
|
Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it. The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer). Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer. If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.
|
N=float(input())
x=N//1.08
if x*1.08<=N<=(x+1)*1.08:
print(x)
else:
print(':(')
|
s159715189
|
Accepted
| 17
| 2,940
| 140
|
N=int(input())
x=int(N/1.08)
a=int(x*1.08)
b=int((x+1)*1.08)
if N==a:
print(x)
exit()
if N==b:
print(x+1)
exit()
else:
print(':(')
|
s507810992
|
p03197
|
u941047297
| 2,000
| 1,048,576
|
Wrong Answer
| 149
| 13,640
| 207
|
There is an apple tree that bears apples of N colors. The N colors of these apples are numbered 1 to N, and there are a_i apples of Color i. You and Lunlun the dachshund alternately perform the following operation (starting from you): * Choose one or more apples from the tree and eat them. Here, the apples chosen at the same time must all have different colors. The one who eats the last apple from the tree will be declared winner. If both you and Lunlun play optimally, which will win?
|
def main():
n = int(input())
A = [int(input()) for _ in range(n)]
if any([a % 2 != 0 for a in A]):
print('second')
else:
print('first')
if __name__ == '__main__':
main()
|
s163552062
|
Accepted
| 146
| 13,848
| 214
|
def main():
n = int(input())
A = [int(input()) for _ in range(n)]
if any([a % 2 != 0 for a in A]):
print('first')
else:
print('second')
if __name__ == '__main__':
main()
|
s739952392
|
p03695
|
u710789518
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 604
|
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
N = int(input())
a = list(map(int, input().split()))
color = [0] * 9
for i in range(N):
if a[i] <= 399:
color[0] += 1
elif a[i] <= 799:
color[1] += 1
elif a[i] <= 1199:
color[2] += 1
elif a[i] <= 1599:
color[3] += 1
elif a[i] <= 1999:
color[4] += 1
elif a[i] <= 2399:
color[5] += 1
elif a[i] <= 2799:
color[6] += 1
elif a[i] <= 3199:
color[7] += 1
else:
color[8] += 1
print(color)
tmp = 0
for i in range(8):
if color[i] > 0:
tmp += 1
color_min = max(tmp, 1)
color_max = tmp + min(8-tmp, color[8])
#print("{0} {1}".format(color_min, color_max))
|
s222333555
|
Accepted
| 17
| 3,064
| 584
|
N = int(input())
a = list(map(int, input().split()))
color = [0] * 9
for i in range(N):
if a[i] < 400:
color[0] += 1
elif a[i] < 800:
color[1] += 1
elif a[i] < 1200:
color[2] += 1
elif a[i] < 1600:
color[3] += 1
elif a[i] < 2000:
color[4] += 1
elif a[i] < 2400:
color[5] += 1
elif a[i] < 2800:
color[6] += 1
elif a[i] < 3200:
color[7] += 1
else:
color[8] += 1
#print(color)
tmp = 0
for i in range(8):
if color[i] > 0:
tmp += 1
color_min = max(tmp, 1)
color_max = tmp + color[8]
print("{0} {1}".format(color_min, color_max))
|
s317400106
|
p03644
|
u383450070
| 2,000
| 262,144
|
Time Limit Exceeded
| 2,104
| 3,060
| 260
|
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())
count = 0
result = 0
if n == 1:
print(1)
else:
for i in range(0, n+1, 2):
counta = 0
num = i
while (num % 2 == 0):
num = num/2
counta += 1
if counta > count:
count = counta
result = i
print(result)
|
s674557006
|
Accepted
| 17
| 2,940
| 74
|
n=int(input())
ans=1
for i in range(7):
if 2**i<=n:ans=2**i
print(ans)
|
s846849935
|
p03351
|
u632557492
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 2,940
| 85
|
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 = [int(i) for i in input().split()]
print('Yes' if (c - a) <= d else 'No')
|
s592055412
|
Accepted
| 17
| 2,940
| 119
|
a, b, c, d = [int(i) for i in input().split()]
print('Yes' if ((abs(b-a)<=d and abs(c-b)<=d)) or abs(c-a)<=d else 'No')
|
s127058177
|
p02394
|
u546968095
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,596
| 345
|
Write a program which reads a rectangle and a circle, and determines whether the circle is arranged inside the rectangle. As shown in the following figures, the upper right coordinate $(W, H)$ of the rectangle and the central coordinate $(x, y)$ and radius $r$ of the circle are given.
