problem_id stringlengths 3 7 | contestId stringclasses 660
values | problem_index stringclasses 27
values | programmingLanguage stringclasses 3
values | testset stringclasses 5
values | incorrect_passedTestCount float64 0 146 | incorrect_timeConsumedMillis float64 15 4.26k | incorrect_memoryConsumedBytes float64 0 271M | incorrect_submission_id stringlengths 7 9 | incorrect_source stringlengths 10 27.7k | correct_passedTestCount float64 2 360 | correct_timeConsumedMillis int64 30 8.06k | correct_memoryConsumedBytes int64 0 475M | correct_submission_id stringlengths 7 9 | correct_source stringlengths 28 21.2k | contest_name stringclasses 664
values | contest_type stringclasses 3
values | contest_start_year int64 2.01k 2.02k | time_limit float64 0.5 15 | memory_limit float64 64 1.02k | title stringlengths 2 54 | description stringlengths 35 3.16k | input_format stringlengths 67 1.76k | output_format stringlengths 18 1.06k ⌀ | interaction_format null | note stringclasses 840
values | examples stringlengths 34 1.16k | rating int64 800 3.4k ⌀ | tags stringclasses 533
values | testset_size int64 2 360 | official_tests stringlengths 44 19.7M | official_tests_complete bool 1
class | input_mode stringclasses 1
value | generated_checker stringclasses 231
values | executable bool 1
class |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
413/C | 413 | C | PyPy 3 | TESTS | 19 | 93 | 0 | 143149190 | n,m = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
c = []
res = 0
for i in b:
c.append(a[i-1])
c = sorted(c,reverse=True)
for i in a:
if i not in c:res+=i
for i in c:
if res>i:res+=res
else:res+=i
print(res) | 67 | 46 | 0 | 178760251 | n, m = list(map(int, input().split()))
arr = list(map(int, input().split()))
auction = set(map(int, input().split()))
non_auction_sum = 0
max_auction = 0
for i in range(n):
if (i+1) in auction:
max_auction = max(max_auction, arr[i])
else:
non_auction_sum += arr[i]
if non_auction_sum < max_auct... | Coder-Strike 2014 - Round 2 | CF | 2,014 | 1 | 256 | Jeopardy! | 'Jeopardy!' is an intellectual game where players answer questions and earn points. Company Q conducts a simplified 'Jeopardy!' tournament among the best IT companies. By a lucky coincidence, the old rivals made it to the finals: company R1 and company R2.
The finals will have n questions, m of them are auction questi... | The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100; m ≤ min(n, 30)) — the total number of questions and the number of auction questions, correspondingly. The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 107) — the prices of the questions. The third line contains m ... | In the single line, print the answer to the problem — the maximum points the R2 company can get if it plays optimally well. It is guaranteed that the answer fits into the integer 64-bit signed type. | null | null | [{"input": "4 1\n1 3 7 5\n3", "output": "18"}, {"input": "3 2\n10 3 8\n2 3", "output": "40"}, {"input": "2 2\n100 200\n1 2", "output": "400"}] | 1,400 | ["greedy", "math"] | 67 | [{"input": "4 1\r\n1 3 7 5\r\n3\r\n", "output": "18\r\n"}, {"input": "3 2\r\n10 3 8\r\n2 3\r\n", "output": "40\r\n"}, {"input": "2 2\r\n100 200\r\n1 2\r\n", "output": "400\r\n"}, {"input": "1 1\r\n1\r\n1\r\n", "output": "1\r\n"}, {"input": "2 2\r\n1 5\r\n1 2\r\n", "output": "10\r\n"}, {"input": "5 3\r\n5 8 7 1 9\r\n2 5... | false | stdio | null | true |
793/C | 793 | C | PyPy 3 | PRETESTS | 6 | 93 | 23,347,200 | 26615105 | # t*v+a = b
# t=(b-a)/v
import sys
#sys.stdin=open("data.txt")
input=sys.stdin.readline
mintime=0
maxtime=10**9
useless=0
n=int(input())
#include <math.h>
# int 'y1' redeclared as different kind of symbol
x1,y1,x2,y2=map(int,input().split())
for _ in range(n):
rx,ry,vx,vy=map(int,input().split())
# x time
... | 80 | 748 | 0 | 44048034 | def comp( x , v , l , r ) :
if x < l and v <= 0 :
return [-1,-10]
if x > r and v >= 0 :
return [-1,-10]
if x >= l and x <= r :
if v == 0 :
if x == l or x == r :
return [-1,-10]
else :
return [0,10**18]
if v > 0 :
... | Tinkoff Challenge - Elimination Round | CF | 2,017 | 2 | 256 | Mice problem | Igor the analyst fell asleep on the work and had a strange dream. In the dream his desk was crowded with computer mice, so he bought a mousetrap to catch them.
The desk can be considered as an infinite plane, then the mousetrap is a rectangle which sides are parallel to the axes, and which opposite sides are located i... | The first line contains single integer n (1 ≤ n ≤ 100 000) — the number of computer mice on the desk.
The second line contains four integers x1, y1, x2 and y2 (0 ≤ x1 ≤ x2 ≤ 100 000), (0 ≤ y1 ≤ y2 ≤ 100 000) — the coordinates of the opposite corners of the mousetrap.
The next n lines contain the information about mic... | In the only line print minimum possible non-negative number t such that if Igor closes the mousetrap at t seconds from the beginning, then all the mice are strictly inside the mousetrap. If there is no such t, print -1.
Your answer is considered correct if its absolute or relative error doesn't exceed 10 - 6.
Formall... | null | Here is a picture of the first sample
Points A, B, C, D - start mice positions, segments are their paths.
Then, at first time when all mice will be in rectangle it will be looks like this:
Here is a picture of the second sample
Points A, D, B will never enter rectangle. | [{"input": "4\n7 7 9 8\n3 5 7 5\n7 5 2 4\n3 3 7 8\n6 6 3 2", "output": "0.57142857142857139685"}, {"input": "4\n7 7 9 8\n0 3 -5 4\n5 0 5 4\n9 9 -1 -6\n10 5 -7 -10", "output": "-1"}] | 2,300 | ["geometry", "implementation", "math", "sortings"] | 80 | [{"input": "4\r\n7 7 9 8\r\n3 5 7 5\r\n7 5 2 4\r\n3 3 7 8\r\n6 6 3 2\r\n", "output": "0.57142857142857139685\r\n"}, {"input": "4\r\n7 7 9 8\r\n0 3 -5 4\r\n5 0 5 4\r\n9 9 -1 -6\r\n10 5 -7 -10\r\n", "output": "-1\r\n"}, {"input": "4\r\n8 42 60 54\r\n9 54 -58 -62\r\n46 47 52 -76\r\n15 50 -37 -40\r\n54 51 78 64\r\n", "outp... | false | stdio | import sys
def main(input_path, output_path, submission_path):
with open(output_path, 'r') as f:
ref_line = f.readline().strip()
with open(submission_path, 'r') as f:
sub_line = f.readline().strip()
if ref_line == '-1':
print(1 if sub_line == '-1' else 0)
return
... | true |
623/A | 623 | A | PyPy 3 | TESTS | 25 | 295 | 10,240,000 | 85870959 | from sys import stdin
n,m = [int(x) for x in stdin.readline().split()]
graph = [set([y for y in range(n)]) for x in range(n)]
for x in range(n):
graph[x].remove(x)
for edge in range(m):
a,b = [int(x) for x in stdin.readline().split()]
a -= 1
b -= 1
graph[a].remove(b)
graph[b].remove(a)
not... | 106 | 514 | 18,329,600 | 52719106 | from math import *
N, M = 510, 3 * 10 ** 5 + 10
g = [[] for _ in range(M)]
n , m = map(int, input().split())
mx = [[False] * N for _ in range(N + 1)]
for _ in range(m):
u, v = map(int, input().split())
mx[u][v] = mx[v][u] = True
all_eq = True
mark = [False] * N
for i in range(1, n + 1):
for j in range(i + 1, n ... | AIM Tech Round (Div. 1) | CF | 2,016 | 2 | 256 | Graph and String | One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
- G has exactly n vertices, numbered from 1 to n.... | The first line of the input contains two integers n and m $$( 1 \leq n \leq 500, 0 \leq m \leq \frac { n ( n - 1 ) } { 2 } )$$ — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It ... | In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the gra... | null | In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to al... | [{"input": "2 1\n1 2", "output": "Yes\naa"}, {"input": "4 3\n1 2\n1 3\n1 4", "output": "No"}] | 1,800 | ["constructive algorithms", "graphs"] | 106 | [{"input": "2 1\r\n1 2\r\n", "output": "Yes\r\naa\r\n"}, {"input": "4 3\r\n1 2\r\n1 3\r\n1 4\r\n", "output": "No\r\n"}, {"input": "4 4\r\n1 2\r\n1 3\r\n1 4\r\n3 4\r\n", "output": "Yes\r\nbacc\r\n"}, {"input": "1 0\r\n", "output": "Yes\r\na\r\n"}, {"input": "8 28\r\n3 2\r\n4 2\r\n7 4\r\n6 3\r\n3 7\r\n8 1\r\n3 4\r\n5 1\r... | false | stdio | import sys
def main():
input_path = sys.argv[1]
ref_output_path = sys.argv[2]
sub_output_path = sys.argv[3]
# Read input
with open(input_path) as f:
n, m = map(int, f.readline().split())
edges = []
for _ in range(m):
u, v = map(int, f.readline().split())
... | true |
623/A | 623 | A | PyPy 3-64 | TESTS | 25 | 140 | 5,529,600 | 148067886 | import sys
input = lambda: sys.stdin.buffer.readline().decode().strip()
n, m = map(int, input().split())
adj, ans = [[False] * n for _ in range(n)], [''] * n
for _ in range(m):
u, v = map(int, input().split())
adj[u - 1][v - 1] = adj[v - 1][u - 1] = True
for i in range(n):
for j in range(n):
if i ... | 106 | 623 | 1,228,800 | 15798582 | n, m = map(int, input().split())
s = [[0] * n for i in range(n)]
for i in range(m):
a, b = map(int, input().split())
s[a - 1][b - 1] = 1
s[b - 1][a - 1] = 1
for i in range(n):
s[i][i] = 1
answer = [0] * n
for i in range(n):
flag = True
for j in range(n):
if s[i][j] == 0:
fla... | AIM Tech Round (Div. 1) | CF | 2,016 | 2 | 256 | Graph and String | One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
- G has exactly n vertices, numbered from 1 to n.... | The first line of the input contains two integers n and m $$( 1 \leq n \leq 500, 0 \leq m \leq \frac { n ( n - 1 ) } { 2 } )$$ — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It ... | In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the gra... | null | In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to al... | [{"input": "2 1\n1 2", "output": "Yes\naa"}, {"input": "4 3\n1 2\n1 3\n1 4", "output": "No"}] | 1,800 | ["constructive algorithms", "graphs"] | 106 | [{"input": "2 1\r\n1 2\r\n", "output": "Yes\r\naa\r\n"}, {"input": "4 3\r\n1 2\r\n1 3\r\n1 4\r\n", "output": "No\r\n"}, {"input": "4 4\r\n1 2\r\n1 3\r\n1 4\r\n3 4\r\n", "output": "Yes\r\nbacc\r\n"}, {"input": "1 0\r\n", "output": "Yes\r\na\r\n"}, {"input": "8 28\r\n3 2\r\n4 2\r\n7 4\r\n6 3\r\n3 7\r\n8 1\r\n3 4\r\n5 1\r... | false | stdio | import sys
def main():
input_path = sys.argv[1]
ref_output_path = sys.argv[2]
sub_output_path = sys.argv[3]
# Read input
with open(input_path) as f:
n, m = map(int, f.readline().split())
edges = []
for _ in range(m):
u, v = map(int, f.readline().split())
... | true |
745/B | 745 | B | Python 3 | TESTS | 36 | 139 | 4,608,000 | 23058646 | n,m = map(int,list(input().split()))
ar = [0]*m
for i in range(n):
s = input()
for j in range(m):
if s[j]=='X':
ar[j]+=1
ss = set(ar)
ss.discard(0)
if len(ss)>1:
print('NO')
else:
print('YES')
# C:\Users\Usuario\HOME2\Programacion\ACM | 77 | 62 | 6,758,400 | 23066037 | def codeforces(piece):
row_number = 0
for row in piece:
num = row.count('X')
if num == 0:
continue
if row_number == 0:
row_number = num
continue
if num != row_number:
return 'NO'
col_number = 0
cols = [[row[i] for row in pi... | Codeforces Round 385 (Div. 2) | CF | 2,016 | 2 | 256 | Hongcow Solves A Puzzle | Hongcow likes solving puzzles.
One day, Hongcow finds two identical puzzle pieces, with the instructions "make a rectangle" next to them. The pieces can be described by an n by m grid of characters, where the character 'X' denotes a part of the puzzle and '.' denotes an empty part of the grid. It is guaranteed that th... | The first line of input will contain two integers n and m (1 ≤ n, m ≤ 500), the dimensions of the puzzle piece.
The next n lines will describe the jigsaw piece. Each line will have length m and will consist of characters '.' and 'X' only. 'X' corresponds to a part of the puzzle piece, '.' is an empty space.
It is gua... | Output "YES" if it is possible for Hongcow to make a rectangle. Output "NO" otherwise. | null | For the first sample, one example of a rectangle we can form is as follows
For the second sample, it is impossible to put two of those pieces without rotating or flipping to form a rectangle.
In the third sample, we can shift the first tile by one to the right, and then compose the following rectangle: | [{"input": "2 3\nXXX\nXXX", "output": "YES"}, {"input": "2 2\n.X\nXX", "output": "NO"}, {"input": "5 5\n.....\n..X..\n.....\n.....\n.....", "output": "YES"}] | 1,400 | ["implementation"] | 77 | [{"input": "2 3\r\nXXX\r\nXXX\r\n", "output": "YES\r\n"}, {"input": "2 2\r\n.X\r\nXX\r\n", "output": "NO\r\n"}, {"input": "5 5\r\n.....\r\n..X..\r\n.....\r\n.....\r\n.....\r\n", "output": "YES\r\n"}, {"input": "1 500\r\n.XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX... | false | stdio | null | true |
515/B | 515 | B | PyPy 3 | TESTS | 40 | 155 | 1,433,600 | 75217147 | I=lambda:list(map(int,input().split()))
n,m=I()
a=I()[1::]
b=I()[1::]
ans=[0]*max(n,m)
bo=[0]*n
gl=[0]*m
for i in a:
bo[i-1]=1
for j in b:
gl[j-1]=1
boy=set()
girl=set()
for i in range(n*m+1):
x=bo[i%n]
y=gl[i%m]
if x|y:
bo[i%n]=1
gl[i%m]=1
if bo.count(1)+gl.count(1)==n+m:
prin... | 56 | 46 | 0 | 218400995 | import sys
sys.setrecursionlimit(100000000)
input=lambda:sys.stdin.readline().strip()
write=lambda x:sys.stdout.write(str(x)+'\n')
# from random import randint
# from copy import deepcopy
# from collections import deque,Counter
# from heapq import heapify,heappush,heappop
# from bisect import bisect_left,bisect,insort... | Codeforces Round 292 (Div. 2) | CF | 2,015 | 2 | 256 | Drazil and His Happy Friends | Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.
There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites $$(i \bmod n)$$-th... | The first line contains two integer n and m (1 ≤ n, m ≤ 100).
The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys.
The third line conatins integer g (0 ≤ ... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | null | By $$i \bmod k$$ we define the remainder of integer division of i by k.
In first sample case:
- On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day.
- On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so no... | [{"input": "2 3\n0\n1 0", "output": "Yes"}, {"input": "2 4\n1 0\n1 2", "output": "No"}, {"input": "2 3\n1 0\n1 1", "output": "Yes"}] | 1,300 | ["brute force", "dsu", "meet-in-the-middle", "number theory"] | 56 | [{"input": "2 3\r\n0\r\n1 0\r\n", "output": "Yes\r\n"}, {"input": "2 4\r\n1 0\r\n1 2\r\n", "output": "No\r\n"}, {"input": "2 3\r\n1 0\r\n1 1\r\n", "output": "Yes\r\n"}, {"input": "16 88\r\n6 5 14 2 0 12 7\r\n30 21 64 35 79 74 39 63 44 81 73 0 27 33 69 12 86 46 20 25 55 52 7 58 23 5 60 32 41 50 82\r\n", "output": "Yes\r... | false | stdio | null | true |
982/A | 982 | A | Python 3 | TESTS | 9 | 124 | 0 | 50425113 | n=int(input())
a=str(input())
b=list(a)
for i in range(n):
b[i]=int(b[i])
k=0
if n==1 and b[0]==0:
k=k+1
if n==2 and b[0]==b[1]==0:
k=k+1
if n>3:
if b[0]==b[1]==0:
k=k+1
if b[n-1]==b[n-2]==0:
k=k+1
if a.find('11')>=0:
k=k+1
if a.find('000')>=0:
k=k+1
if k==0:
print('Yes')
... | 55 | 46 | 0 | 173770461 | import sys
def read_input(input_path=None):
if input_path is None:
f = sys.stdin
else:
f = open(input_path, 'r')
n = int(f.readline())
chairs = f.readline().strip()
return n, chairs
def sol(n, chairs):
# chairs = 'o' + chairs + 'o'
#
# return ["YES" if "xx" not in c... | Codeforces Round 484 (Div. 2) | CF | 2,018 | 1 | 256 | Row | You're given a row with $$$n$$$ chairs. We call a seating of people "maximal" if the two following conditions hold:
1. There are no neighbors adjacent to anyone seated.
2. It's impossible to seat one more person without violating the first rule.
The seating is given as a string consisting of zeros and ones ($$$0$$$ m... | The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 1000$$$) — the number of chairs.
The next line contains a string of $$$n$$$ characters, each of them is either zero or one, describing the seating. | Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".
You are allowed to print letters in whatever case you'd like (uppercase or lowercase). | null | In sample case one the given seating is maximal.
In sample case two the person at chair three has a neighbour to the right.
In sample case three it is possible to seat yet another person into chair three. | [{"input": "3\n101", "output": "Yes"}, {"input": "4\n1011", "output": "No"}, {"input": "5\n10001", "output": "No"}] | 1,200 | ["brute force", "constructive algorithms"] | 55 | [{"input": "3\r\n101\r\n", "output": "Yes\r\n"}, {"input": "4\r\n1011\r\n", "output": "No\r\n"}, {"input": "5\r\n10001\r\n", "output": "No\r\n"}, {"input": "1\r\n0\r\n", "output": "No\r\n"}, {"input": "1\r\n1\r\n", "output": "Yes\r\n"}, {"input": "100\r\n01010010101010010100100101010010101001010010010010100101010100101... | false | stdio | null | true |
982/A | 982 | A | Python 3 | TESTS | 49 | 124 | 0 | 44689901 | n = int(input())
s = input()
first = -1
second = -1
minimum = 9999999
maximum = 0
def check_val(value):
if(value==2 or value==3):
return True
else:
return False
possible = ['1','01','10','010']
if(s in possible):
print('YES')
else:
for i in range(n):
if(int(s[i])==1... | 55 | 46 | 0 | 183823666 | n=input()
s='0'+input()+'0'
print('No' if '000' in s or '11' in s else 'Yes') | Codeforces Round 484 (Div. 2) | CF | 2,018 | 1 | 256 | Row | You're given a row with $$$n$$$ chairs. We call a seating of people "maximal" if the two following conditions hold:
1. There are no neighbors adjacent to anyone seated.
2. It's impossible to seat one more person without violating the first rule.
The seating is given as a string consisting of zeros and ones ($$$0$$$ m... | The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 1000$$$) — the number of chairs.
The next line contains a string of $$$n$$$ characters, each of them is either zero or one, describing the seating. | Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".
You are allowed to print letters in whatever case you'd like (uppercase or lowercase). | null | In sample case one the given seating is maximal.
In sample case two the person at chair three has a neighbour to the right.
In sample case three it is possible to seat yet another person into chair three. | [{"input": "3\n101", "output": "Yes"}, {"input": "4\n1011", "output": "No"}, {"input": "5\n10001", "output": "No"}] | 1,200 | ["brute force", "constructive algorithms"] | 55 | [{"input": "3\r\n101\r\n", "output": "Yes\r\n"}, {"input": "4\r\n1011\r\n", "output": "No\r\n"}, {"input": "5\r\n10001\r\n", "output": "No\r\n"}, {"input": "1\r\n0\r\n", "output": "No\r\n"}, {"input": "1\r\n1\r\n", "output": "Yes\r\n"}, {"input": "100\r\n01010010101010010100100101010010101001010010010010100101010100101... | false | stdio | null | true |
446/A | 446 | A | PyPy 3 | TESTS | 20 | 218 | 10,342,400 | 47505771 | R = lambda: map(int, input().split())
n = int(input())
a = list(R())
dp = [1] * n
res = 1
for i in range(n - 2, -1 , -1):
dp[i] = dp[i + 1] + 1 if a[i] < a[i + 1] else 1
cnt = 1
for i in range(1, n - 1):
if a[i + 1] > a[i - 1] + 1:
res = max(res, cnt + dp[i + 1] + 1)
else:
res = max(res, cnt... | 92 | 186 | 19,660,800 | 143453579 | n=int(input())
a=list(map(int,input().split()))
l=[1]
r=[1]
for i in range(1,n):
if a[i]>a[i-1]:
l.append(l[-1]+1)
else:
l.append(1)
for i in range(n-2,-1,-1):
if a[i]<a[i+1]:
r.append(r[-1]+1)
else :
r.append(1)
r=r[::-1]
mx=max(max(l),max(r))
if mx!=n:
mx+=1
for i i... | Codeforces Round #FF (Div. 1) | CF | 2,014 | 1 | 256 | DZY Loves Sequences | DZY has a sequence a, consisting of n integers.
We'll call a sequence ai, ai + 1, ..., aj (1 ≤ i ≤ j ≤ n) a subsegment of the sequence a. The value (j - i + 1) denotes the length of the subsegment.
Your task is to find the longest subsegment of a, such that it is possible to change at most one number (change one numb... | The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109). | In a single line print the answer to the problem — the maximum length of the required subsegment. | null | You can choose subsegment a2, a3, a4, a5, a6 and change its 3rd element (that is a4) to 4. | [{"input": "6\n7 2 3 1 5 6", "output": "5"}] | 1,600 | ["dp", "implementation", "two pointers"] | 92 | [{"input": "6\r\n7 2 3 1 5 6\r\n", "output": "5\r\n"}, {"input": "10\r\n424238336 649760493 681692778 714636916 719885387 804289384 846930887 957747794 596516650 189641422\r\n", "output": "9\r\n"}, {"input": "50\r\n804289384 846930887 681692778 714636916 957747794 424238336 719885387 649760493 596516650 189641422 25202... | false | stdio | null | true |
550/A | 550 | A | Python 3 | TESTS | 50 | 93 | 8,396,800 | 214120912 | string = input()
ans = []
count = 0
i = 1
sett =set()
for i in range(1,len(string)):
if (string[i-1] == "A" and string[i] == "B") or (string[i-1] == "B" and string[i] == "A"):
ans.append([string[i-1]+string[i],i-1,i])
# print(ans)
x =True
result = "NO"
a,b= -1,-1
for i in ans:
if i[0]=="AB":
... | 85 | 46 | 0 | 187124557 | a=input()
n=a.split("AB",1)
r=a.split("BA",1)
if len(n)>1 and "BA" in n[1]:
print ("YES")
elif len(r)>1 and "AB" in r[1]:
print ("YES")
else:
print ("NO") | Codeforces Round 306 (Div. 2) | CF | 2,015 | 2 | 256 | Two Substrings | You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order). | The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters. | Print "YES" (without the quotes), if string s contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise. | null | In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | [{"input": "ABA", "output": "NO"}, {"input": "BACFAB", "output": "YES"}, {"input": "AXBYBXA", "output": "NO"}] | 1,500 | ["brute force", "dp", "greedy", "implementation", "strings"] | 85 | [{"input": "ABA\r\n", "output": "NO\r\n"}, {"input": "BACFAB\r\n", "output": "YES\r\n"}, {"input": "AXBYBXA\r\n", "output": "NO\r\n"}, {"input": "ABABAB\r\n", "output": "YES\r\n"}, {"input": "BBBBBBBBBB\r\n", "output": "NO\r\n"}, {"input": "ABBA\r\n", "output": "YES\r\n"}, {"input": "ABAXXXAB\r\n", "output": "YES\r\n"}... | false | stdio | null | true |
550/A | 550 | A | Python 3 | TESTS | 36 | 1,981 | 8,294,400 | 202949868 | s = str(input())
ans = 'NO'
if 'AB' not in s and 'BA' not in s:
ans = 'NO'
else:
cur = 0
while cur < len(s):
try:
cur = s.index('AB', cur)
s2 = s[0:cur]+s[cur+2::]
if 'BA' in s2:
ans = 'YES'
break
else:
c... | 85 | 46 | 0 | 189781943 | s=input()
u=s.find('AB')
v=s.find('BA')
print('YES'if(u+1 and s.find('BA',u+2)+1)or(v+1 and s.find('AB',v+2)+1)else'NO') | Codeforces Round 306 (Div. 2) | CF | 2,015 | 2 | 256 | Two Substrings | You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order). | The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters. | Print "YES" (without the quotes), if string s contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise. | null | In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | [{"input": "ABA", "output": "NO"}, {"input": "BACFAB", "output": "YES"}, {"input": "AXBYBXA", "output": "NO"}] | 1,500 | ["brute force", "dp", "greedy", "implementation", "strings"] | 85 | [{"input": "ABA\r\n", "output": "NO\r\n"}, {"input": "BACFAB\r\n", "output": "YES\r\n"}, {"input": "AXBYBXA\r\n", "output": "NO\r\n"}, {"input": "ABABAB\r\n", "output": "YES\r\n"}, {"input": "BBBBBBBBBB\r\n", "output": "NO\r\n"}, {"input": "ABBA\r\n", "output": "YES\r\n"}, {"input": "ABAXXXAB\r\n", "output": "YES\r\n"}... | false | stdio | null | true |
550/A | 550 | A | PyPy 3-64 | TESTS | 63 | 62 | 1,945,600 | 207965257 | #import io, os, sys
#input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
#print = lambda x: sys.stdout.write(str(x) + "\n")
II = lambda: int(input())
MII = lambda: map(int, input().split())
LMII = lambda: list(MII())
#SLMII = lambda: sorted(LMII())
s = input()
n = len(s)
sABBA = ['AB', 'BA']
i = j = 0
sB... | 85 | 46 | 0 | 190125776 | str=input()
x1=str.find("AB")
x2=str.find("BA",x1+2)
y1=str.find("BA")
y2=str.find("AB",y1+2)
if (x1!=-1 and x2!=-1) or (y1!=-1 and y2!=-1):
print("YES")
else:
print("NO") | Codeforces Round 306 (Div. 2) | CF | 2,015 | 2 | 256 | Two Substrings | You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order). | The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters. | Print "YES" (without the quotes), if string s contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise. | null | In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | [{"input": "ABA", "output": "NO"}, {"input": "BACFAB", "output": "YES"}, {"input": "AXBYBXA", "output": "NO"}] | 1,500 | ["brute force", "dp", "greedy", "implementation", "strings"] | 85 | [{"input": "ABA\r\n", "output": "NO\r\n"}, {"input": "BACFAB\r\n", "output": "YES\r\n"}, {"input": "AXBYBXA\r\n", "output": "NO\r\n"}, {"input": "ABABAB\r\n", "output": "YES\r\n"}, {"input": "BBBBBBBBBB\r\n", "output": "NO\r\n"}, {"input": "ABBA\r\n", "output": "YES\r\n"}, {"input": "ABAXXXAB\r\n", "output": "YES\r\n"}... | false | stdio | null | true |
446/A | 446 | A | PyPy 3 | TESTS | 20 | 202 | 12,390,400 | 110830424 | n=int(input())
arr=list(map(int,input().split()))
pre,pre1=[1],[1]
for i in range(1,n):
if arr[i]>arr[i-1]:
pre.append(pre[-1]+1)
else:
pre.append(1)
arr.reverse()
for i in range(1,n):
if arr[i]<arr[i-1]:
pre1.append(pre1[-1]+1)
else:
pre1.append(1)
pre1.reverse()
arr.rev... | 92 | 187 | 15,257,600 | 115861712 | n=int(input())
a=[0]+list(map(int,input().split()))+[0]
l=[]
st=1
for i in range(2,n+2):
if a[i]<=a[i-1]:
l.append([st,i-1])
st=i
if len(l)==1:
print(n)
exit()
mx=0
for i in range(1,len(l)):
lf=l[i-1][1]
rt=l[i][0]
ln1=l[i][1]-l[i][0]+1
ln2=l[i-1][1]-l[i-1][0]+1
if a[rt... | Codeforces Round #FF (Div. 1) | CF | 2,014 | 1 | 256 | DZY Loves Sequences | DZY has a sequence a, consisting of n integers.
We'll call a sequence ai, ai + 1, ..., aj (1 ≤ i ≤ j ≤ n) a subsegment of the sequence a. The value (j - i + 1) denotes the length of the subsegment.
Your task is to find the longest subsegment of a, such that it is possible to change at most one number (change one numb... | The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109). | In a single line print the answer to the problem — the maximum length of the required subsegment. | null | You can choose subsegment a2, a3, a4, a5, a6 and change its 3rd element (that is a4) to 4. | [{"input": "6\n7 2 3 1 5 6", "output": "5"}] | 1,600 | ["dp", "implementation", "two pointers"] | 92 | [{"input": "6\r\n7 2 3 1 5 6\r\n", "output": "5\r\n"}, {"input": "10\r\n424238336 649760493 681692778 714636916 719885387 804289384 846930887 957747794 596516650 189641422\r\n", "output": "9\r\n"}, {"input": "50\r\n804289384 846930887 681692778 714636916 957747794 424238336 719885387 649760493 596516650 189641422 25202... | false | stdio | null | true |
837/D | 837 | D | PyPy 3 | TESTS | 23 | 124 | 20,070,400 | 132496966 | def main():
n,k = map(int, input().split())
A = list(map(int,input().split()))
A2 = [0]*n
A5 = [0]*n
for i, a in enumerate(A):
while a%2 == 0:
A2[i] += 1
a //= 2
while a%5 == 0:
A5[i] += 1
a //= 5
#print(A2)
#print(A5)
INF... | 99 | 623 | 11,161,600 | 172361559 | import sys
input = sys.stdin.readline
n, k = map(int, input().split())
a = list(map(int, input().split()))
sc5, c2, c5 = 0, [], []
for i in range(n):
num, c = a[i], 0
while num % 5 == 0:
num //= 5
c += 1
sc5 += c
c5.append(c)
c = 0
while num % 2 == 0:
num //= 2
c ... | Educational Codeforces Round 26 | ICPC | 2,017 | 2 | 256 | Round Subset | Let's call the roundness of the number the number of zeros to which it ends.
You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum possible. | The first line contains two integer numbers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 1018). | Print maximal roundness of product of the chosen subset of length k. | null | In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.
In the second example subset [15, 16, 25] has product 6000, roundness 3.
In the third example all subsets has product with roundness 0. | [{"input": "3 2\n50 4 20", "output": "3"}, {"input": "5 3\n15 16 3 25 9", "output": "3"}, {"input": "3 3\n9 77 13", "output": "0"}] | 2,100 | ["dp", "math"] | 99 | [{"input": "3 2\r\n50 4 20\r\n", "output": "3\r\n"}, {"input": "5 3\r\n15 16 3 25 9\r\n", "output": "3\r\n"}, {"input": "3 3\r\n9 77 13\r\n", "output": "0\r\n"}, {"input": "1 1\r\n200000000\r\n", "output": "8\r\n"}, {"input": "1 1\r\n3\r\n", "output": "0\r\n"}, {"input": "3 1\r\n1000000000000000000 800000000000000000 6... | false | stdio | null | true |
847/E | 847 | E | Python 3 | TESTS | 16 | 124 | 0 | 44568676 | from math import ceil
def solve(field,size,answer):
left = 0
star_pos = -1
rstar_pos = -1
p_pos = -1
distance = 0
cost = 0
for i in range(size):
#print(p_pos,star_pos)
if field[i] == "*":
if star_pos != -1:
if p_pos != -1:
if st... | 77 | 124 | 7,372,800 | 121279454 | '''
pacman->food
pacman->food->food :hmmge:
larges smaller val
smallest larger val
'''
def check(mid,p,f):
u=0
for i in range(len(p)):
left=p[i]-f[u]
left=max(0,left)
if mid<left:
return 0
right=max(mid-2*left,0)
right=max(right,(mid-left)//2)
while u<len(f) and (f[u]<=p[i]+right): #bich mai fas ra
... | 2017-2018 ACM-ICPC, NEERC, Southern Subregional Contest, qualification stage (Online Mirror, ACM-ICPC Rules, Teams Preferred) | ICPC | 2,017 | 1 | 256 | Packmen | A game field is a strip of 1 × n square cells. In some cells there are Packmen, in some cells — asterisks, other cells are empty.
Packman can move to neighboring cell in 1 time unit. If there is an asterisk in the target cell then Packman eats it. Packman doesn't spend any time to eat an asterisk.
In the initial mome... | The first line contains a single integer n (2 ≤ n ≤ 105) — the length of the game field.
The second line contains the description of the game field consisting of n symbols. If there is symbol '.' in position i — the cell i is empty. If there is symbol '*' in position i — in the cell i contains an asterisk. If there is... | Print minimum possible time after which Packmen can eat all asterisks. | null | In the first example Packman in position 4 will move to the left and will eat asterisk in position 1. He will spend 3 time units on it. During the same 3 time units Packman in position 6 will eat both of neighboring with it asterisks. For example, it can move to the left and eat asterisk in position 5 (in 1 time unit) ... | [{"input": "7\n*..P*P*", "output": "3"}, {"input": "10\n.**PP.*P.*", "output": "2"}] | 1,800 | ["binary search", "dp"] | 77 | [{"input": "7\r\n*..P*P*\r\n", "output": "3\r\n"}, {"input": "10\r\n.**PP.*P.*\r\n", "output": "2\r\n"}, {"input": "19\r\n**P.*..*..P..*.*P**\r\n", "output": "7\r\n"}, {"input": "12\r\nP**.*P*P*P**\r\n", "output": "3\r\n"}, {"input": "17\r\n.*P*P**P**.**P...\r\n", "output": "3\r\n"}, {"input": "58\r\n..P.P*.P*.P...PPP.... | false | stdio | null | true |
837/D | 837 | D | PyPy 3 | TESTS | 13 | 233 | 5,632,000 | 57330897 | n, k = map(int, input().split())
a = list(map(int, input().split()))
b = [[0 for i in range(2)] for j in range(n)]
for i in range(n):
while a[i] % 2 == 0:
b[i][0] += 1
a[i] >>= 1
while a[i] % 5 == 0:
b[i][1] += 1
a[i] //= 5
cur2, cur5 = 0, 0
for i in range(k):
cur2 += b[i][0]... | 99 | 857 | 11,878,400 | 172414804 | import sys
import math
import collections
import random
from heapq import heappush, heappop
from functools import reduce
input = sys.stdin.readline
ints = lambda: list(map(int, input().split()))
def printQry(a, b) -> None:
sa = str(a)
sb = str(b)
print(f"? {sa} {sb}", flush = True)
def printAns(ans) -> N... | Educational Codeforces Round 26 | ICPC | 2,017 | 2 | 256 | Round Subset | Let's call the roundness of the number the number of zeros to which it ends.
You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum possible. | The first line contains two integer numbers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 1018). | Print maximal roundness of product of the chosen subset of length k. | null | In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.
In the second example subset [15, 16, 25] has product 6000, roundness 3.
In the third example all subsets has product with roundness 0. | [{"input": "3 2\n50 4 20", "output": "3"}, {"input": "5 3\n15 16 3 25 9", "output": "3"}, {"input": "3 3\n9 77 13", "output": "0"}] | 2,100 | ["dp", "math"] | 99 | [{"input": "3 2\r\n50 4 20\r\n", "output": "3\r\n"}, {"input": "5 3\r\n15 16 3 25 9\r\n", "output": "3\r\n"}, {"input": "3 3\r\n9 77 13\r\n", "output": "0\r\n"}, {"input": "1 1\r\n200000000\r\n", "output": "8\r\n"}, {"input": "1 1\r\n3\r\n", "output": "0\r\n"}, {"input": "3 1\r\n1000000000000000000 800000000000000000 6... | false | stdio | null | true |
681/A | 681 | A | Python 3 | TESTS | 8 | 62 | 0 | 116228551 | n=int(input())
k=[]
z=0
for i in range(n):
k+=input().split()
for i in range(1,len(k)):
if (k[i])>="2400" and i%3!=0:
z+=1
if z>=2 and k[i-1]<k[i]:
print("YES")
break
else:
z=0
else:
print("NO") | 60 | 31 | 0 | 149208261 | con=True
for _ in [0]*int(input()):
a=[*map(int,input().split()[1:])]
if a[0]>=2400 and a[1]>a[0] and con:print('YES');con=False
if con:print('NO') | Codeforces Round 357 (Div. 2) | CF | 2,016 | 1 | 256 | A Good Contest | Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance.