|
def cinr(i):
tmp = (i.split(" "))
W = int(tmp[0])
H = int(tmp[1])
x = int(tmp[2])
y = int(tmp[3])
r = int(tmp[4])
print(W, H, x, y, r)
if x >= 0 + r and x <= W - r and y >= 0 + r and y <= H - r:
print("YES")
else:
print("NO")
return 0
if __name__ == "__main__":
ret = cinr(input())
|
s554721398
|
Accepted
| 20
| 5,600
| 320
|
def cinr(i):
tmp = (i.split(" "))
W = int(tmp[0])
H = int(tmp[1])
x = int(tmp[2])
y = int(tmp[3])
r = int(tmp[4])
if x >= 0 + r and x <= W - r and y >= 0 + r and y <= H - r:
print("Yes")
else:
print("No")
return 0
if __name__ == "__main__":
ret = cinr(input())
|
s308849409
|
p03861
|
u069170167
| 2,000
| 262,144
|
Wrong Answer
| 2,152
| 898,968
| 81
|
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 = list(map(int, input().split()))
print(sum([c%x for c in range(a,b+1)]))
|
s960017588
|
Accepted
| 29
| 3,068
| 62
|
a, b, x = list(map(int, input().split()))
print(b//x-(a-1)//x)
|
s760538808
|
p03609
|
u597047658
| 2,000
| 262,144
|
Wrong Answer
| 18
| 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(t)
else:
print(X)
|
s508746166
|
Accepted
| 17
| 2,940
| 86
|
X, t = map(int, input().split())
if X >= t:
print(X - t)
else:
print(0)
|
s001611939
|
p03455
|
u481197205
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,064
| 106
|
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())
ab = a * b
if ab%2 ==0:
print('even')
else:
print('odd')
|
s825985213
|
Accepted
| 17
| 2,940
| 106
|
a, b = map(int, input().split())
ab = a * b
if ab%2 ==0:
print('Even')
else:
print('Odd')
|
s494954042
|
p03575
|
u370331385
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 590
|
You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges. The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i. An edge whose removal disconnects the graph is called a _bridge_. Find the number of the edges that are bridges among the M edges.
|
N,M = map(int,input().split())
graph = [[0]*N for _ in range(N)]
Edge = []
for i in range(M):
a,b = map(int,input().split())
graph[a-1][b-1] = 1
graph[b-1][a-1] = 1
Edge.append([a-1,b-1])
def dfs(x):
if(vis[x] == 1): return
vis[x] = 1
for i in range(N):
if(graph[x][i] == 1):
dfs(0)
ans = 0
for i in range(M):
graph[Edge[i][0]][Edge[i][1]] = 0
graph[Edge[i][1]][Edge[i][0]] = 0
vis = [0]*60
dfs(0)
for j in range(N):
if(vis[j] == 0):
ans += 1
break
graph[Edge[i][0]][Edge[i][1]] = 1
graph[Edge[i][1]][Edge[i][0]] = 1
print(ans)
|
s977402496
|
Accepted
| 22
| 3,064
| 2,056
|
class UnionFind():
def __init__(self, n):
self.n = n
self.root = [-1]*(n+1)
self.rnk = [0]*(n+1)
def Find_Root(self, x):
if(self.root[x] < 0):
return x
else:
self.root[x] = self.Find_Root(self.root[x])
return self.root[x]
def Unite(self, x, y):
x = self.Find_Root(x)
y = self.Find_Root(y)
if(x == y):
return
elif(self.rnk[x] > self.rnk[y]):
self.root[x] += self.root[y]
self.root[y] = x
else:
self.root[y] += self.root[x]
self.root[x] = y
if(self.rnk[x] == self.rnk[y]):
self.rnk[y] += 1
def isSameGroup(self, x, y):
return self.Find_Root(x) == self.Find_Root(y)
def Count(self, x):
return -self.root[self.Find_Root(x)]
N,M = map(int,input().split())
Edge = []
for i in range(M):
a,b = map(int,input().split())
Edge.append([a-1,b-1])
ans = 0
for i in range(M):
graph = UnionFind(N)
for j in range(M):
if(i != j):
graph.Unite(Edge[j][0],Edge[j][1])
if(graph.Count(0) != N): ans += 1
print(ans)
|
s990379085
|
p03852
|
u865413330
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 122
|
Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: `a`, `e`, `i`, `o` and `u`.