Anton wants the color of hi... | The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the number of participants Anton has outscored in this contest .
The next n lines describe participants results: the i-th of them consists of a participant handle namei and two integers beforei and afteri ( - 4000 ≤ beforei, afteri ≤ 4000) — parti... | Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise. | null | In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest.
In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before... | [{"input": "3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749", "output": "YES"}, {"input": "3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450", "output": "NO"}] | 800 | ["implementation"] | 60 | [{"input": "3\r\nBurunduk1 2526 2537\r\nBudAlNik 2084 2214\r\nsubscriber 2833 2749\r\n", "output": "YES"}, {"input": "3\r\nApplejack 2400 2400\r\nFluttershy 2390 2431\r\nPinkie_Pie -2500 -2450\r\n", "output": "NO"}, {"input": "1\r\nDb -3373 3591\r\n", "output": "NO"}, {"input": "5\r\nQ2bz 960 2342\r\nhmX 2710 -1348\r\n... | false | stdio | null | true |
681/A | 681 | A | Python 3 | TESTS | 8 | 108 | 0 | 90044033 | n = int(input())
a = [input().split() for i in range(n)]
k = 0
for j in range(n):
if int(a[j][1]) >= 2400 and a[j][2] > a[j][1]:
k += 1
break
if k > 0:
print('YES')
else:
print('NO') | 60 | 31 | 0 | 221573962 | ctr=0
for i in range(int(input())):
cid,bef,aft=map(str,input().split())
bef=int(bef)
aft=int(aft)
if bef>=2400:
if aft-bef>0:
ctr+=1
if ctr>0:
print("YES")
else:
print("NO") | Codeforces Round 357 (Div. 2) | CF | 2,016 | 1 | 256 | A Good Contest | Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance.
Anton wants the color of hi... | The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the number of participants Anton has outscored in this contest .
The next n lines describe participants results: the i-th of them consists of a participant handle namei and two integers beforei and afteri ( - 4000 ≤ beforei, afteri ≤ 4000) — parti... | Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise. | null | In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest.
In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before... | [{"input": "3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749", "output": "YES"}, {"input": "3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450", "output": "NO"}] | 800 | ["implementation"] | 60 | [{"input": "3\r\nBurunduk1 2526 2537\r\nBudAlNik 2084 2214\r\nsubscriber 2833 2749\r\n", "output": "YES"}, {"input": "3\r\nApplejack 2400 2400\r\nFluttershy 2390 2431\r\nPinkie_Pie -2500 -2450\r\n", "output": "NO"}, {"input": "1\r\nDb -3373 3591\r\n", "output": "NO"}, {"input": "5\r\nQ2bz 960 2342\r\nhmX 2710 -1348\r\n... | false | stdio | null | true |
681/A | 681 | A | Python 3 | TESTS | 8 | 93 | 0 | 58093743 | n=int(input())
f=0
for i in range(n):
l=list(map(str,input().split()))
if l[1]>='2400' and l[2]>l[1]:
f=1
break
if f==1:
print("YES")
else:
print("NO") | 60 | 46 | 0 | 136167929 | n = int(input())
result = 0
for i in range(n):
a, b, c = input().split()
if int(c) > int(b) >= 2400:
result = 1
print("YES" if result == 1 else "NO") | Codeforces Round 357 (Div. 2) | CF | 2,016 | 1 | 256 | A Good Contest | Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance.
Anton wants the color of hi... | The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the number of participants Anton has outscored in this contest .
The next n lines describe participants results: the i-th of them consists of a participant handle namei and two integers beforei and afteri ( - 4000 ≤ beforei, afteri ≤ 4000) — parti... | Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise. | null | In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest.
In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before... | [{"input": "3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749", "output": "YES"}, {"input": "3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450", "output": "NO"}] | 800 | ["implementation"] | 60 | [{"input": "3\r\nBurunduk1 2526 2537\r\nBudAlNik 2084 2214\r\nsubscriber 2833 2749\r\n", "output": "YES"}, {"input": "3\r\nApplejack 2400 2400\r\nFluttershy 2390 2431\r\nPinkie_Pie -2500 -2450\r\n", "output": "NO"}, {"input": "1\r\nDb -3373 3591\r\n", "output": "NO"}, {"input": "5\r\nQ2bz 960 2342\r\nhmX 2710 -1348\r\n... | false | stdio | null | true |
681/A | 681 | A | PyPy 3 | TESTS | 8 | 140 | 0 | 71018318 | k=0
for i in range(int(input())):
a,x,y=input().split()
if int(x)>=2400 and y>x:
k=1
if k==1:
print("YES")
else:
print("NO") | 60 | 46 | 0 | 139355211 | t=int(input())
check=1
for i in range(t):
s=input()
a=s.split(' ')
if int(a[1])>=2400 and int(a[1])<int(a[2]):
check=0
if check==1:
print('NO')
else:
print('YES') | Codeforces Round 357 (Div. 2) | CF | 2,016 | 1 | 256 | A Good Contest | Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance.
Anton wants the color of hi... | The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the number of participants Anton has outscored in this contest .
The next n lines describe participants results: the i-th of them consists of a participant handle namei and two integers beforei and afteri ( - 4000 ≤ beforei, afteri ≤ 4000) — parti... | Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise. | null | In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest.
In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before... | [{"input": "3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749", "output": "YES"}, {"input": "3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450", "output": "NO"}] | 800 | ["implementation"] | 60 | [{"input": "3\r\nBurunduk1 2526 2537\r\nBudAlNik 2084 2214\r\nsubscriber 2833 2749\r\n", "output": "YES"}, {"input": "3\r\nApplejack 2400 2400\r\nFluttershy 2390 2431\r\nPinkie_Pie -2500 -2450\r\n", "output": "NO"}, {"input": "1\r\nDb -3373 3591\r\n", "output": "NO"}, {"input": "5\r\nQ2bz 960 2342\r\nhmX 2710 -1348\r\n... | false | stdio | null | true |
681/A | 681 | A | Python 3 | TESTS | 8 | 108 | 0 | 77606288 | for _ in range(int(input())):
name, before, after = input().split()
if int(before) >=2400:
if after > before:
print("YES")
exit(0)
print('NO') | 60 | 46 | 0 | 142667952 | """
123456789101112131415
"""
import sys
import math
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(str,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,in... | Codeforces Round 357 (Div. 2) | CF | 2,016 | 1 | 256 | A Good Contest | Codeforces user' handle color depends on his rating — it is red if his rating is greater or equal to 2400; it is orange if his rating is less than 2400 but greater or equal to 2200, etc. Each time participant takes part in a rated contest, his rating is changed depending on his performance.
Anton wants the color of hi... | The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the number of participants Anton has outscored in this contest .
The next n lines describe participants results: the i-th of them consists of a participant handle namei and two integers beforei and afteri ( - 4000 ≤ beforei, afteri ≤ 4000) — parti... | Print «YES» (quotes for clarity), if Anton has performed good in the contest and «NO» (quotes for clarity) otherwise. | null | In the first sample, Anton has outscored user with handle Burunduk1, whose handle was colored red before the contest and his rating has increased after the contest.
In the second sample, Applejack's rating has not increased after the contest, while both Fluttershy's and Pinkie_Pie's handles were not colored red before... | [{"input": "3\nBurunduk1 2526 2537\nBudAlNik 2084 2214\nsubscriber 2833 2749", "output": "YES"}, {"input": "3\nApplejack 2400 2400\nFluttershy 2390 2431\nPinkie_Pie -2500 -2450", "output": "NO"}] | 800 | ["implementation"] | 60 | [{"input": "3\r\nBurunduk1 2526 2537\r\nBudAlNik 2084 2214\r\nsubscriber 2833 2749\r\n", "output": "YES"}, {"input": "3\r\nApplejack 2400 2400\r\nFluttershy 2390 2431\r\nPinkie_Pie -2500 -2450\r\n", "output": "NO"}, {"input": "1\r\nDb -3373 3591\r\n", "output": "NO"}, {"input": "5\r\nQ2bz 960 2342\r\nhmX 2710 -1348\r\n... | false | stdio | null | true |
676/C | 676 | C | PyPy 3 | TESTS | 37 | 155 | 4,403,200 | 118349399 | def find(a, pref, k):
n = len(a)
a1 = []
for i in range(n):
if i % 2 == 1:
a1.append(0)
else:
a1.append(a[i])
_sum = 0
_max = -1
for i in range(1, n, 2):
if a[i] > _max:
_max = a[i]
r = 1
s1 = a1[0]
for l in range(n):
... | 117 | 62 | 2,764,800 | 166309064 | def solve(c,s):
ln=len(s)
dp=[0]*(ln+1)
for i in range(ln):
if s[i]=='a':
dp[i+1]=dp[i]+1
else:
dp[i+1]=dp[i]
lft=0
rt=ln
while lft<rt:
md=(lft+rt+1)//2
f=0
for i in range(md, ln+1):
tmp=dp[i]-dp[i-md]
if tmp... | Codeforces Round 354 (Div. 2) | CF | 2,016 | 1 | 256 | Vasya and String | High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than k characters of the original string. W... | The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only. | Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than k characters. | null | In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". | [{"input": "4 2\nabba", "output": "4"}, {"input": "8 1\naabaabaa", "output": "5"}] | 1,500 | ["binary search", "dp", "strings", "two pointers"] | 117 | [{"input": "4 2\r\nabba\r\n", "output": "4\r\n"}, {"input": "8 1\r\naabaabaa\r\n", "output": "5\r\n"}, {"input": "1 0\r\na\r\n", "output": "1\r\n"}, {"input": "1 1\r\nb\r\n", "output": "1\r\n"}, {"input": "1 0\r\nb\r\n", "output": "1\r\n"}, {"input": "1 1\r\na\r\n", "output": "1\r\n"}, {"input": "10 10\r\nbbbbbbbbbb\r\... | false | stdio | null | true |
416/D | 416 | D | Python 3 | TESTS | 78 | 451 | 16,588,800 | 6364119 | n = int(input())
a = list(map(int, input().split()))
i, j, k = 0, -1, -1
ans = 1
for l in range(len(a)):
if a[l] == -1:
continue
if j == -1:
j = l
elif k == -1:
if (a[l] - a[j]) % (l - j) == 0 and a[j] - (a[l] - a[j]) / (l - j) * (j - i) > 0:
k = l
else:
i, j = l, l
ans += 1
el... | 84 | 171 | 27,648,000 | 302828817 | import sys,math,heapq
from collections import Counter
#sys.setrecursionlimit(200000)
#fin = open('input.txt', 'r')
#input = lambda: fin.readline().strip()
input=lambda: sys.stdin.readline().strip()
import random
RANDOM = random.getrandbits(20)
class Wrapper(int):
def __init__(self, x):
int.__init__(x)
... | Codeforces Round 241 (Div. 2) | CF | 2,014 | 1 | 256 | Population Size | Polycarpus develops an interesting theory about the interrelation of arithmetic progressions with just everything in the world. His current idea is that the population of the capital of Berland changes over time like an arithmetic progression. Well, or like multiple arithmetic progressions.
Polycarpus believes that if... | The first line of the input contains integer n (1 ≤ n ≤ 2·105) — the number of elements in the sequence. The second line contains integer values a1, a2, ..., an separated by a space (1 ≤ ai ≤ 109 or ai = - 1). | Print the minimum number of arithmetic progressions that you need to write one after another to get sequence a. The positions marked as -1 in a can be represented by any positive integers. | null | null | [{"input": "9\n8 6 4 2 1 4 7 10 2", "output": "3"}, {"input": "9\n-1 6 -1 2 -1 4 7 -1 2", "output": "3"}, {"input": "5\n-1 -1 -1 -1 -1", "output": "1"}, {"input": "7\n-1 -1 4 5 1 2 3", "output": "2"}] | 2,400 | ["greedy", "implementation", "math"] | 84 | [{"input": "9\r\n8 6 4 2 1 4 7 10 2\r\n", "output": "3\r\n"}, {"input": "9\r\n-1 6 -1 2 -1 4 7 -1 2\r\n", "output": "3\r\n"}, {"input": "5\r\n-1 -1 -1 -1 -1\r\n", "output": "1\r\n"}, {"input": "7\r\n-1 -1 4 5 1 2 3\r\n", "output": "2\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}, {"input": "1\r\n65\r\n", "output": ... | false | stdio | null | true |
600/C | 600 | C | Python 3 | TESTS | 6 | 62 | 204,800 | 111955339 | counts = [0]*26
for c in input():
counts[ord(c)-97] += 1
# print(*counts)
ans = ""
i = 26
while i:
i -= 1
if counts[i]&1:
# Find min thing with odd count and increase that
j = 0
while j < 26 and counts[j]&1 < 1:
j += 1
if j < 26 and counts[j]&1:
counts[i] -= 1
counts[j] += 1
break
# print(ans)
... | 66 | 62 | 2,150,400 | 215107109 | def rl():
return list(map(int,input().split()))
def ri():
return int(input())
def rs():
return input()
def rm():
return map(int,input().split())
def main():
s=rs()
cnt=[0]*26
for c in s:
cnt[ord(c)-97]+=1
od=ev=0
for i in range(26):
od+=cnt[i]%2==1
ev+=cnt[i]... | Educational Codeforces Round 2 | ICPC | 2,015 | 2 | 256 | Make Palindrome | A string is called palindrome if it reads the same from left to right and from right to left. For example "kazak", "oo", "r" and "mikhailrubinchikkihcniburliahkim" are palindroms, but strings "abb" and "ij" are not.
You are given string s consisting of lowercase Latin letters. At once you can choose any position in th... | The only line contains string s (1 ≤ |s| ≤ 2·105) consisting of only lowercase Latin letters. | Print the lexicographically smallest palindrome that can be obtained with the minimal number of changes. | null | null | [{"input": "aabc", "output": "abba"}, {"input": "aabcd", "output": "abcba"}] | 1,800 | ["constructive algorithms", "greedy", "strings"] | 66 | [{"input": "aabc\r\n", "output": "abba\r\n"}, {"input": "aabcd\r\n", "output": "abcba\r\n"}, {"input": "u\r\n", "output": "u\r\n"}, {"input": "ttttt\r\n", "output": "ttttt\r\n"}, {"input": "xxxvvvxxvv\r\n", "output": "vvvxxxxvvv\r\n"}, {"input": "wrwrwfrrfrffrrwwwffffwrfrrwfrrfrwwfwfrwfwfwffwrrwfrrrwwwfrrrwfrrfwrwwrwrr... | false | stdio | null | true |
837/D | 837 | D | PyPy 3 | TESTS | 24 | 514 | 5,120,000 | 131910761 | import sys
input = sys.stdin.readline
n, k = map(int, input().split())
m = 18 * n
a = list(map(int, input().split()))
inf = 114514
dp = [[-inf] * (m + 1) for _ in range(n + 1)]
dp[0][0] = 0
c = 0
for i in range(n):
ai = a[i]
x, y = 0, 0
while not ai % 2:
x += 1
ai //= 2
while not ai % 5... | 99 | 1,076 | 7,987,200 | 131912158 | import sys
input = sys.stdin.readline
n, k = map(int, input().split())
m = 30 * n
a = list(map(int, input().split()))
inf = 114514
dp = [[-inf] * (m + 1) for _ in range(n + 1)]
dp[0][0] = 0
dp0 = [0] * (n + 1)
for i in range(n):
ai = a[i]
x, y = 0, 0
while not ai % 2:
y += 1
ai //= 2
wh... | Educational Codeforces Round 26 | ICPC | 2,017 | 2 | 256 | Round Subset | Let's call the roundness of the number the number of zeros to which it ends.
You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum possible. | The first line contains two integer numbers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 1018). | Print maximal roundness of product of the chosen subset of length k. | null | In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.
In the second example subset [15, 16, 25] has product 6000, roundness 3.
In the third example all subsets has product with roundness 0. | [{"input": "3 2\n50 4 20", "output": "3"}, {"input": "5 3\n15 16 3 25 9", "output": "3"}, {"input": "3 3\n9 77 13", "output": "0"}] | 2,100 | ["dp", "math"] | 99 | [{"input": "3 2\r\n50 4 20\r\n", "output": "3\r\n"}, {"input": "5 3\r\n15 16 3 25 9\r\n", "output": "3\r\n"}, {"input": "3 3\r\n9 77 13\r\n", "output": "0\r\n"}, {"input": "1 1\r\n200000000\r\n", "output": "8\r\n"}, {"input": "1 1\r\n3\r\n", "output": "0\r\n"}, {"input": "3 1\r\n1000000000000000000 800000000000000000 6... | false | stdio | null | true |
799/D | 799 | D | PyPy 3-64 | TESTS | 26 | 265 | 20,684,800 | 204499804 | from math import *
a,b,w,h,n=map(int,input().split())
w,h=ceil(a/w),ceil(b/h)
MAX=100000
MAX=min(MAX,max(w,h))
if w<=1 and h<=1:
print(0)
else:
l=list(map(int,input().split()))
l.sort()
l.reverse()
l=l[:34]
dp=[0 for i in range(MAX+1)] #max width for height i
dp[0]=MAX*MAX
dp[1]=1
gg... | 103 | 62 | 2,560,000 | 230893177 | f = lambda: map(int, input().split())
a, b, h, w, n = f()
c = sorted(list(f()), key=lambda x: -x)
d = {(h, w), (w, h)}
for i, q in enumerate([1] + c):
for u, v in d.copy():
h, w = u, v * q
if a <= w and b <= h or a <= h and b <= w:
print(i)
exit()
d.add((h, w))
... | Playrix Codescapes Cup (Codeforces Round 413, rated, Div. 1 + Div. 2) | CF | 2,017 | 1 | 256 | Field expansion | In one of the games Arkady is fond of the game process happens on a rectangular field. In the game process Arkady can buy extensions for his field, each extension enlarges one of the field sizes in a particular number of times. Formally, there are n extensions, the i-th of them multiplies the width or the length (by Ar... | The first line contains five integers a, b, h, w and n (1 ≤ a, b, h, w, n ≤ 100 000) — the sizes of the rectangle needed to be placed, the initial sizes of the field and the number of available extensions.
The second line contains n integers a1, a2, ..., an (2 ≤ ai ≤ 100 000), where ai equals the integer a side multip... | Print the minimum number of extensions needed to reach Arkady's goal. If it is not possible to place the rectangle on the field with all extensions, print -1. If the rectangle can be placed on the initial field, print 0. | null | In the first example it is enough to use any of the extensions available. For example, we can enlarge h in 5 times using the second extension. Then h becomes equal 10 and it is now possible to place the rectangle on the field. | [{"input": "3 3 2 4 4\n2 5 4 10", "output": "1"}, {"input": "3 3 3 3 5\n2 3 5 4 2", "output": "0"}, {"input": "5 5 1 2 3\n2 2 3", "output": "-1"}, {"input": "3 4 1 1 3\n2 3 2", "output": "3"}] | 2,100 | ["brute force", "dp", "meet-in-the-middle"] | 103 | [{"input": "3 3 2 4 4\r\n2 5 4 10\r\n", "output": "1\r\n"}, {"input": "3 3 3 3 5\r\n2 3 5 4 2\r\n", "output": "0\r\n"}, {"input": "5 5 1 2 3\r\n2 2 3\r\n", "output": "-1\r\n"}, {"input": "3 4 1 1 3\r\n2 3 2\r\n", "output": "3\r\n"}, {"input": "572 540 6 2 12\r\n2 3 2 2 2 3 3 3 2 2 2 2\r\n", "output": "-1\r\n"}, {"input... | false | stdio | null | true |
638/B | 638 | B | Python 3 | PRETESTS | 9 | 62 | 4,608,000 | 16843287 | n = int(input())
arr = []
arr2 = ""
for i in range(n):
arr.append(str(input()))
result = arr[0]
z = 0
while len(arr) != 0:
for i in range(len(arr)):
if result.find(arr[i-z]) != -1:
del arr[i-z]
z+=1
continue
elif arr[i-z].find(result)!= -1:
result ... | 67 | 62 | 4,812,800 | 16855472 | n=int(input())
s=set()
for i in range(n):
p=input()
if i!=n-1:
s.add(p)
n-=1
w={}
for i in s:
w.update([(i,set(i))])
w.update([(p,set(p))])
y=''
while n>0:
u=0
for i in s:
u+=1
if p[0] in w[i] and p[-1] in w[i]:
p=i
s.discard(i)
break
elif p[0] in w[i]:
p=i[:i.find(p[0])]+p
s.discard(i)
b... | VK Cup 2016 - Qualification Round 2 | CF | 2,016 | 1 | 256 | Making Genome in Berland | Berland scientists face a very important task - given the parts of short DNA fragments, restore the dinosaur DNA! The genome of a berland dinosaur has noting in common with the genome that we've used to: it can have 26 distinct nucleotide types, a nucleotide of each type can occur at most once. If we assign distinct En... | The first line of the input contains a positive integer n (1 ≤ n ≤ 100) — the number of genome fragments.
Each of the next lines contains one descriptions of a fragment. Each fragment is a non-empty string consisting of distinct small letters of the English alphabet. It is not guaranteed that the given fragments are d... | In the single line of the output print the genome of the minimum length that contains all the given parts. All the nucleotides in the genome must be distinct. If there are multiple suitable strings, print the string of the minimum length. If there also are multiple suitable strings, you can print any of them. | null | null | [{"input": "3\nbcd\nab\ncdef", "output": "abcdef"}, {"input": "4\nx\ny\nz\nw", "output": "xyzw"}] | 1,500 | ["*special", "dfs and similar", "strings"] | 67 | [{"input": "3\r\nbcd\r\nab\r\ncdef\r\n", "output": "abcdef\r\n"}, {"input": "4\r\nx\r\ny\r\nz\r\nw\r\n", "output": "xyzw\r\n"}, {"input": "25\r\nef\r\nfg\r\ngh\r\nhi\r\nij\r\njk\r\nkl\r\nlm\r\nmn\r\nno\r\nab\r\nbc\r\ncd\r\nde\r\nop\r\npq\r\nqr\r\nrs\r\nst\r\ntu\r\nuv\r\nvw\r\nwx\r\nxy\r\nyz\r\n", "output": "abcdefghijk... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path) as f:
n = int(f.readline().strip())
fragments = [line.strip() for line in f]
with open(output_path) as f:
correct_output = f.read().strip()
... | true |
638/B | 638 | B | Python 3 | PRETESTS | 9 | 46 | 4,915,200 | 16839935 | def lfs(new):
for i in range(len(new)):
flag = True
for j in range(len(new)):
if j != i:
if new[i][0] in new[j]:
flag = False
if flag:
return i
def vlastsovetam(data):
new = []
for i in range(len(data)):
flag = True... | 67 | 62 | 4,915,200 | 16839269 | #!/usr/bin/env python3
n = eval(input())
substrings = [input() for line in range(n)]
letters = {}
for word in substrings:
for letter in word:
if not letter in letters:
letters[letter] = [word[:(word.index(letter))],word[(word.index(letter)+1):]]
else:
if len(letters[letter]... | VK Cup 2016 - Qualification Round 2 | CF | 2,016 | 1 | 256 | Making Genome in Berland | Berland scientists face a very important task - given the parts of short DNA fragments, restore the dinosaur DNA! The genome of a berland dinosaur has noting in common with the genome that we've used to: it can have 26 distinct nucleotide types, a nucleotide of each type can occur at most once. If we assign distinct En... | The first line of the input contains a positive integer n (1 ≤ n ≤ 100) — the number of genome fragments.
Each of the next lines contains one descriptions of a fragment. Each fragment is a non-empty string consisting of distinct small letters of the English alphabet. It is not guaranteed that the given fragments are d... | In the single line of the output print the genome of the minimum length that contains all the given parts. All the nucleotides in the genome must be distinct. If there are multiple suitable strings, print the string of the minimum length. If there also are multiple suitable strings, you can print any of them. | null | null | [{"input": "3\nbcd\nab\ncdef", "output": "abcdef"}, {"input": "4\nx\ny\nz\nw", "output": "xyzw"}] | 1,500 | ["*special", "dfs and similar", "strings"] | 67 | [{"input": "3\r\nbcd\r\nab\r\ncdef\r\n", "output": "abcdef\r\n"}, {"input": "4\r\nx\r\ny\r\nz\r\nw\r\n", "output": "xyzw\r\n"}, {"input": "25\r\nef\r\nfg\r\ngh\r\nhi\r\nij\r\njk\r\nkl\r\nlm\r\nmn\r\nno\r\nab\r\nbc\r\ncd\r\nde\r\nop\r\npq\r\nqr\r\nrs\r\nst\r\ntu\r\nuv\r\nvw\r\nwx\r\nxy\r\nyz\r\n", "output": "abcdefghijk... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path) as f:
n = int(f.readline().strip())
fragments = [line.strip() for line in f]
with open(output_path) as f:
correct_output = f.read().strip()
... | true |
515/B | 515 | B | Python 3 | TESTS | 19 | 62 | 204,800 | 114852125 | from collections import deque
n, m=[int(k) for k in input().split()]
w=[int(k) for k in input().split()][1:]
q=[int(k) for k in input().split()][1:]
x, y=len(w), len(q)
qwe={}
zxc={}
for i in range(n*m+11):
if i%n not in qwe:
qwe[i%n]=[i%m]
elif i%m not in qwe[i%n]:
qwe[i%n].append(i%m)
if ... | 56 | 46 | 0 | 218402049 | import sys
sys.setrecursionlimit(100000000)
input=lambda:sys.stdin.readline().strip()
write=lambda x:sys.stdout.write(str(x)+'\n')
# from random import randint
# from copy import deepcopy
# from collections import deque,Counter
# from heapq import heapify,heappush,heappop
# from bisect import bisect_left,bisect,insort... | Codeforces Round 292 (Div. 2) | CF | 2,015 | 2 | 256 | Drazil and His Happy Friends | Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.
There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites $$(i \bmod n)$$-th... | The first line contains two integer n and m (1 ≤ n, m ≤ 100).
The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys.
The third line conatins integer g (0 ≤ ... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | null | By $$i \bmod k$$ we define the remainder of integer division of i by k.
In first sample case:
- On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day.
- On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so no... | [{"input": "2 3\n0\n1 0", "output": "Yes"}, {"input": "2 4\n1 0\n1 2", "output": "No"}, {"input": "2 3\n1 0\n1 1", "output": "Yes"}] | 1,300 | ["brute force", "dsu", "meet-in-the-middle", "number theory"] | 56 | [{"input": "2 3\r\n0\r\n1 0\r\n", "output": "Yes\r\n"}, {"input": "2 4\r\n1 0\r\n1 2\r\n", "output": "No\r\n"}, {"input": "2 3\r\n1 0\r\n1 1\r\n", "output": "Yes\r\n"}, {"input": "16 88\r\n6 5 14 2 0 12 7\r\n30 21 64 35 79 74 39 63 44 81 73 0 27 33 69 12 86 46 20 25 55 52 7 58 23 5 60 32 41 50 82\r\n", "output": "Yes\r... | false | stdio | null | true |
725/A | 725 | A | Python 3 | TESTS | 8 | 46 | 614,400 | 30551023 | n=int(input())
s=input()
k=s.count('><')
if k==0:
pass
elif k==1:
k=s.find('><')
for i in range(k+1,n):
if s[i]=='<':
n-=1
else:
break
for i in range(k,-1,-1):
if s[i]=='>':
n-=1
else:
break
else:
k=s.rfind('><')
for... | 78 | 46 | 102,400 | 146871880 | def solve():
n = int(input())
s = input()
ans = 0
for i in s:
if i == "<":
ans += 1
else:
break
for i in s[::-1]:
if i == ">":
ans += 1
else:
break
print(ans)
solve() | Canada Cup 2016 | CF | 2,016 | 2 | 256 | Jumping Ball | In a new version of the famous Pinball game, one of the most important parts of the game field is a sequence of n bumpers. The bumpers are numbered with integers from 1 to n from left to right. There are two types of bumpers. They are denoted by the characters '<' and '>'. When the ball hits the bumper at position i it... | The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the length of the sequence of bumpers. The second line contains the string, which consists of the characters '<' and '>'. The character at the i-th position of this string corresponds to the type of the i-th bumper. | Print one integer — the number of positions in the sequence such that the ball will eventually fall from the game field if it starts at that position. | null | In the first sample, the ball will fall from the field if starts at position 1 or position 2.
In the second sample, any starting position will result in the ball falling from the field. | [{"input": "4\n<<><", "output": "2"}, {"input": "5\n>>>>>", "output": "5"}, {"input": "4\n>><<", "output": "0"}] | 1,000 | ["implementation"] | 78 | [{"input": "4\r\n<<><\r\n", "output": "2\r\n"}, {"input": "5\r\n>>>>>\r\n", "output": "5\r\n"}, {"input": "4\r\n>><<\r\n", "output": "0\r\n"}, {"input": "3\r\n<<>\r\n", "output": "3\r\n"}, {"input": "3\r\n<<<\r\n", "output": "3\r\n"}, {"input": "3\r\n><<\r\n", "output": "0\r\n"}, {"input": "1\r\n<\r\n", "output": "1\r\... | false | stdio | null | true |
45/I | 45 | I | PyPy 3-64 | TESTS | 13 | 124 | 0 | 167949969 | import sys
input = sys.stdin.readline
n = int(input())
w = sorted(map(int, input().split()))
d = [i for i in w if i < 0]
e = [i for i in w if i > 0]
if len(d) == len(e) == 0:
print(0)
exit(0)
if len(e) == 0 and len(d) == 1:
print(d[0])
exit(0)
if len(d) % 2:
d.sort()
d = d[:-1]
e.extend(d)
pr... | 65 | 154 | 6,963,200 | 125022759 | n = int(input())
a = list(map(int, input().split()))
pos = []
neg = []
z = 0
for i in range(n):
if a[i] > 0:
pos.append(a[i])
elif a[i] < 0:
neg.append(a[i])
else:
z += 1
m = len(neg)
neg.sort()
if m % 2:
m -= 1
for i in range(m):
pos.append(neg[i])
if len(pos) > 0:
... | School Team Contest 3 (Winter Computer School 2010/11) | ICPC | 2,010 | 2 | 256 | TCMCF+++ | Vasya has gotten interested in programming contests in TCMCF+++ rules. On the contest n problems were suggested and every problem had a cost — a certain integral number of points (perhaps, negative or even equal to zero). According to TCMCF+++ rules, only accepted problems can earn points and the overall number of poin... | The first line contains an integer n (1 ≤ n ≤ 100) — the number of the suggested problems. The next line contains n space-separated integers ci ( - 100 ≤ ci ≤ 100) — the cost of the i-th task. The tasks' costs may coinсide. | Print space-separated the costs of the problems that needed to be solved to get the maximal possible number of points. Do not forget, please, that it was necessary to solve at least one problem. If there are several solutions to that problem, print any of them. | null | null | [{"input": "5\n1 2 -3 3 3", "output": "3 1 2 3"}, {"input": "13\n100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "100 100 100 100 100 100 100 100 100 100 100 100 100"}, {"input": "4\n-2 -2 -2 -2", "output": "-2 -2 -2 -2"}] | 1,400 | ["greedy"] | 65 | [{"input": "5\r\n1 2 -3 3 3\r\n", "output": "3 1 2 3 \r\n"}, {"input": "13\r\n100 100 100 100 100 100 100 100 100 100 100 100 100\r\n", "output": "100 100 100 100 100 100 100 100 100 100 100 100 100 \r\n"}, {"input": "4\r\n-2 -2 -2 -2\r\n", "output": "-2 -2 -2 -2 \r\n"}, {"input": "1\r\n1\r\n", "output": "1 \r\n"}, {"i... | false | stdio | import sys
from collections import Counter
def compute_max_product(ci):
if all(x == 0 for x in ci):
return 0
positives = [x for x in ci if x > 0]
negatives = [x for x in ci if x < 0]
zeros = [x for x in ci if x == 0]
non_zero = positives + negatives
if not non_zero:
return 0
... | true |
550/A | 550 | A | PyPy 3-64 | TESTS | 51 | 62 | 0 | 198567618 | def solve():
s = input()
abi0 = s.find('AB', 0)
abi1 = s.find('AB', abi0+1) if abi0 != -1 else -1
abi2 = s.find('AB', abi1+1) if abi1 != -1 else -1
bai0 = s.find('BA')
if bai0 == -1:
return False
if abi0 != -1 and abi0 != bai0-1 and abi0 != bai0+1:
return True
... | 85 | 46 | 0 | 195984665 | s = input().strip()
if 'AB' in s and 'BA' in s[s.index('AB')+2:]:
print('YES')
elif 'BA' in s and 'AB' in s[s.index('BA')+2:]:
print('YES')
else:
print('NO') | Codeforces Round 306 (Div. 2) | CF | 2,015 | 2 | 256 | Two Substrings | You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order). | The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters. | Print "YES" (without the quotes), if string s contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise. | null | In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | [{"input": "ABA", "output": "NO"}, {"input": "BACFAB", "output": "YES"}, {"input": "AXBYBXA", "output": "NO"}] | 1,500 | ["brute force", "dp", "greedy", "implementation", "strings"] | 85 | [{"input": "ABA\r\n", "output": "NO\r\n"}, {"input": "BACFAB\r\n", "output": "YES\r\n"}, {"input": "AXBYBXA\r\n", "output": "NO\r\n"}, {"input": "ABABAB\r\n", "output": "YES\r\n"}, {"input": "BBBBBBBBBB\r\n", "output": "NO\r\n"}, {"input": "ABBA\r\n", "output": "YES\r\n"}, {"input": "ABAXXXAB\r\n", "output": "YES\r\n"}... | false | stdio | null | true |
45/I | 45 | I | Python 3 | TESTS | 9 | 218 | 307,200 | 75714701 | def TCMCF():
num = int(input())
costs = [int(i) for i in input().split()]
if num == 1:
print(costs[0])
elif num == 2:
print(max(costs[0]*costs[1],costs[0],costs[1]))
else:
negative_numbers = []
for i in costs:
if i < 0:
negative_numbers.append(i)
if costs.count(0) == len(cost... | 65 | 218 | 307,200 | 93067158 | from bisect import bisect_left as bl;n=int(input())
if n==1:
print(input())
else:
a=list(map(int,input().split()));a.sort()
if bl(a,0)%2==1:
if a[bl(a,0)-1]<0:a.remove(a[bl(a,0)-1])
q=len(a)
while 0 in a and q>1:a.remove(0);q-=1
print(*a) | School Team Contest 3 (Winter Computer School 2010/11) | ICPC | 2,010 | 2 | 256 | TCMCF+++ | Vasya has gotten interested in programming contests in TCMCF+++ rules. On the contest n problems were suggested and every problem had a cost — a certain integral number of points (perhaps, negative or even equal to zero). According to TCMCF+++ rules, only accepted problems can earn points and the overall number of poin... | The first line contains an integer n (1 ≤ n ≤ 100) — the number of the suggested problems. The next line contains n space-separated integers ci ( - 100 ≤ ci ≤ 100) — the cost of the i-th task. The tasks' costs may coinсide. | Print space-separated the costs of the problems that needed to be solved to get the maximal possible number of points. Do not forget, please, that it was necessary to solve at least one problem. If there are several solutions to that problem, print any of them. | null | null | [{"input": "5\n1 2 -3 3 3", "output": "3 1 2 3"}, {"input": "13\n100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "100 100 100 100 100 100 100 100 100 100 100 100 100"}, {"input": "4\n-2 -2 -2 -2", "output": "-2 -2 -2 -2"}] | 1,400 | ["greedy"] | 65 | [{"input": "5\r\n1 2 -3 3 3\r\n", "output": "3 1 2 3 \r\n"}, {"input": "13\r\n100 100 100 100 100 100 100 100 100 100 100 100 100\r\n", "output": "100 100 100 100 100 100 100 100 100 100 100 100 100 \r\n"}, {"input": "4\r\n-2 -2 -2 -2\r\n", "output": "-2 -2 -2 -2 \r\n"}, {"input": "1\r\n1\r\n", "output": "1 \r\n"}, {"i... | false | stdio | import sys
from collections import Counter
def compute_max_product(ci):
if all(x == 0 for x in ci):
return 0
positives = [x for x in ci if x > 0]
negatives = [x for x in ci if x < 0]
zeros = [x for x in ci if x == 0]
non_zero = positives + negatives
if not non_zero:
return 0
... | true |
550/A | 550 | A | Python 3 | TESTS | 50 | 77 | 1,843,200 | 225913270 | s = str(input())
a = 0
b = 0
i = 0
ans = []
while i < (len(s)-1) :
if s[i] == "A" and s[i+1] == "B" :
a += 1
ans.append(i)
i += 1
elif s[i] == "B" and s[i+1] == "A" :
b += 1
ans.append(i)
i += 2
else:
i += 1
if a>=2 and b>=2 :
print("YES")
elif a<... | 85 | 46 | 0 | 196334991 | s=input();u=s.find('AB');v=s.find('BA')
print('YES'if(u+1and s.find('BA',u+2)+1)or(v+1and s.find('AB',v+2)+1)else'NO') | Codeforces Round 306 (Div. 2) | CF | 2,015 | 2 | 256 | Two Substrings | You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order). | The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters. | Print "YES" (without the quotes), if string s contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise. | null | In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | [{"input": "ABA", "output": "NO"}, {"input": "BACFAB", "output": "YES"}, {"input": "AXBYBXA", "output": "NO"}] | 1,500 | ["brute force", "dp", "greedy", "implementation", "strings"] | 85 | [{"input": "ABA\r\n", "output": "NO\r\n"}, {"input": "BACFAB\r\n", "output": "YES\r\n"}, {"input": "AXBYBXA\r\n", "output": "NO\r\n"}, {"input": "ABABAB\r\n", "output": "YES\r\n"}, {"input": "BBBBBBBBBB\r\n", "output": "NO\r\n"}, {"input": "ABBA\r\n", "output": "YES\r\n"}, {"input": "ABAXXXAB\r\n", "output": "YES\r\n"}... | false | stdio | null | true |
550/A | 550 | A | PyPy 3-64 | TESTS | 51 | 77 | 6,553,600 | 220380432 | a = list(input())
k = False
k1 = False
i = 0
while i < len(a):
if a[i:i+2] == ['B', 'A']:
k = True
a[i] = '/'
a[i+1] = '/'
break
i += 1
i = 0
while i < len(a):
if a[i:i+2] == ['A', 'B']:
k1 = True
a[i] = '/'
a[i+1] = '/'
break
i += 1
if k... | 85 | 46 | 0 | 197319059 | z=input()
u=z.find('AB')
v=z.find('BA')
print('YES'if(u+1and z.find('BA',u+2)+1)or(v+1and z.find('AB',v+2)+1)else'NO') | Codeforces Round 306 (Div. 2) | CF | 2,015 | 2 | 256 | Two Substrings | You are given string s. Your task is to determine if the given string s contains two non-overlapping substrings "AB" and "BA" (the substrings can go in any order). | The only line of input contains a string s of length between 1 and 105 consisting of uppercase Latin letters. | Print "YES" (without the quotes), if string s contains two non-overlapping substrings "AB" and "BA", and "NO" otherwise. | null | In the first sample test, despite the fact that there are substrings "AB" and "BA", their occurrences overlap, so the answer is "NO".