|
import sys
input = sys.stdin.readline
vowels = "aiueo"
c = input()
print ("vowel") if c in vowels else print("consonant")
|
s874832965
|
Accepted
| 18
| 2,940
| 83
|
vowels = "aiueo"
c = input()
print("vowel") if c in vowels else print("consonant")
|
s000489748
|
p03140
|
u966601619
| 2,000
| 1,048,576
|
Wrong Answer
| 18
| 2,940
| 154
|
You are given three strings A, B and C. Each of these is a string of length N consisting of lowercase English letters. Our objective is to make all these three strings equal. For that, you can repeatedly perform the following operation: * Operation: Choose one of the strings A, B and C, and specify an integer i between 1 and N (inclusive). Change the i-th character from the beginning of the chosen string to some other lowercase English letter. What is the minimum number of operations required to achieve the objective?
|
N = int(input())
A = input()
B = input()
C = input()
count = 0
for num in range(N) :
count += len(list(set(list( [ A[num], B[num], C[num] ] ) ) )) - 1
|
s078573546
|
Accepted
| 18
| 3,060
| 167
|
N = int(input())
A = input()
B = input()
C = input()
count = 0
for num in range(N) :
count += len(list(set(list( [ A[num], B[num], C[num] ] ) ) )) - 1
print(count)
|
s676612399
|
p03130
|
u185243955
| 2,000
| 1,048,576
|
Wrong Answer
| 17
| 3,060
| 207
|
There are four towns, numbered 1,2,3 and 4. Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally. No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads. Determine if we can visit all the towns by traversing each of the roads exactly once.
|
city = [0, 0, 0, 0]
for i in range(3):
a, b = ([int(x)-1 for x in input().split()])
city[a] += 1
city[b] += 1
if city[a] > 2 or city[b] > 2:
print('No')
exit()
print('Yes')
|
s754369268
|
Accepted
| 17
| 2,940
| 206
|
city = [0, 0, 0, 0]
for i in range(3):
a, b = ([int(x)-1 for x in input().split()])
city[a] += 1
city[b] += 1
if city[a] > 2 or city[b] > 2:
print('NO')
exit()
print('YES')
|
s131260168
|
p00423
|
u243053043
| 1,000
| 131,072
|
Wrong Answer
| 120
| 7,572
| 315
|
A と B の 2 人のプレーヤーが， 0 から 9 までの数字が書かれたカードを使ってゲームを行う．最初に， 2 人は与えられた n 枚ずつのカードを，裏向きにして横一列に並べる．その後， 2 人は各自の左から 1 枚ずつカードを表向きにしていき，書かれた数字が大きい方のカードの持ち主が，その 2 枚のカードを取る．このとき，その 2 枚のカードに書かれた数字の合計が，カードを取ったプレーヤーの得点となるものとする．ただし，開いた 2 枚のカードに同じ数字が書かれているときには，引き分けとし，各プレーヤーが自分のカードを 1 枚ずつ取るものとする． 例えば， A，B の持ち札が，以下の入力例 1 から 3 のように並べられている場合を考えよう．ただし，入力ファイルは n + 1 行からなり， 1 行目には各プレーヤのカード枚数 n が書かれており， i + 1 行目（i = 1，2，... ，n）には A の左から i 枚目のカードの数字と B の左から i 枚目の カードの数字が，空白を区切り文字としてこの順で書かれている．すなわち，入力ファイルの 2 行目以降は，左側の列が A のカードの並びを，右側の列が B のカードの並びを，それぞれ表している．このとき，ゲーム終了後の A と B の得点は，それぞれ，対応する出力例に示したものとなる． 入力ファイルに対応するゲームが終了したときの A の得点と B の得点を，この順に空白を区切り文字として 1 行に出力するプログラムを作成しなさい．ただし， n ≤ 10000 とする． 入力例１ | 入力例２ | 入力例３ ---|---|--- 3| 3| 3 9 1| 9 1| 9 1 5 4| 5 4| 5 5 0 8| 1 0| 1 8 出力例１ | 出力例２ | 出力例３ 19 8| 20 0| 15 14
|
while True:
n_line = input()
if n_line == "0":
break
score_a, score_b = (0, 0)
for i in range(int(n_line)):
a, b = [int(x) for x in input().split(" ")]
if a >= b:
score_a += a
if a <= b:
score_b += b
print("{} {}".format(score_a, score_b))
|
s880956391