In the second sample test there are the following occurrences of the substrings: BACFAB.
In the third sample test there is no substring "AB" nor substring "BA". | [{"input": "ABA", "output": "NO"}, {"input": "BACFAB", "output": "YES"}, {"input": "AXBYBXA", "output": "NO"}] | 1,500 | ["brute force", "dp", "greedy", "implementation", "strings"] | 85 | [{"input": "ABA\r\n", "output": "NO\r\n"}, {"input": "BACFAB\r\n", "output": "YES\r\n"}, {"input": "AXBYBXA\r\n", "output": "NO\r\n"}, {"input": "ABABAB\r\n", "output": "YES\r\n"}, {"input": "BBBBBBBBBB\r\n", "output": "NO\r\n"}, {"input": "ABBA\r\n", "output": "YES\r\n"}, {"input": "ABAXXXAB\r\n", "output": "YES\r\n"}... | false | stdio | null | true |
581/C | 581 | C | Python 3 | TESTS | 7 | 46 | 0 | 201076889 | n,k=map(int,input().split()) ; arr=list(map(int,input().split())) ; ans=min(n*100 , sum(arr)+k ) ; print(ans//10) | 98 | 109 | 5,120,000 | 13267634 | n, k = map(int, input().split(' ')[:2])
a = list(map(int, input().split(' ')[:n]))
b = [0 for i in range(101)]
rating = 0
for i in a:
b[i]+=1
rating += i // 10
for last_dig in range(9, 0, -1):
for first_dig in range(10):
v = first_dig * 10 + last_dig
need_for_one = 10 - last_dig
c... | Codeforces Round 322 (Div. 2) | CF | 2,015 | 1 | 256 | Developing Skills | Petya loves computer games. Finally a game that he's been waiting for so long came out!
The main character of this game has n different skills, each of which is characterized by an integer ai from 0 to 100. The higher the number ai is, the higher is the i-th skill of the character. The total rating of the character is... | The first line of the input contains two positive integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 107) — the number of skills of the character and the number of units of improvements at Petya's disposal.
The second line of the input contains a sequence of n integers ai (0 ≤ ai ≤ 100), where ai characterizes the level of the i-... | The first line of the output should contain a single non-negative integer — the maximum total rating of the character that Petya can get using k or less improvement units. | null | In the first test case the optimal strategy is as follows. Petya has to improve the first skill to 10 by spending 3 improvement units, and the second skill to 10, by spending one improvement unit. Thus, Petya spends all his improvement units and the total rating of the character becomes equal to lfloor frac{100}{10} r... | [{"input": "2 4\n7 9", "output": "2"}, {"input": "3 8\n17 15 19", "output": "5"}, {"input": "2 2\n99 100", "output": "20"}] | 1,400 | ["implementation", "math", "sortings"] | 98 | [{"input": "2 4\r\n7 9\r\n", "output": "2\r\n"}, {"input": "3 8\r\n17 15 19\r\n", "output": "5\r\n"}, {"input": "2 2\r\n99 100\r\n", "output": "20\r\n"}, {"input": "100 10000\r\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ... | false | stdio | null | true |
413/C | 413 | C | Python 3 | TESTS | 32 | 124 | 0 | 48322600 | n, m = input().split()
znac=input().split()
nay=input().split()
ayc=[]
summ=0
obi = []
for i in range(int(n)):
if str(i+1) in nay:
ayc.append(znac[i])
else:
obi.append(znac[i])
nayk=reversed(sorted(ayc))
for i in obi:
summ+=int(i)
for i in nayk:
if summ<int(i):
summ+=int(i)
e... | 67 | 46 | 0 | 205670441 | n, m = map(int, input().split())
a = list(map(int, input().split()))
b = [a[i - 1] for i in map(int, input().split())]
b.sort(reverse = True)
s = sum(a) - sum(b)
for i in b: s += s if i < s else i
print(s) | Coder-Strike 2014 - Round 2 | CF | 2,014 | 1 | 256 | Jeopardy! | 'Jeopardy!' is an intellectual game where players answer questions and earn points. Company Q conducts a simplified 'Jeopardy!' tournament among the best IT companies. By a lucky coincidence, the old rivals made it to the finals: company R1 and company R2.
The finals will have n questions, m of them are auction questi... | The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100; m ≤ min(n, 30)) — the total number of questions and the number of auction questions, correspondingly. The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 107) — the prices of the questions. The third line contains m ... | In the single line, print the answer to the problem — the maximum points the R2 company can get if it plays optimally well. It is guaranteed that the answer fits into the integer 64-bit signed type. | null | null | [{"input": "4 1\n1 3 7 5\n3", "output": "18"}, {"input": "3 2\n10 3 8\n2 3", "output": "40"}, {"input": "2 2\n100 200\n1 2", "output": "400"}] | 1,400 | ["greedy", "math"] | 67 | [{"input": "4 1\r\n1 3 7 5\r\n3\r\n", "output": "18\r\n"}, {"input": "3 2\r\n10 3 8\r\n2 3\r\n", "output": "40\r\n"}, {"input": "2 2\r\n100 200\r\n1 2\r\n", "output": "400\r\n"}, {"input": "1 1\r\n1\r\n1\r\n", "output": "1\r\n"}, {"input": "2 2\r\n1 5\r\n1 2\r\n", "output": "10\r\n"}, {"input": "5 3\r\n5 8 7 1 9\r\n2 5... | false | stdio | null | true |
638/B | 638 | B | Python 3 | PRETESTS | 9 | 62 | 5,120,000 | 16843600 | from sys import exit
n = int(input())
a = []
for i in range(n):
s = input()
for j in range(len(a)):
check = a[j]
if s in check:
break
if check in s:
a[j] = s
break
else:
a.append(s)
br_flag = False
while True:
for j in range(len(a)):
... | 67 | 62 | 4,915,200 | 16840651 | parts = []
empty = True
def chrx(x):
return chr(x+97)
def ordx(x):
return ord(x)-97
include = [ [False for _2 in range(26)] for _1 in range(100)]
exist = [False]*100
#"""
n = int(input())
for _ in range(n):
parts.append(input())
exist[_] = True
for c in parts[-1]:
include[_][ordx(c)] = ... | VK Cup 2016 - Qualification Round 2 | CF | 2,016 | 1 | 256 | Making Genome in Berland | Berland scientists face a very important task - given the parts of short DNA fragments, restore the dinosaur DNA! The genome of a berland dinosaur has noting in common with the genome that we've used to: it can have 26 distinct nucleotide types, a nucleotide of each type can occur at most once. If we assign distinct En... | The first line of the input contains a positive integer n (1 ≤ n ≤ 100) — the number of genome fragments.
Each of the next lines contains one descriptions of a fragment. Each fragment is a non-empty string consisting of distinct small letters of the English alphabet. It is not guaranteed that the given fragments are d... | In the single line of the output print the genome of the minimum length that contains all the given parts. All the nucleotides in the genome must be distinct. If there are multiple suitable strings, print the string of the minimum length. If there also are multiple suitable strings, you can print any of them. | null | null | [{"input": "3\nbcd\nab\ncdef", "output": "abcdef"}, {"input": "4\nx\ny\nz\nw", "output": "xyzw"}] | 1,500 | ["*special", "dfs and similar", "strings"] | 67 | [{"input": "3\r\nbcd\r\nab\r\ncdef\r\n", "output": "abcdef\r\n"}, {"input": "4\r\nx\r\ny\r\nz\r\nw\r\n", "output": "xyzw\r\n"}, {"input": "25\r\nef\r\nfg\r\ngh\r\nhi\r\nij\r\njk\r\nkl\r\nlm\r\nmn\r\nno\r\nab\r\nbc\r\ncd\r\nde\r\nop\r\npq\r\nqr\r\nrs\r\nst\r\ntu\r\nuv\r\nvw\r\nwx\r\nxy\r\nyz\r\n", "output": "abcdefghijk... | false | stdio | import sys
def main():
input_path = sys.argv[1]
output_path = sys.argv[2]
submission_path = sys.argv[3]
with open(input_path) as f:
n = int(f.readline().strip())
fragments = [line.strip() for line in f]
with open(output_path) as f:
correct_output = f.read().strip()
... | true |
733/C | 733 | C | PyPy 3 | TESTS | 9 | 140 | 0 | 77664040 | '''R = lambda: map(int, input().split())
n = int(input())
a = list(R())
m = int(input())
b = list(R())
st = i = j = 0
res = ''
while i < m:
psum = 0
while j < n and psum < b[i]:
psum += a[j]
j += 1
if psum != b[i]:
print('NO')
exit(0)
print(j,st,a[st],a[j-1])
mi = st ... | 139 | 46 | 0 | 219702093 | R = lambda: map(int, input().split())
n = int(input())
a = list(R())
m = int(input())
b = list(R())
st = i = j = 0
res = ''
while i < m:
psum = 0
while j < n and psum < b[i]:
psum += a[j]
j += 1
if psum != b[i]:
print('NO')
exit(0)
mi = st if a[st] > a[j - 1] else j - 1
... | Codeforces Round 378 (Div. 2) | CF | 2,016 | 1 | 256 | Epidemic in Monstropolis | There was an epidemic in Monstropolis and all monsters became sick. To recover, all monsters lined up in queue for an appointment to the only doctor in the city.
Soon, monsters became hungry and began to eat each other.
One monster can eat other monster if its weight is strictly greater than the weight of the monster... | The first line contains single integer n (1 ≤ n ≤ 500) — the number of monsters in the initial queue.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106) — the initial weights of the monsters.
The third line contains single integer k (1 ≤ k ≤ n) — the number of monsters in the queue after the joke.
Th... | In case if no actions could lead to the final queue, print "NO" (without quotes) in the only line.
Otherwise print "YES" (without quotes) in the first line. In the next n - k lines print actions in the chronological order. In each line print x — the index number of the monster in the current queue which eats and, sepa... | null | In the first example, initially there were n = 6 monsters, their weights are [1, 2, 2, 2, 1, 2] (in order of queue from the first monster to the last monster). The final queue should be [5, 5]. The following sequence of eatings leads to the final queue:
- the second monster eats the monster to the left (i.e. the first... | [{"input": "6\n1 2 2 2 1 2\n2\n5 5", "output": "YES\n2 L\n1 R\n4 L\n3 L"}, {"input": "5\n1 2 3 4 5\n1\n15", "output": "YES\n5 L\n4 L\n3 L\n2 L"}, {"input": "5\n1 1 1 3 3\n3\n2 1 6", "output": "NO"}] | 1,800 | ["constructive algorithms", "dp", "greedy", "two pointers"] | 139 | [{"input": "6\r\n1 2 2 2 1 2\r\n2\r\n5 5\r\n", "output": "YES\r\n2 L\r\n1 R\r\n4 L\r\n3 L\r\n"}, {"input": "5\r\n1 2 3 4 5\r\n1\r\n15\r\n", "output": "YES\r\n5 L\r\n4 L\r\n3 L\r\n2 L\r\n"}, {"input": "5\r\n1 1 1 3 3\r\n3\r\n2 1 6\r\n", "output": "NO"}, {"input": "5\r\n1 1 1 1 2\r\n3\r\n1 1 4\r\n", "output": "YES\r\n5 L... | false | stdio | import sys
def read_file(path):
with open(path, 'r') as f:
return f.read().splitlines()
input_path, output_path, submission_path = sys.argv[1], sys.argv[2], sys.argv[3]
input_lines = read_file(input_path)
n = int(input_lines[0])
a = list(map(int, input_lines[1].split()))
k = int(input_lines[2])
b = list(... | true |
982/A | 982 | A | Python 3 | TESTS | 27 | 124 | 0 | 40407719 | n = int(input())
flag = True
a = input()
for i in range(1,n-1):
if a[i] == '1':
if a[i-1] == '1' or a[i+1] == '1':
print("No")
flag = False
break
if a[i] == '0':
if a[i - 1] == '0' and a[i + 1] == '0':
print("No")
flag = False
... | 55 | 46 | 0 | 192764494 | # LUOGU_RID: 101739723
n = int(input())
s = input()
print((s.find('11') != -1 or any(s[i] == '0' and (i == 0 or s[i - 1] == '0') and (i == n - 1 or s[i + 1] == '0') for i in range(n))) and 'No' or 'Yes') | Codeforces Round 484 (Div. 2) | CF | 2,018 | 1 | 256 | Row | You're given a row with $$$n$$$ chairs. We call a seating of people "maximal" if the two following conditions hold:
1. There are no neighbors adjacent to anyone seated.
2. It's impossible to seat one more person without violating the first rule.
The seating is given as a string consisting of zeros and ones ($$$0$$$ m... | The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 1000$$$) — the number of chairs.
The next line contains a string of $$$n$$$ characters, each of them is either zero or one, describing the seating. | Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".
You are allowed to print letters in whatever case you'd like (uppercase or lowercase). | null | In sample case one the given seating is maximal.
In sample case two the person at chair three has a neighbour to the right.
In sample case three it is possible to seat yet another person into chair three. | [{"input": "3\n101", "output": "Yes"}, {"input": "4\n1011", "output": "No"}, {"input": "5\n10001", "output": "No"}] | 1,200 | ["brute force", "constructive algorithms"] | 55 | [{"input": "3\r\n101\r\n", "output": "Yes\r\n"}, {"input": "4\r\n1011\r\n", "output": "No\r\n"}, {"input": "5\r\n10001\r\n", "output": "No\r\n"}, {"input": "1\r\n0\r\n", "output": "No\r\n"}, {"input": "1\r\n1\r\n", "output": "Yes\r\n"}, {"input": "100\r\n01010010101010010100100101010010101001010010010010100101010100101... | false | stdio | null | true |
762/C | 762 | C | Python 3 | TESTS | 17 | 92 | 1,331,200 | 109096293 | import sys
input = sys.stdin.readline
def solve():
a = input().strip()
b = input().strip()
n = len(a)
m = len(b)
c = [0]*m
d = [0]*m
p = 0
for i in range(m):
for j in range(p, n):
if a[j] == b[i]:
c[i] = j
p = j + 1
break
else:
for j in range(i,m):
c[j] = n
break
p = n-1
for i... | 99 | 109 | 6,963,200 | 168750210 | def solve(s1,s2):
n=len(s1)
m=len(s2)
s1=' ' +s1 + ' '
s2=' ' +s2 + ' '
tmp=0
v1=[0]*100005
v2=[0]*100005
for i in range(1,n+1):
if s1[i]==s2[tmp+1]:
tmp+=1
v1[i]=tmp
tmp=m+1
v2[n+1]=tmp
for i in range(n,0,-1):
if s1[i]==s2[tmp-1]:
... | Educational Codeforces Round 17 | ICPC | 2,017 | 2 | 256 | Two strings | You are given two strings a and b. You have to remove the minimum possible number of consecutive (standing one after another) characters from string b in such a way that it becomes a subsequence of string a. It can happen that you will not need to remove any characters at all, or maybe you will have to remove all of th... | The first line contains string a, and the second line — string b. Both of these strings are nonempty and consist of lowercase letters of English alphabet. The length of each string is no bigger than 105 characters. | On the first line output a subsequence of string a, obtained from b by erasing the minimum number of consecutive characters.
If the answer consists of zero characters, output «-» (a minus sign). | null | In the first example strings a and b don't share any symbols, so the longest string that you can get is empty.
In the second example ac is a subsequence of a, and at the same time you can obtain it by erasing consecutive symbols cepted from string b. | [{"input": "hi\nbob", "output": "-"}, {"input": "abca\naccepted", "output": "ac"}, {"input": "abacaba\nabcdcba", "output": "abcba"}] | 2,100 | ["binary search", "hashing", "strings", "two pointers"] | 99 | [{"input": "hi\r\nbob\r\n", "output": "-\r\n"}, {"input": "abca\r\naccepted\r\n", "output": "ac\r\n"}, {"input": "abacaba\r\nabcdcba\r\n", "output": "abcba\r\n"}, {"input": "lo\r\neuhaqdhhzlnkmqnakgwzuhurqlpmdm\r\n", "output": "-\r\n"}, {"input": "aaeojkdyuilpdvyewjfrftkpcobhcumwlaoiocbfdtvjkhgda\r\nmlmarpivirqbxcyhyer... | false | stdio | null | true |
982/A | 982 | A | PyPy 3 | TESTS | 9 | 140 | 0 | 90464444 | lenght=input()
lenght=int(lenght)
word=input()
if lenght == 1:
if "0" in word:
print("No")
else:
print("Yes")
exit()
last_chars = word[lenght-2:]
if "00" in last_chars:
print("No")
exit()
if "000" in word:
print("No")
exit()
if "11" in word:
print("No")
... | 55 | 46 | 0 | 192774037 | n = int(input())
s = input()
if len(s)==1:
print('No') if s=='0' else print('Yes')
elif len(s)==2:
print('No') if s=='00' or s=='11' else print('Yes')
else:
if s.count('11')>0:
print('No')
else:
for i in range(n):
if s[i]=='0':
s = list(s)
s[... | Codeforces Round 484 (Div. 2) | CF | 2,018 | 1 | 256 | Row | You're given a row with $$$n$$$ chairs. We call a seating of people "maximal" if the two following conditions hold:
1. There are no neighbors adjacent to anyone seated.
2. It's impossible to seat one more person without violating the first rule.
The seating is given as a string consisting of zeros and ones ($$$0$$$ m... | The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 1000$$$) — the number of chairs.
The next line contains a string of $$$n$$$ characters, each of them is either zero or one, describing the seating. | Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".
You are allowed to print letters in whatever case you'd like (uppercase or lowercase). | null | In sample case one the given seating is maximal.
In sample case two the person at chair three has a neighbour to the right.
In sample case three it is possible to seat yet another person into chair three. | [{"input": "3\n101", "output": "Yes"}, {"input": "4\n1011", "output": "No"}, {"input": "5\n10001", "output": "No"}] | 1,200 | ["brute force", "constructive algorithms"] | 55 | [{"input": "3\r\n101\r\n", "output": "Yes\r\n"}, {"input": "4\r\n1011\r\n", "output": "No\r\n"}, {"input": "5\r\n10001\r\n", "output": "No\r\n"}, {"input": "1\r\n0\r\n", "output": "No\r\n"}, {"input": "1\r\n1\r\n", "output": "Yes\r\n"}, {"input": "100\r\n01010010101010010100100101010010101001010010010010100101010100101... | false | stdio | null | true |
985/B | 985 | B | PyPy 3-64 | TESTS | 26 | 311 | 93,593,600 | 224853496 | class Switch:
def __init__(self, s):
self.data = set()
for i, ch in enumerate(s):
if ch == '1':
self.data.add(i)
def __len__(self):
return len(self.data)
def __ior__(self, other):
self.data |= other.data
return self
def main():
n, m... | 67 | 109 | 921,600 | 38563718 | def main():
n, m = map(int, input().split())
switches = [int(input(), 2) for _ in range(n)]
lambs_on = 0
duplicate_control = 0
for lambs_turn_on in switches:
duplicate_control |= lambs_on & lambs_turn_on
lambs_on |= lambs_turn_on
if any(lambs_turn_on & lambs_on & ~duplicate_con... | Educational Codeforces Round 44 (Rated for Div. 2) | ICPC | 2,018 | 3 | 256 | Switches and Lamps | You are given n switches and m lamps. The i-th switch turns on some subset of the lamps. This information is given as the matrix a consisting of n rows and m columns where ai, j = 1 if the i-th switch turns on the j-th lamp and ai, j = 0 if the i-th switch is not connected to the j-th lamp.
Initially all m lamps are t... | The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2000) — the number of the switches and the number of the lamps.
The following n lines contain m characters each. The character ai, j is equal to '1' if the i-th switch turns on the j-th lamp and '0' otherwise.
It is guaranteed that if you press all... | Print "YES" if there is a switch that if you will ignore it and press all the other n - 1 switches then all m lamps will be turned on. Print "NO" if there is no such switch. | null | null | [{"input": "4 5\n10101\n01000\n00111\n10000", "output": "YES"}, {"input": "4 5\n10100\n01000\n00110\n00101", "output": "NO"}] | 1,200 | ["implementation"] | 67 | [{"input": "4 5\r\n10101\r\n01000\r\n00111\r\n10000\r\n", "output": "YES\r\n"}, {"input": "4 5\r\n10100\r\n01000\r\n00110\r\n00101\r\n", "output": "NO\r\n"}, {"input": "1 5\r\n11111\r\n", "output": "NO\r\n"}, {"input": "10 1\r\n1\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n1\r\n", "output": "YES\r\n"}, {"input": "1 1\r\... | false | stdio | null | true |
985/B | 985 | B | Python 3 | TESTS | 26 | 140 | 819,200 | 39063467 | n=list(map(int,input().split()))
l,r=n[0],n[1]
arr=[]
gf=0
for x in range(l):
arr.append(int(input(),2))
for x in range(r):
gf=gf+pow(2,x)
mx=max(arr)
k=0
for x in range(0,l):
if(bin(mx)!=bin(mx|arr[x])):
mx=(mx|arr[x])
k=k+1
k=k+1
print('YES') if(k<l and mx==gf ) else print('NO') | 67 | 109 | 5,222,400 | 38520749 | input()
a = tuple(map(lambda s: int(s.strip(), 2), __import__('sys').stdin.readlines()))
t = T = 0
for x in a:
t |= T & x
T |= x
print(['YES', 'NO'][all(x & (x ^ t) for x in a)]) | Educational Codeforces Round 44 (Rated for Div. 2) | ICPC | 2,018 | 3 | 256 | Switches and Lamps | You are given n switches and m lamps. The i-th switch turns on some subset of the lamps. This information is given as the matrix a consisting of n rows and m columns where ai, j = 1 if the i-th switch turns on the j-th lamp and ai, j = 0 if the i-th switch is not connected to the j-th lamp.
Initially all m lamps are t... | The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2000) — the number of the switches and the number of the lamps.
The following n lines contain m characters each. The character ai, j is equal to '1' if the i-th switch turns on the j-th lamp and '0' otherwise.
It is guaranteed that if you press all... | Print "YES" if there is a switch that if you will ignore it and press all the other n - 1 switches then all m lamps will be turned on. Print "NO" if there is no such switch. | null | null | [{"input": "4 5\n10101\n01000\n00111\n10000", "output": "YES"}, {"input": "4 5\n10100\n01000\n00110\n00101", "output": "NO"}] | 1,200 | ["implementation"] | 67 | [{"input": "4 5\r\n10101\r\n01000\r\n00111\r\n10000\r\n", "output": "YES\r\n"}, {"input": "4 5\r\n10100\r\n01000\r\n00110\r\n00101\r\n", "output": "NO\r\n"}, {"input": "1 5\r\n11111\r\n", "output": "NO\r\n"}, {"input": "10 1\r\n1\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n1\r\n", "output": "YES\r\n"}, {"input": "1 1\r\... | false | stdio | null | true |
985/B | 985 | B | Python 3 | TESTS | 26 | 2,605 | 74,956,800 | 38547910 | a=[int(i) for i in input().split()]
n=a[0]
m=a[1]
a=[]
for i in range(n):
s=str(input())
y=[]
for j in range(m):
if(s[j]=='1'):
y.append(j)
a.append(y)
a.sort(key=len,reverse=True)
b=[False]*(m)
i=0
count=0
while(i<=n-1):
flag=0
for j in range(len(a[i])):
if(b[a[i][j]]==False):
b[a[i][j]]=True
flag=1... | 67 | 124 | 716,800 | 38491775 | a = []
was_already = was_twice = 0
n = int(input().split()[0])
for _ in range(n):
ai = int(input(), base=2)
a.append(ai)
was_twice |= was_already & ai
was_already |= ai
for ai in a:
if ai & (ai ^ was_twice):
continue
print("YES")
break
else:
print("NO") | Educational Codeforces Round 44 (Rated for Div. 2) | ICPC | 2,018 | 3 | 256 | Switches and Lamps | You are given n switches and m lamps. The i-th switch turns on some subset of the lamps. This information is given as the matrix a consisting of n rows and m columns where ai, j = 1 if the i-th switch turns on the j-th lamp and ai, j = 0 if the i-th switch is not connected to the j-th lamp.
Initially all m lamps are t... | The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2000) — the number of the switches and the number of the lamps.
The following n lines contain m characters each. The character ai, j is equal to '1' if the i-th switch turns on the j-th lamp and '0' otherwise.
It is guaranteed that if you press all... | Print "YES" if there is a switch that if you will ignore it and press all the other n - 1 switches then all m lamps will be turned on. Print "NO" if there is no such switch. | null | null | [{"input": "4 5\n10101\n01000\n00111\n10000", "output": "YES"}, {"input": "4 5\n10100\n01000\n00110\n00101", "output": "NO"}] | 1,200 | ["implementation"] | 67 | [{"input": "4 5\r\n10101\r\n01000\r\n00111\r\n10000\r\n", "output": "YES\r\n"}, {"input": "4 5\r\n10100\r\n01000\r\n00110\r\n00101\r\n", "output": "NO\r\n"}, {"input": "1 5\r\n11111\r\n", "output": "NO\r\n"}, {"input": "10 1\r\n1\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n0\r\n1\r\n", "output": "YES\r\n"}, {"input": "1 1\r\... | false | stdio | null | true |
812/B | 812 | B | Python 3 | TESTS | 42 | 124 | 307,200 | 76990537 | n,m=list(map(int,input().split()))
Li=[]
for i in range(n):
Li.append(input())
on=[]
L=[]
R=[]
for s in Li[::-1]:
c=0
count=1
p=-1
for i in s:
if c==0 and i=='1':
c=1
L.append(count)
on.append(1)
if i=='1':
p=count
count+=1
... | 72 | 46 | 0 | 186754062 | n, m = map(int, input().split())
p = [input() for y in range(n)][::-1]
l = r = d = 0
i = j = 0
for y, t in enumerate(p):
if '1' in t:
l, r = min(l - i, r - j) + 2 * m + 2, min(l + i, r + j)
i, j = t.find('1'), t.rfind('1')
l, r, d = l - i, r + j, y
print(min(l, r) + d) | Codeforces Round 417 (Div. 2) | CF | 2,017 | 1 | 256 | Sagheer, the Hausmeister | Some people leave the lights at their workplaces on when they leave that is a waste of resources. As a hausmeister of DHBW, Sagheer waits till all students and professors leave the university building, then goes and turns all the lights off.
The building consists of n floors with stairs at the left and the right sides... | The first line contains two integers n and m (1 ≤ n ≤ 15 and 1 ≤ m ≤ 100) — the number of floors and the number of rooms in each floor, respectively.
The next n lines contains the building description. Each line contains a binary string of length m + 2 representing a floor (the left stairs, then m rooms, then the righ... | Print a single integer — the minimum total time needed to turn off all the lights. | null | In the first example, Sagheer will go to room 1 in the ground floor, then he will go to room 2 in the second floor using the left or right stairs.
In the second example, he will go to the fourth room in the ground floor, use right stairs, go to the fourth room in the second floor, use right stairs again, then go to th... | [{"input": "2 2\n0010\n0100", "output": "5"}, {"input": "3 4\n001000\n000010\n000010", "output": "12"}, {"input": "4 3\n01110\n01110\n01110\n01110", "output": "18"}] | 1,600 | ["bitmasks", "brute force", "dp"] | 72 | [{"input": "2 2\r\n0010\r\n0100\r\n", "output": "5\r\n"}, {"input": "3 4\r\n001000\r\n000010\r\n000010\r\n", "output": "12\r\n"}, {"input": "4 3\r\n01110\r\n01110\r\n01110\r\n01110\r\n", "output": "18\r\n"}, {"input": "3 2\r\n0000\r\n0100\r\n0100\r\n", "output": "4\r\n"}, {"input": "1 89\r\n0000000000000000000000000000... | false | stdio | null | true |
812/B | 812 | B | PyPy 3 | TESTS | 42 | 202 | 23,347,200 | 28807908 | n, m = list(map(int, input().split()))
house = []
def look_right(position, house):
x, y = position
l = len(house[y])
count = l - x
ps = 0
while ps == 0:
pos = [l-1, y - 1]
for i in range(l-1, -1, -1):
if house[y-1][i] == 1:
pos[0] = i
ps ... | 72 | 46 | 0 | 230134682 | n_and_m = input()
n, m = n_and_m.split() #n - кол-во целых чисел, k - кол-во элем. заданной посл-ти меньше либо равны x(результат программы)
n = int(n)
m = int(m)
p = [input() for y in range(n)][::-1]
l = r = d = 0
i = j = 0
for y, t in enumerate(p):
if '1' in t:
l, r = min(l - i, r - j) + 2 * m + 2, min(... | Codeforces Round 417 (Div. 2) | CF | 2,017 | 1 | 256 | Sagheer, the Hausmeister | Some people leave the lights at their workplaces on when they leave that is a waste of resources. As a hausmeister of DHBW, Sagheer waits till all students and professors leave the university building, then goes and turns all the lights off.
The building consists of n floors with stairs at the left and the right sides... | The first line contains two integers n and m (1 ≤ n ≤ 15 and 1 ≤ m ≤ 100) — the number of floors and the number of rooms in each floor, respectively.
The next n lines contains the building description. Each line contains a binary string of length m + 2 representing a floor (the left stairs, then m rooms, then the righ... | Print a single integer — the minimum total time needed to turn off all the lights. | null | In the first example, Sagheer will go to room 1 in the ground floor, then he will go to room 2 in the second floor using the left or right stairs.
In the second example, he will go to the fourth room in the ground floor, use right stairs, go to the fourth room in the second floor, use right stairs again, then go to th... | [{"input": "2 2\n0010\n0100", "output": "5"}, {"input": "3 4\n001000\n000010\n000010", "output": "12"}, {"input": "4 3\n01110\n01110\n01110\n01110", "output": "18"}] | 1,600 | ["bitmasks", "brute force", "dp"] | 72 | [{"input": "2 2\r\n0010\r\n0100\r\n", "output": "5\r\n"}, {"input": "3 4\r\n001000\r\n000010\r\n000010\r\n", "output": "12\r\n"}, {"input": "4 3\r\n01110\r\n01110\r\n01110\r\n01110\r\n", "output": "18\r\n"}, {"input": "3 2\r\n0000\r\n0100\r\n0100\r\n", "output": "4\r\n"}, {"input": "1 89\r\n0000000000000000000000000000... | false | stdio | null | true |
628/D | 628 | D | PyPy 3 | TESTS | 20 | 1,855 | 38,707,200 | 114079273 | # by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO,IOBase
def solve(x,m,d,mod,par):
dp = [[[0]*m for _ in range(2)] for _ in range(len(x))]
# remainder of number formed till now ; smaller/equal ; place value
zz = int(x[0])
for _ in range(1,zz):
if... | 66 | 217 | 3,379,200 | 189367480 | from sys import stdin
mod = 10**9 + 7
_all = map(str, stdin.read().split())
m, d, a, b = int(next(_all)), int(next(_all)), next(_all), next(_all)
n = len(a)
din, din1 = [0]*m, [0]*m
eqa, eqb, eqab = 1, 1, 1
xa, xb = 0, 0
for i in range(n):
eqa1, eqb1, eqab1 = 0, 0, 0
din1 = [0]*m
for dig in range(10):
if i % 2 =... | Educational Codeforces Round 8 | ICPC | 2,016 | 2 | 256 | Magic Numbers | Consider the decimal presentation of an integer. Let's call a number d-magic if digit d appears in decimal presentation of the number on even positions and nowhere else.
For example, the numbers 1727374, 17, 1 are 7-magic but 77, 7, 123, 34, 71 are not 7-magic. On the other hand the number 7 is 0-magic, 123 is 2-magic... | The first line contains two integers m, d (1 ≤ m ≤ 2000, 0 ≤ d ≤ 9) — the parameters from the problem statement.
The second line contains positive integer a in decimal presentation (without leading zeroes).
The third line contains positive integer b in decimal presentation (without leading zeroes).
It is guaranteed ... | Print the only integer a — the remainder after dividing by 109 + 7 of the number of d-magic numbers in segment [a, b] that are multiple of m. | null | The numbers from the answer of the first example are 16, 26, 36, 46, 56, 76, 86 and 96.
The numbers from the answer of the second example are 2, 4, 6 and 8.
The numbers from the answer of the third example are 1767, 2717, 5757, 6707, 8797 and 9747. | [{"input": "2 6\n10\n99", "output": "8"}, {"input": "2 0\n1\n9", "output": "4"}, {"input": "19 7\n1000\n9999", "output": "6"}] | 2,200 | ["dp"] | 66 | [{"input": "2 6\r\n10\r\n99\r\n", "output": "8\r\n"}, {"input": "2 0\r\n1\r\n9\r\n", "output": "4\r\n"}, {"input": "19 7\r\n1000\r\n9999\r\n", "output": "6\r\n"}, {"input": "9 4\r\n33\r\n52\r\n", "output": "0\r\n"}, {"input": "10 8\r\n18\r\n59\r\n", "output": "0\r\n"}, {"input": "43 3\r\n587\r\n850\r\n", "output": "1\r... | false | stdio | null | true |
812/B | 812 | B | Python 3 | TESTS | 42 | 264 | 307,200 | 27527822 | # 1000
n_m = [int(x) for x in input().split()]
n_m[1] += 1
floors = []
rooms = []
position = [0, 0]
def path_cost():
s = []
MIN=0
index = 0
y = rooms[0][1]
for i in range(len(rooms)):
if rooms[i][1]>y:
continue
if position[1]-rooms[i][1]==0 :
s.append(abs(r... | 72 | 62 | 0 | 194156841 | import sys
input = sys.stdin.readline
n, m = map(int, input().split())
a = 0
b = m+1
c = [0, 0]
g = [input()[:-1] for _ in range(n)]
for i in range(n-1, 0, -1):
if '1' in g[i]:
c = [g[i].rindex('1') if a + g[i].rindex('1')*2 < b+m+1 else g[i].index('1'), m+1-g[i].index('1') if b + (m+1-g[i].index('1'))*2 <... | Codeforces Round 417 (Div. 2) | CF | 2,017 | 1 | 256 | Sagheer, the Hausmeister | Some people leave the lights at their workplaces on when they leave that is a waste of resources. As a hausmeister of DHBW, Sagheer waits till all students and professors leave the university building, then goes and turns all the lights off.
The building consists of n floors with stairs at the left and the right sides... | The first line contains two integers n and m (1 ≤ n ≤ 15 and 1 ≤ m ≤ 100) — the number of floors and the number of rooms in each floor, respectively.
The next n lines contains the building description. Each line contains a binary string of length m + 2 representing a floor (the left stairs, then m rooms, then the righ... | Print a single integer — the minimum total time needed to turn off all the lights. | null | In the first example, Sagheer will go to room 1 in the ground floor, then he will go to room 2 in the second floor using the left or right stairs.