|
Accepted
| 110
| 7,612
| 387
|
while True:
n_line = input()
if n_line == "0":
break
score_a, score_b = (0, 0)
for i in range(int(n_line)):
a, b = [int(x) for x in input().split(" ")]
if a > b:
score_a += a + b
elif a < b:
score_b += a + b
else:
score_a += a
score_b += b
print("{} {}".format(score_a, score_b))
|
s921012528
|
p03548
|
u985702482
| 2,000
| 262,144
|
Wrong Answer
| 18
| 3,064
| 177
|
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 isu(x, y, z):
n = int(x / (y + z))
if x - n * (y + z) - z < 0:
return int(n - 1)
return int(n)
x, y, z = map(int, input().strip().split())
isu(x, y, z)
|
s329642805
|
Accepted
| 17
| 2,940
| 184
|
def isu(x, y, z):
n = int(x / (y + z))
if x - n * (y + z) - z < 0:
return int(n - 1)
return int(n)
x, y, z = map(int, input().strip().split())
print(isu(x, y, z))
|
s412644590
|
p03860
|
u951684192
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 46
|
Snuke is going to open a contest named "AtCoder s Contest". Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters. Snuke has decided to abbreviate the name of the contest as "AxC". Here, x is the uppercase English letter at the beginning of s. Given the name of the contest, print the abbreviation of the name.
|
a,b,c = input().split()
print(a[0],b[0],c[0])
|
s411120358
|
Accepted
| 17
| 2,940
| 44
|
a,b,c= input().split()
print(a[0]+b[0]+c[0])
|
s119219622
|
p02392
|
u050103511
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,512
| 93
|
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
a,b,c = map(int, input().split())
if a < b and a < c and b < c:
print("Yes")
print("No")
|
s963194665
|
Accepted
| 20
| 7,660
| 124
|
a,b,c = map(int, input().split())
if a < b and b < c:
print("Yes")
else:
print("No")
|
s695253675
|
p03457
|
u877283726
| 2,000
| 262,144
|
Wrong Answer
| 494
| 32,800
| 499
|
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
scedule = []
scedule.append([0, 0, 0])
for i in range(N):
scedule.append(list(map(int,input().split())))
print(scedule)
for i in range(N):
timelimit = scedule[i+1][0] - scedule[i][0]
x_distance = abs(scedule[i+1][1] - scedule[i][1])
y_distance = abs(scedule[i+1][2] - scedule[i][2])
if timelimit < (x_distance + y_distance):
print("No")
exit()
if timelimit % (x_distance + y_distance) != 0:
print("No")
exit()
print("Yes")
|
s841097318
|
Accepted
| 434
| 27,300
| 483
|
N = int(input())
scedule = []
scedule.append([0, 0, 0])
for i in range(N):
scedule.append(list(map(int,input().split())))
for i in range(N):
timelimit = scedule[i+1][0] - scedule[i][0]
x_distance = abs(scedule[i+1][1] - scedule[i][1])
y_distance = abs(scedule[i+1][2] - scedule[i][2])
if timelimit < (x_distance + y_distance):
print("No")
exit()
if timelimit % (x_distance + y_distance) != 0:
print("No")
exit()
print("Yes")
|
s672444894
|
p00004
|
u525269094
| 1,000
| 131,072
|
Wrong Answer
| 30
| 7,760
| 279
|
Write a program which solve a simultaneous equation: ax + by = c dx + ey = f The program should print x and y for given a, b, c, d, e and f (-1,000 ≤ a, b, c, d, e, f ≤ 1,000). You can suppose that given equation has a unique solution.