In the second example, he will go to the fourth room in the ground floor, use right stairs, go to the fourth room in the second floor, use right stairs again, then go to th... | [{"input": "2 2\n0010\n0100", "output": "5"}, {"input": "3 4\n001000\n000010\n000010", "output": "12"}, {"input": "4 3\n01110\n01110\n01110\n01110", "output": "18"}] | 1,600 | ["bitmasks", "brute force", "dp"] | 72 | [{"input": "2 2\r\n0010\r\n0100\r\n", "output": "5\r\n"}, {"input": "3 4\r\n001000\r\n000010\r\n000010\r\n", "output": "12\r\n"}, {"input": "4 3\r\n01110\r\n01110\r\n01110\r\n01110\r\n", "output": "18\r\n"}, {"input": "3 2\r\n0000\r\n0100\r\n0100\r\n", "output": "4\r\n"}, {"input": "1 89\r\n0000000000000000000000000000... | false | stdio | null | true |
812/B | 812 | B | Python 3 | TESTS | 42 | 46 | 0 | 231542755 | n_and_m = input()
n, m = n_and_m.split() #n - кол-во целых чисел, k - кол-во элем. заданной посл-ти меньше либо равны x(результат программы)
n = int(n)
m = int(m)
p = [input() for y in range(n)][::-1]
l = r = d = 0
i = j = 0
pos = 0
put = 0
for y, t in enumerate(p):
if '1' in t:
d = y
i = t.find(... | 72 | 62 | 307,200 | 27491979 | n, m = map(int, input().split())
a = []
for i in range(n):
a.append(input())
# print(a)
l = 0
r = INF = 100000
h = 0
while h < n:
if a[h] == '0' * (m + 2):
h += 1
else:
break
if h != n:
for i in range(n - 1, h, -1):
x = INF
y = 0
for j in range(m + 2):
... | Codeforces Round 417 (Div. 2) | CF | 2,017 | 1 | 256 | Sagheer, the Hausmeister | Some people leave the lights at their workplaces on when they leave that is a waste of resources. As a hausmeister of DHBW, Sagheer waits till all students and professors leave the university building, then goes and turns all the lights off.
The building consists of n floors with stairs at the left and the right sides... | The first line contains two integers n and m (1 ≤ n ≤ 15 and 1 ≤ m ≤ 100) — the number of floors and the number of rooms in each floor, respectively.
The next n lines contains the building description. Each line contains a binary string of length m + 2 representing a floor (the left stairs, then m rooms, then the righ... | Print a single integer — the minimum total time needed to turn off all the lights. | null | In the first example, Sagheer will go to room 1 in the ground floor, then he will go to room 2 in the second floor using the left or right stairs.
In the second example, he will go to the fourth room in the ground floor, use right stairs, go to the fourth room in the second floor, use right stairs again, then go to th... | [{"input": "2 2\n0010\n0100", "output": "5"}, {"input": "3 4\n001000\n000010\n000010", "output": "12"}, {"input": "4 3\n01110\n01110\n01110\n01110", "output": "18"}] | 1,600 | ["bitmasks", "brute force", "dp"] | 72 | [{"input": "2 2\r\n0010\r\n0100\r\n", "output": "5\r\n"}, {"input": "3 4\r\n001000\r\n000010\r\n000010\r\n", "output": "12\r\n"}, {"input": "4 3\r\n01110\r\n01110\r\n01110\r\n01110\r\n", "output": "18\r\n"}, {"input": "3 2\r\n0000\r\n0100\r\n0100\r\n", "output": "4\r\n"}, {"input": "1 89\r\n0000000000000000000000000000... | false | stdio | null | true |
812/B | 812 | B | Python 3 | TESTS | 42 | 46 | 102,400 | 230367730 | n, m = list(map(int, input().split()))
floor = []
count = 0
for i in range(n):
str = input()
floor.append(list(map(int, list(str))))
if 1 in floor[i]:
count += 1
i = n - 1
j = 0
time = 0
while count > 0:
if 1 in floor[i]:
if j == 0:
while 1 in floor[i]:
j += 1... | 72 | 62 | 307,200 | 27493791 | def booly(s):
return bool(int(s))
n, m = map(int, input().split());
nothing = "0"*(m+2);
a = [];
for i in range(n):
a.append(input())
for i in range(n):
if a[i] != nothing:
break;
a = a[i:]
a.reverse();
n = len(a)
#print(a)
leftA = [i.find("1") for i in a]
rightA = [i.rfind("1") for i in a]
... | Codeforces Round 417 (Div. 2) | CF | 2,017 | 1 | 256 | Sagheer, the Hausmeister | Some people leave the lights at their workplaces on when they leave that is a waste of resources. As a hausmeister of DHBW, Sagheer waits till all students and professors leave the university building, then goes and turns all the lights off.
The building consists of n floors with stairs at the left and the right sides... | The first line contains two integers n and m (1 ≤ n ≤ 15 and 1 ≤ m ≤ 100) — the number of floors and the number of rooms in each floor, respectively.
The next n lines contains the building description. Each line contains a binary string of length m + 2 representing a floor (the left stairs, then m rooms, then the righ... | Print a single integer — the minimum total time needed to turn off all the lights. | null | In the first example, Sagheer will go to room 1 in the ground floor, then he will go to room 2 in the second floor using the left or right stairs.
In the second example, he will go to the fourth room in the ground floor, use right stairs, go to the fourth room in the second floor, use right stairs again, then go to th... | [{"input": "2 2\n0010\n0100", "output": "5"}, {"input": "3 4\n001000\n000010\n000010", "output": "12"}, {"input": "4 3\n01110\n01110\n01110\n01110", "output": "18"}] | 1,600 | ["bitmasks", "brute force", "dp"] | 72 | [{"input": "2 2\r\n0010\r\n0100\r\n", "output": "5\r\n"}, {"input": "3 4\r\n001000\r\n000010\r\n000010\r\n", "output": "12\r\n"}, {"input": "4 3\r\n01110\r\n01110\r\n01110\r\n01110\r\n", "output": "18\r\n"}, {"input": "3 2\r\n0000\r\n0100\r\n0100\r\n", "output": "4\r\n"}, {"input": "1 89\r\n0000000000000000000000000000... | false | stdio | null | true |
812/B | 812 | B | Python 3 | TESTS | 42 | 62 | 0 | 27490195 | read = lambda: map(int, input().split())
n, m = read()
a = [input() for i in range(n)]
l = [-1] * n
r = [-1] * n
for i in range(n):
ind = n - i - 1
fst = lst = -1
for j in range(1, m + 1):
if a[ind][j] == '1':
if fst == -1:
fst = j
lst = j
l[i], r[i] = fst... | 72 | 62 | 307,200 | 27494695 | ans = 0
n, m = map(int, input().split())
arr = [""] + [input() for i in range(n)][::-1]
dp = [[-1, 10 ** 9, 0] for i in range(n + 1)]
z = 0
for i in range(1, 1 + n):
z += arr[i].count("1")
for i in range(1, n + 1):
l = arr[i].find("1")
r = arr[i].rfind("1")
z -= arr[i].count("1")
if l != -1:
... | Codeforces Round 417 (Div. 2) | CF | 2,017 | 1 | 256 | Sagheer, the Hausmeister | Some people leave the lights at their workplaces on when they leave that is a waste of resources. As a hausmeister of DHBW, Sagheer waits till all students and professors leave the university building, then goes and turns all the lights off.
The building consists of n floors with stairs at the left and the right sides... | The first line contains two integers n and m (1 ≤ n ≤ 15 and 1 ≤ m ≤ 100) — the number of floors and the number of rooms in each floor, respectively.
The next n lines contains the building description. Each line contains a binary string of length m + 2 representing a floor (the left stairs, then m rooms, then the righ... | Print a single integer — the minimum total time needed to turn off all the lights. | null | In the first example, Sagheer will go to room 1 in the ground floor, then he will go to room 2 in the second floor using the left or right stairs.
In the second example, he will go to the fourth room in the ground floor, use right stairs, go to the fourth room in the second floor, use right stairs again, then go to th... | [{"input": "2 2\n0010\n0100", "output": "5"}, {"input": "3 4\n001000\n000010\n000010", "output": "12"}, {"input": "4 3\n01110\n01110\n01110\n01110", "output": "18"}] | 1,600 | ["bitmasks", "brute force", "dp"] | 72 | [{"input": "2 2\r\n0010\r\n0100\r\n", "output": "5\r\n"}, {"input": "3 4\r\n001000\r\n000010\r\n000010\r\n", "output": "12\r\n"}, {"input": "4 3\r\n01110\r\n01110\r\n01110\r\n01110\r\n", "output": "18\r\n"}, {"input": "3 2\r\n0000\r\n0100\r\n0100\r\n", "output": "4\r\n"}, {"input": "1 89\r\n0000000000000000000000000000... | false | stdio | null | true |
926/H | 926 | H | Python 3 | TESTS | 22 | 265 | 26,419,200 | 226617490 | n, k = [int(i) for i in input().split()]
scores = [int(i) for i in input().split()]
colors = input().strip()
scores_r = [scores[i] for i in range(n) if colors[i] == 'R']
scores_o = [scores[i] for i in range(n) if colors[i] == 'O']
scores_w = [scores[i] for i in range(n) if colors[i] == 'W']
scores_r.sort(reverse ... | 98 | 265 | 19,251,200 | 197516180 | import sys
def solve(first_color, second_color, need):
if need < 2 or min(len(first_color), len(second_color)) < 1 or len(first_color) + len(second_color) < need:
return -1
total = -first_color[0] - second_color[0]
temp = []
for cur_iter in range(1, len(first_color)):
temp.append(first... | VK Cup 2018 - Wild-card Round 1 | ICPC | 2,018 | 1 | 256 | Endless Roses Most Beautiful | Arkady decided to buy roses for his girlfriend.
A flower shop has white, orange and red roses, and the total amount of them is n. Arkady thinks that red roses are not good together with white roses, so he won't buy a bouquet containing both red and white roses. Also, Arkady won't buy a bouquet where all roses have the... | The first line contains two integers n and k (1 ≤ k ≤ n ≤ 200 000) — the number of roses in the show and the number of roses Arkady wants to buy.
The second line contains a sequence of integers b1, b2, ..., bn (1 ≤ bi ≤ 10 000), where bi equals the beauty of the i-th rose.
The third line contains a string c of length... | Print the maximum possible total beauty of a bouquet of k roses that satisfies the constraints above. If it is not possible to make a single such bouquet, print -1. | null | In the first example Arkady wants to buy 3 roses. He can, for example, buy both red roses (their indices are 1 and 2, and their total beauty is 7) and the only orange rose (its index is 3, its beauty is 4). This way the total beauty of the bouquet is 11.
In the second example Arkady can not buy a bouquet because all r... | [{"input": "5 3\n4 3 4 1 6\nRROWW", "output": "11"}, {"input": "5 2\n10 20 14 20 11\nRRRRR", "output": "-1"}, {"input": "11 5\n5 6 3 2 3 4 7 5 4 5 6\nRWOORWORROW", "output": "28"}] | 2,200 | [] | 98 | [{"input": "5 3\r\n4 3 4 1 6\r\nRROWW\r\n", "output": "11\r\n"}, {"input": "5 2\r\n10 20 14 20 11\r\nRRRRR\r\n", "output": "-1\r\n"}, {"input": "11 5\r\n5 6 3 2 3 4 7 5 4 5 6\r\nRWOORWORROW\r\n", "output": "28\r\n"}, {"input": "15 10\r\n8560 6244 9607 5137 7187 3217 5527 9919 282 8748 3529 6110 5767 521 3393\r\nOWRWOOR... | false | stdio | null | true |
733/C | 733 | C | PyPy 3 | TESTS | 9 | 140 | 0 | 75972933 | n = int(input())
scum = list(map(int,input().split()))
k = int(input())
bag = list(map(int,input().split()))
hoot = []
www = 0
w = 0
getsome = []
for i in range(k):
c = 0
for j in range(w,n):
c+=scum[j]
if c == bag[i]:
hoot.append(j+1)
w = j+1
break
if... | 139 | 46 | 0 | 230899366 | f = lambda: list(map(int, input().split()))
def g(i, j, k):
z = y = i
for x in range(i, j):
if a[x] > a[y]: y = x
if y > i: z = y
else:
while a[y] == a[i]:
y += 1
if y == j: return []
y -= 1
l = [str(k + y - x) + ' L' for x in range(i, y)]
r = [st... | Codeforces Round 378 (Div. 2) | CF | 2,016 | 1 | 256 | Epidemic in Monstropolis | There was an epidemic in Monstropolis and all monsters became sick. To recover, all monsters lined up in queue for an appointment to the only doctor in the city.
Soon, monsters became hungry and began to eat each other.
One monster can eat other monster if its weight is strictly greater than the weight of the monster... | The first line contains single integer n (1 ≤ n ≤ 500) — the number of monsters in the initial queue.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106) — the initial weights of the monsters.
The third line contains single integer k (1 ≤ k ≤ n) — the number of monsters in the queue after the joke.
Th... | In case if no actions could lead to the final queue, print "NO" (without quotes) in the only line.
Otherwise print "YES" (without quotes) in the first line. In the next n - k lines print actions in the chronological order. In each line print x — the index number of the monster in the current queue which eats and, sepa... | null | In the first example, initially there were n = 6 monsters, their weights are [1, 2, 2, 2, 1, 2] (in order of queue from the first monster to the last monster). The final queue should be [5, 5]. The following sequence of eatings leads to the final queue:
- the second monster eats the monster to the left (i.e. the first... | [{"input": "6\n1 2 2 2 1 2\n2\n5 5", "output": "YES\n2 L\n1 R\n4 L\n3 L"}, {"input": "5\n1 2 3 4 5\n1\n15", "output": "YES\n5 L\n4 L\n3 L\n2 L"}, {"input": "5\n1 1 1 3 3\n3\n2 1 6", "output": "NO"}] | 1,800 | ["constructive algorithms", "dp", "greedy", "two pointers"] | 139 | [{"input": "6\r\n1 2 2 2 1 2\r\n2\r\n5 5\r\n", "output": "YES\r\n2 L\r\n1 R\r\n4 L\r\n3 L\r\n"}, {"input": "5\r\n1 2 3 4 5\r\n1\r\n15\r\n", "output": "YES\r\n5 L\r\n4 L\r\n3 L\r\n2 L\r\n"}, {"input": "5\r\n1 1 1 3 3\r\n3\r\n2 1 6\r\n", "output": "NO"}, {"input": "5\r\n1 1 1 1 2\r\n3\r\n1 1 4\r\n", "output": "YES\r\n5 L... | false | stdio | import sys
def read_file(path):
with open(path, 'r') as f:
return f.read().splitlines()
input_path, output_path, submission_path = sys.argv[1], sys.argv[2], sys.argv[3]
input_lines = read_file(input_path)
n = int(input_lines[0])
a = list(map(int, input_lines[1].split()))
k = int(input_lines[2])
b = list(... | true |
416/D | 416 | D | Python 3 | TESTS | 77 | 358 | 16,588,800 | 6610195 | import sys
import math
n = int(sys.stdin.readline())
if n <= 2:
print(1)
sys.exit(0)
a = [int(s) for s in sys.stdin.readline().split()]
st = -1 # index of first positive number in current subset of a
ed = -1 # index last positive number in current subset of a
# differation is (a[ed] - a[st])/(ed - st)
l... | 84 | 373 | 16,384,000 | 6954252 | n = int(input())
a = list(map(int, input().split()))
i = 0
ans = 0
while i < n:
ans += 1
i1 = i
while i1 < n and a[i1] == -1:
i1 += 1
if i1 == n:
break
i2 = i1 + 1
while i2 < n and a[i2] == -1:
i2 += 1
if i2 == n:
break
dist = i2 - i1
step = (a[i2] - a... | Codeforces Round 241 (Div. 2) | CF | 2,014 | 1 | 256 | Population Size | Polycarpus develops an interesting theory about the interrelation of arithmetic progressions with just everything in the world. His current idea is that the population of the capital of Berland changes over time like an arithmetic progression. Well, or like multiple arithmetic progressions.
Polycarpus believes that if... | The first line of the input contains integer n (1 ≤ n ≤ 2·105) — the number of elements in the sequence. The second line contains integer values a1, a2, ..., an separated by a space (1 ≤ ai ≤ 109 or ai = - 1). | Print the minimum number of arithmetic progressions that you need to write one after another to get sequence a. The positions marked as -1 in a can be represented by any positive integers. | null | null | [{"input": "9\n8 6 4 2 1 4 7 10 2", "output": "3"}, {"input": "9\n-1 6 -1 2 -1 4 7 -1 2", "output": "3"}, {"input": "5\n-1 -1 -1 -1 -1", "output": "1"}, {"input": "7\n-1 -1 4 5 1 2 3", "output": "2"}] | 2,400 | ["greedy", "implementation", "math"] | 84 | [{"input": "9\r\n8 6 4 2 1 4 7 10 2\r\n", "output": "3\r\n"}, {"input": "9\r\n-1 6 -1 2 -1 4 7 -1 2\r\n", "output": "3\r\n"}, {"input": "5\r\n-1 -1 -1 -1 -1\r\n", "output": "1\r\n"}, {"input": "7\r\n-1 -1 4 5 1 2 3\r\n", "output": "2\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}, {"input": "1\r\n65\r\n", "output": ... | false | stdio | null | true |
581/C | 581 | C | PyPy 3-64 | TESTS | 95 | 155 | 14,848,000 | 231484514 | from os import path
from sys import stdin, stdout
filename = "../templates/input.txt"
if path.exists(filename):
stdin = open(filename, 'r')
def input():
return stdin.readline().rstrip()
def print(*args, sep=' ', end='\n'):
stdout.write(sep.join(map(str, args)))
stdout.write(end)
def solution():
... | 98 | 109 | 13,414,400 | 229663137 | n, k = map(int, input().split())
a = list(map(int, input().split()))
s = 0
r = [(10 - x % 10) % 10 for x in a]
s += sum(x // 10 for x in a)
r.sort()
for i in range(n):
if r[i]:
if r[i] <= k:
k -= r[i]
s += 1
s += k // 10
s = min(s, n * 10)
print(s)# 1698223430.592529 | Codeforces Round 322 (Div. 2) | CF | 2,015 | 1 | 256 | Developing Skills | Petya loves computer games. Finally a game that he's been waiting for so long came out!
The main character of this game has n different skills, each of which is characterized by an integer ai from 0 to 100. The higher the number ai is, the higher is the i-th skill of the character. The total rating of the character is... | The first line of the input contains two positive integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ 107) — the number of skills of the character and the number of units of improvements at Petya's disposal.