|
import sys
import math
r = sys.stdin.readlines()
r.pop(0)
n = [[int(i) for i in (j.split())] for j in r] #n is list of each lines
for l in n:
y = (l[2]*l[3]-l[0]*l[5])/(l[1]*l[3]-l[0]*l[4])
x = (l[2]-l[1]*y)/l[0]
print("{:.3f} {:.3f} ".format(x,y))
|
s398602737
|
Accepted
| 20
| 7,508
| 273
|
import sys
import math
r = sys.stdin.readlines()
n = [[float(i) for i in (j.split())] for j in r]
for l in n:
y = (l[2]*l[3]-l[0]*l[5])/(l[1]*l[3]-l[0]*l[4])+0
x = (l[2]*l[4]-l[5]*l[1])/(l[0]*l[4]-l[3]*l[1])+0
print("%.3f %.3f" %(x, y))
|
s956442980
|
p04043
|
u077080573
| 2,000
| 262,144
|
Wrong Answer
| 37
| 3,068
| 13
|
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.
|
print("hoge")
|
s999993951
|
Accepted
| 1,859
| 7,020
| 279
|
A,B,C = input().split()
array =[A,B,C]
array.sort()
if "5" in array[0]:
del array[0]
if "5" in array[0]:
del array[0]
if "7" in array[0]:
print("YES")
else:
print("NO")
else:
print("NO")
else:
print("NO")
|
s601190435
|
p02417
|
u641082901
| 1,000
| 131,072
|
Wrong Answer
| 20
| 5,552
| 126
|
Write a program which counts and reports the number of each alphabetical letter. Ignore the case of characters.
|
letters = 'abcdefghijklmnopqrstuvwxyz'
s = input()
for i in range(26) :
print(letters[i] + ' : %d' % s.count(letters[i]))
|
s966246026
|
Accepted
| 20
| 5,556
| 154
|
import sys
letters = 'abcdefghijklmnopqrstuvwxyz'
s = sys.stdin.read().lower()
for i in range(26) :
print(letters[i] + ' : %d' % s.count(letters[i]))
|
s629638404
|
p03693
|
u252964975
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 84
|
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?
|
a,b,c=map(int, input().split())
if 10*b+c%4 == 0:
print('YES')
else:
print('NO')
|
s318068331
|
Accepted
| 17
| 2,940
| 87
|
a,b,c=map(int, input().split())
if (10*b+c)%4 == 0:
print('YES')
else:
print('NO')
|
s627586738
|
p02748
|
u399481362
| 2,000
| 1,048,576
|
Wrong Answer
| 440
| 50,424
| 558
|
You are visiting a large electronics store to buy a refrigerator and a microwave. The store sells A kinds of refrigerators and B kinds of microwaves. The i-th refrigerator ( 1 \le i \le A ) is sold at a_i yen (the currency of Japan), and the j-th microwave ( 1 \le j \le B ) is sold at b_j yen. You have M discount tickets. With the i-th ticket ( 1 \le i \le M ), you can get a discount of c_i yen from the total price when buying the x_i-th refrigerator and the y_i-th microwave together. Only one ticket can be used at a time. You are planning to buy one refrigerator and one microwave. Find the minimum amount of money required.
|
s = input().split(' ')
n = 0
if int(s[0]) > 0:
n += 1
if int(s[1]) > 0:
n += 1
if int(s[2]) > 0:
n += int(s[2])
coupon = []
for times in range(1, n+1):
if times == 1:
ref = input().split(' ')
if times == 2:
micro = input().split(' ')
if times > 2:
coupon.append(input().split(' '))
price = int(ref[0]) + int(micro[0])
minPrice = price
for discount in coupon:
price = int(ref[int(discount[0])-1]) + int(micro[int(discount[1])-1]) - int(discount[2])
minPrice = min(minPrice, price)
print("$$$", minPrice)
|
s505320761
|
Accepted
| 554
| 50,804
| 643
|
s = input().split(' ')
n = 0
if int(s[0]) > 0:
n += 1
if int(s[1]) > 0:
n += 1
if int(s[2]) > 0:
n += int(s[2])
coupon = []
for times in range(1, n+1):
if times == 1:
ref = input().split(' ')
if times == 2:
micro = input().split(' ')
if times > 2:
coupon.append(input().split(' '))
price = int(ref[0]) + int(micro[0])
minPrice = price
for discount in coupon:
price = int(ref[int(discount[0])-1]) + int(micro[int(discount[1])-1]) - int(discount[2])
minPrice = min(minPrice, price)
ref.sort()
micro.sort()
price = int(ref[0]) + int(micro[0])
minPrice = min(minPrice, price)
print(minPrice)
|
s526932917
|
p02262
|
u612243550
| 6,000
| 131,072
|
Wrong Answer
| 20
| 7,688
| 548
|
Shell Sort is a generalization of [Insertion Sort](http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A) to arrange a list of $n$ elements $A$. 1 insertionSort(A, n, g) 2 for i = g to n-1 3 v = A[i] 4 j = i - g 5 while j >= 0 && A[j] > v 6 A[j+g] = A[j] 7 j = j - g 8 cnt++ 9 A[j+g] = v 10 11 shellSort(A, n) 12 cnt = 0 13 m = ? 14 G[] = {?, ?,..., ?} 15 for i = 0 to m-1 16 insertionSort(A, n, G[i]) A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every $g$-th elements. Beginning with large values of $g$, it repeats the insertion sort with smaller $g$. Your task is to complete the above program by filling ?. Write a program which reads an integer $n$ and a sequence $A$, and prints $m$, $G_i (i = 0, 1, ..., m − 1)$ in the pseudo code and the sequence $A$ in ascending order. The output of your program must meet the following requirements: * $1 \leq m \leq 100$ * $0 \leq G_i \leq n$ * cnt does not exceed $\lceil n^{1.5}\rceil$
|
def insertionSort(A, n, g):
global cnt
for i in range(g, n):
v = A[i]
j = i - g
while j >= 0 and int(A[j]) > int(v):
A[j + g] = A[j]
j = j - g
cnt += 1
A[j + g] = v
def shellSort(A, n):
global m, G
for i in range(0, m):
insertionSort(A, n, G[i])
cnt = 0
n = int(input())
A = []
for i in range(0, n):
A.append(int(input()))
m = 3
G = (4,3,1)
shellSort(A, n)
print(m)
print(" ".join(map(str, G)))
print(cnt)
for i in range(0, n):
print(A[i])
|
s779091403
|
Accepted
| 21,860
| 47,520
| 683
|
def insertionSort(A, n, g):
global cnt
for i in range(g, n):
v = A[i]
j = i - g
while j >= 0 and A[j] > v:
A[j + g] = A[j]
j = j - g
cnt += 1
A[j + g] = v
def shellSort(A, n):
global m, G
for i in range(0, m):
insertionSort(A, n, G[i])
n = int(input())
A = []
for i in range(0, n):
A.append(int(input()))
G = []
h = 1
for i in range(1, 999999):
if h > n:
break
G.append(h)
h = 3 * h + 1
G.reverse()
cnt = 0
m = len(G)
shellSort(A, n)
'''
m = 2
G = (4,1)
shellSort(A, n)
'''
print(m)
print(" ".join(map(str, G)))
print(cnt)
for i in range(0, n):
print(A[i])
|
s437388183
|
p04043
|
u600261652
| 2,000
| 262,144
|
Wrong Answer
| 17
| 2,940
| 81
|
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("YES" if A.count(5) == 2 and A.count(7) == 1 else "NO")
|
s220905398
|
Accepted
| 17
| 2,940
| 85
|
A = input().split()
print("YES" if A.count("5") == 2 and A.count("7") == 1 else "NO")
|
s043697602
|
p03672
|
u362829196
| 2,000
| 262,144
|
Wrong Answer
| 17
| 3,060
| 255
|
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.
|
#!/usr/bin/env python
# -*- coding: utf-8 -*-
import sys
S = sys.stdin.readline()
for i in range(1, len(S)):
S2 = S[0:-i]
if len(S2)%2 != 0:
continue
S3 = S2[0:int(len(S2)/2)]
S4 = S2[int(len(S2)/2):]
if S3 == S4:
print(str(len(S2)))
break
|
s566378936
|
Accepted
| 17
| 3,060
| 303
|
S = str(input())
S = S[:-1]
hantei = False
while hantei == False:
if len(S) % 2 == 1:
S = S[:-1]
else:
half = int(len(S)/2)
last = int(len(S))
if S[0:half] == S[half:last]:
print(last)
hantei = True
else:
S = S[0:-1]
|
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