The second line of the input contains a sequence of n integers ai (0 ≤ ai ≤ 100), where ai characterizes the level of the i-... | The first line of the output should contain a single non-negative integer — the maximum total rating of the character that Petya can get using k or less improvement units. | null | In the first test case the optimal strategy is as follows. Petya has to improve the first skill to 10 by spending 3 improvement units, and the second skill to 10, by spending one improvement unit. Thus, Petya spends all his improvement units and the total rating of the character becomes equal to lfloor frac{100}{10} r... | [{"input": "2 4\n7 9", "output": "2"}, {"input": "3 8\n17 15 19", "output": "5"}, {"input": "2 2\n99 100", "output": "20"}] | 1,400 | ["implementation", "math", "sortings"] | 98 | [{"input": "2 4\r\n7 9\r\n", "output": "2\r\n"}, {"input": "3 8\r\n17 15 19\r\n", "output": "5\r\n"}, {"input": "2 2\r\n99 100\r\n", "output": "20\r\n"}, {"input": "100 10000\r\n0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ... | false | stdio | null | true |
798/B | 798 | B | Python 3 | TESTS | 3 | 30 | 0 | 203867335 | def analiza(x):
arr=[]
mov=0
for i in x:
for j in x:
if i != j:
mov+=comprob(i,j)
arr.append(mov)
mov=0
return min(arr)
def comprob(x,y):
m=0
x=list(x)
y=list(y)
p=list(y)
while(True):
if(x==y):
return 0
... | 99 | 62 | 5,529,600 | 26544102 | import sys
n = int(input())
S = []
for _ in range(n):
s = input()
S.append(s+s)
k = len(S[0])//2
mn = 1e10
for i in range(k):
s = S[0][i:i+k]
sm = 0
for x in S:
try:
sm += x.index(s)
except ValueError:
print(-1)
sys.exit(0)
mn = min(mn, sm)
pri... | Codeforces Round 410 (Div. 2) | CF | 2,017 | 2 | 256 | Mike and strings | Mike has n strings s1, s2, ..., sn each consisting of lowercase English letters. In one move he can choose a string si, erase the first character and append it to the end of the string. For example, if he has the string "coolmike", in one move he can transform it into the string "oolmikec".
Now Mike asks himself: what... | The first line contains integer n (1 ≤ n ≤ 50) — the number of strings.
This is followed by n lines which contain a string each. The i-th line corresponding to string si. Lengths of strings are equal. Lengths of each string is positive and don't exceed 50. | Print the minimal number of moves Mike needs in order to make all the strings equal or print - 1 if there is no solution. | null | In the first sample testcase the optimal scenario is to perform operations in such a way as to transform all strings into "zwoxz". | [{"input": "4\nxzzwo\nzwoxz\nzzwox\nxzzwo", "output": "5"}, {"input": "2\nmolzv\nlzvmo", "output": "2"}, {"input": "3\nkc\nkc\nkc", "output": "0"}, {"input": "3\naa\naa\nab", "output": "-1"}] | 1,300 | ["brute force", "dp", "strings"] | 99 | [{"input": "4\r\nxzzwo\r\nzwoxz\r\nzzwox\r\nxzzwo\r\n", "output": "5\r\n"}, {"input": "2\r\nmolzv\r\nlzvmo\r\n", "output": "2\r\n"}, {"input": "3\r\nkc\r\nkc\r\nkc\r\n", "output": "0\r\n"}, {"input": "3\r\naa\r\naa\r\nab\r\n", "output": "-1\r\n"}, {"input": "3\r\nkwkb\r\nkbkw\r\nbkwk\r\n", "output": "3\r\n"}, {"input":... | false | stdio | null | true |
812/B | 812 | B | Python 3 | TESTS | 42 | 233 | 307,200 | 27529899 | # 1000
n_m = [int(x) for x in input().split()]
n_m[1] += 1
floors = []
rooms = []
position = [0, 0]
def judge (a,b,rooms):
#for a
temp=rooms[:]
temp_room=temp.pop(a)
temp_cost=do(temp,temp_room)
# for b
temp1 = rooms[:]
temp1_room = temp1.pop(b)
temp1_cost = do(temp1,temp1_room)
... | 72 | 62 | 512,000 | 27522047 | from collections import Counter
import sys
def left_time(floor):
light_counter = Counter(floor)
# print(light_counter)
# print(light_counter['1'])
if light_counter['1'] == 0:
return 0
else:
last_light_search = 1
while last_light_search <= len(floor):
# print(last... | Codeforces Round 417 (Div. 2) | CF | 2,017 | 1 | 256 | Sagheer, the Hausmeister | Some people leave the lights at their workplaces on when they leave that is a waste of resources. As a hausmeister of DHBW, Sagheer waits till all students and professors leave the university building, then goes and turns all the lights off.
The building consists of n floors with stairs at the left and the right sides... | The first line contains two integers n and m (1 ≤ n ≤ 15 and 1 ≤ m ≤ 100) — the number of floors and the number of rooms in each floor, respectively.
The next n lines contains the building description. Each line contains a binary string of length m + 2 representing a floor (the left stairs, then m rooms, then the righ... | Print a single integer — the minimum total time needed to turn off all the lights. | null | In the first example, Sagheer will go to room 1 in the ground floor, then he will go to room 2 in the second floor using the left or right stairs.
In the second example, he will go to the fourth room in the ground floor, use right stairs, go to the fourth room in the second floor, use right stairs again, then go to th... | [{"input": "2 2\n0010\n0100", "output": "5"}, {"input": "3 4\n001000\n000010\n000010", "output": "12"}, {"input": "4 3\n01110\n01110\n01110\n01110", "output": "18"}] | 1,600 | ["bitmasks", "brute force", "dp"] | 72 | [{"input": "2 2\r\n0010\r\n0100\r\n", "output": "5\r\n"}, {"input": "3 4\r\n001000\r\n000010\r\n000010\r\n", "output": "12\r\n"}, {"input": "4 3\r\n01110\r\n01110\r\n01110\r\n01110\r\n", "output": "18\r\n"}, {"input": "3 2\r\n0000\r\n0100\r\n0100\r\n", "output": "4\r\n"}, {"input": "1 89\r\n0000000000000000000000000000... | false | stdio | null | true |
812/B | 812 | B | Python 3 | TESTS | 42 | 234 | 307,200 | 27531582 | # 1000
n_m = [int(x) for x in input().split()]
n_m[1] += 1
floors = []
rooms = []
position = [0, 0]
def judge(a, b, rooms):
# for a
temp = rooms[:]
temp_room = temp.pop(a)
temp_cost = do(temp, temp_room)
# for b
temp1 = rooms[:]
temp1_room = temp1.pop(b)
temp1_cost = do(temp1, temp1_... | 72 | 62 | 1,536,000 | 209607846 | import sys
input = lambda: sys.stdin.readline().rstrip()
N,M = map(int, input().split())
S = []
for _ in range(N):
S.append(input())
S = S[::-1]
while S and not '1' in S[-1]:
S.pop()
if not S:
exit(print(0))
N = len(S)
INF=float("inf")
dp = [[INF,INF] for _ in range(N+1)]
dp[0][0] = -1
for i in rang... | Codeforces Round 417 (Div. 2) | CF | 2,017 | 1 | 256 | Sagheer, the Hausmeister | Some people leave the lights at their workplaces on when they leave that is a waste of resources. As a hausmeister of DHBW, Sagheer waits till all students and professors leave the university building, then goes and turns all the lights off.
The building consists of n floors with stairs at the left and the right sides... | The first line contains two integers n and m (1 ≤ n ≤ 15 and 1 ≤ m ≤ 100) — the number of floors and the number of rooms in each floor, respectively.
The next n lines contains the building description. Each line contains a binary string of length m + 2 representing a floor (the left stairs, then m rooms, then the righ... | Print a single integer — the minimum total time needed to turn off all the lights. | null | In the first example, Sagheer will go to room 1 in the ground floor, then he will go to room 2 in the second floor using the left or right stairs.
In the second example, he will go to the fourth room in the ground floor, use right stairs, go to the fourth room in the second floor, use right stairs again, then go to th... | [{"input": "2 2\n0010\n0100", "output": "5"}, {"input": "3 4\n001000\n000010\n000010", "output": "12"}, {"input": "4 3\n01110\n01110\n01110\n01110", "output": "18"}] | 1,600 | ["bitmasks", "brute force", "dp"] | 72 | [{"input": "2 2\r\n0010\r\n0100\r\n", "output": "5\r\n"}, {"input": "3 4\r\n001000\r\n000010\r\n000010\r\n", "output": "12\r\n"}, {"input": "4 3\r\n01110\r\n01110\r\n01110\r\n01110\r\n", "output": "18\r\n"}, {"input": "3 2\r\n0000\r\n0100\r\n0100\r\n", "output": "4\r\n"}, {"input": "1 89\r\n0000000000000000000000000000... | false | stdio | null | true |
812/B | 812 | B | Python 3 | TESTS | 42 | 62 | 307,200 | 27607513 | def left_traverse(a,m,n,l):
for i in range(0,n):
for j in range(1,m+1):
if a[i][j]==1:
l[i]=j
break
def right_traverse(a,m,n,r):
for i in range(0,n):
for j in range(m,0,-1):
if a[i][j]==1:
r[i]=j
break
def rk... | 72 | 62 | 1,638,400 | 27495577 | import sys
def main():
n,m = map(int,sys.stdin.readline().split())
m+=2
z = []
for i in range(n):
z.append(sys.stdin.readline().rstrip())
ans = 0
y = n-1
x = 0
q = [[x,y,ans]]
for i in range(n-1,-1,-1):
first =-1
last = -1
for j in range(m):
... | Codeforces Round 417 (Div. 2) | CF | 2,017 | 1 | 256 | Sagheer, the Hausmeister | Some people leave the lights at their workplaces on when they leave that is a waste of resources. As a hausmeister of DHBW, Sagheer waits till all students and professors leave the university building, then goes and turns all the lights off.
The building consists of n floors with stairs at the left and the right sides... | The first line contains two integers n and m (1 ≤ n ≤ 15 and 1 ≤ m ≤ 100) — the number of floors and the number of rooms in each floor, respectively.
The next n lines contains the building description. Each line contains a binary string of length m + 2 representing a floor (the left stairs, then m rooms, then the righ... | Print a single integer — the minimum total time needed to turn off all the lights. | null | In the first example, Sagheer will go to room 1 in the ground floor, then he will go to room 2 in the second floor using the left or right stairs.
In the second example, he will go to the fourth room in the ground floor, use right stairs, go to the fourth room in the second floor, use right stairs again, then go to th... | [{"input": "2 2\n0010\n0100", "output": "5"}, {"input": "3 4\n001000\n000010\n000010", "output": "12"}, {"input": "4 3\n01110\n01110\n01110\n01110", "output": "18"}] | 1,600 | ["bitmasks", "brute force", "dp"] | 72 | [{"input": "2 2\r\n0010\r\n0100\r\n", "output": "5\r\n"}, {"input": "3 4\r\n001000\r\n000010\r\n000010\r\n", "output": "12\r\n"}, {"input": "4 3\r\n01110\r\n01110\r\n01110\r\n01110\r\n", "output": "18\r\n"}, {"input": "3 2\r\n0000\r\n0100\r\n0100\r\n", "output": "4\r\n"}, {"input": "1 89\r\n0000000000000000000000000000... | false | stdio | null | true |
676/C | 676 | C | Python 3 | TESTS | 67 | 124 | 1,536,000 | 106708516 | #Created By Minsol Jeong
def I3():
n, k = [int(x) for x in input().split()]
s = input()
a='a'
b='b'
maxBeauty = 1
a_max = s.count(a)
b_max = s.count(b)
if a_max <= k or b_max <= k:
maxBeauty = n
elif a_max >= b_max:
foundB = [i for i, x in enumerate(s) if x == b]
... | 117 | 62 | 3,686,400 | 147689765 | def check(n,k,s,x):
cnt, ans, j = 0, 0, 0
for i in range(n):
if s[i] == x:
cnt += 1
if cnt > k:
while cnt > k:
if s[j] == x:
cnt -= 1
j += 1
ans = max(ans, i - j + 1)
return ans
def solve():
n, k = map(int, input().split())
s = input()
print(max(check(n, k, s, 'a'), check(n, k, s, 'b')... | Codeforces Round 354 (Div. 2) | CF | 2,016 | 1 | 256 | Vasya and String | High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than k characters of the original string. W... | The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only. | Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than k characters. | null | In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". | [{"input": "4 2\nabba", "output": "4"}, {"input": "8 1\naabaabaa", "output": "5"}] | 1,500 | ["binary search", "dp", "strings", "two pointers"] | 117 | [{"input": "4 2\r\nabba\r\n", "output": "4\r\n"}, {"input": "8 1\r\naabaabaa\r\n", "output": "5\r\n"}, {"input": "1 0\r\na\r\n", "output": "1\r\n"}, {"input": "1 1\r\nb\r\n", "output": "1\r\n"}, {"input": "1 0\r\nb\r\n", "output": "1\r\n"}, {"input": "1 1\r\na\r\n", "output": "1\r\n"}, {"input": "10 10\r\nbbbbbbbbbb\r\... | false | stdio | null | true |
618/C | 618 | C | PyPy 3 | TESTS | 14 | 998 | 33,382,400 | 28419123 | n = int(input())
pos = []
for i in range(0, n):
tmp = input()
tmp = tmp.split()
pos.append([int(tmp[0]), int(tmp[1])])
x1 = pos[0][0]
y1 = pos[0][1]
x2 = pos[1][0]
y2 = pos[1][1]
tar = 1
for i in range(2, n):
if (pos[i][0] - x1) * (y2 - y1) == (x2 - x1) * (pos[i][1] - y1)\
and abs(pos[i][0] - x1... | 98 | 343 | 28,979,200 | 175796040 | import sys
input = sys.stdin.buffer.readline
def gcd(a, b):
if a > b:
a, b = b, a
if b % a==0:
return a
return gcd(b % a, a)
def find_slope(p1, p2):
x1, y1 = p1
x2, y2 = p2
if x1==x2:
return (False, None, None)
num1 = (y2-y1)
den1 = (x2-x1)
if den1 < 0:
... | Wunder Fund Round 2016 (Div. 1 + Div. 2 combined) | CF | 2,016 | 2 | 256 | Constellation | Cat Noku has obtained a map of the night sky. On this map, he found a constellation with n stars numbered from 1 to n. For each i, the i-th star is located at coordinates (xi, yi). No two stars are located at the same position.
In the evening Noku is going to take a look at the night sky. He would like to find three d... | The first line of the input contains a single integer n (3 ≤ n ≤ 100 000).
Each of the next n lines contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109).
It is guaranteed that no two stars lie at the same point, and there does not exist a line such that all stars lie on that line. | Print three distinct integers on a single line — the indices of the three points that form a triangle that satisfies the conditions stated in the problem.
If there are multiple possible answers, you may print any of them. | null | In the first sample, we can print the three indices in any order.
In the second sample, we have the following picture.
Note that the triangle formed by starts 1, 4 and 3 doesn't satisfy the conditions stated in the problem, as point 5 is not strictly outside of this triangle (it lies on it's border). | [{"input": "3\n0 1\n1 0\n1 1", "output": "1 2 3"}, {"input": "5\n0 0\n0 2\n2 0\n2 2\n1 1", "output": "1 3 5"}] | 1,600 | ["geometry", "implementation"] | 98 | [{"input": "3\r\n0 1\r\n1 0\r\n1 1\r\n", "output": "1 2 3\r\n"}, {"input": "5\r\n0 0\r\n0 2\r\n2 0\r\n2 2\r\n1 1\r\n", "output": "1 3 5\r\n"}, {"input": "3\r\n819934317 939682125\r\n487662889 8614219\r\n-557136619 382982369\r\n", "output": "1 3 2\r\n"}, {"input": "10\r\n25280705 121178189\r\n219147240 -570920213\r\n-82... | false | stdio | import sys
def main(input_path, output_path, submission_path):
# Read input points
with open(input_path, 'r') as f:
n = int(f.readline().strip())
points = []
for _ in range(n):
x, y = map(int, f.readline().strip().split())
points.append((x, y))
# Read su... | true |
676/C | 676 | C | Python 3 | TESTS | 67 | 78 | 204,800 | 218432683 | n,k = map(int,input().split())
word = str(input())
len_a = word.count('a')
len_b = word.count('b')
if len_a >= len_b:
change = 'b'
else:
change = 'a'
left = 0
count = 0
max_length = 0
for i in range(n):
if word[i] == change:
count += 1
while count > k:
if word[left] == change:
... | 117 | 77 | 102,400 | 218459242 | def max_beauty(n, k, s):
maxBeauty = 0
countA = 0
countB = 0
left = 0
for right in range(n):
if s[right] == 'a':
countA += 1
else:
countB += 1
window_length = right - left + 1
if min(countA, countB) <= k:
maxBeauty = max(m... | Codeforces Round 354 (Div. 2) | CF | 2,016 | 1 | 256 | Vasya and String | High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than k characters of the original string. W... | The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only. | Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than k characters. | null | In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". | [{"input": "4 2\nabba", "output": "4"}, {"input": "8 1\naabaabaa", "output": "5"}] | 1,500 | ["binary search", "dp", "strings", "two pointers"] | 117 | [{"input": "4 2\r\nabba\r\n", "output": "4\r\n"}, {"input": "8 1\r\naabaabaa\r\n", "output": "5\r\n"}, {"input": "1 0\r\na\r\n", "output": "1\r\n"}, {"input": "1 1\r\nb\r\n", "output": "1\r\n"}, {"input": "1 0\r\nb\r\n", "output": "1\r\n"}, {"input": "1 1\r\na\r\n", "output": "1\r\n"}, {"input": "10 10\r\nbbbbbbbbbb\r\... | false | stdio | null | true |
799/D | 799 | D | PyPy 3-64 | TESTS | 26 | 280 | 20,684,800 | 204500174 | from math import *
a,b,w0,h0,n=map(int,input().split())
w,h=ceil(a/w0),ceil(b/h0)
x,y=ceil(b/h0),ceil(a/w0)
MAX=100000
MAX=min(MAX,max(w,h)+888)
if w<=1 and h<=1:
print(0)
else:
l=list(map(int,input().split()))
l.sort()
l.reverse()
l=l[:34]
dp=[0 for i in range(MAX+1)] #max width for height i
... | 103 | 826 | 2,150,400 | 27034917 | import sys
MAXV = 100010
d = [0] * MAXV
a, b, h, w, n = map(int,input().split())
arr = input().split()
for it in range(n):
arr[it] = int(arr[it])
# print(arr)
# print(a, b, h, w, n)
def solve(a, b, h, w, z, product, it):
# print(">", a, b, h, w, z, product, it)
k = 0
if a % h:
k = a // h... | Playrix Codescapes Cup (Codeforces Round 413, rated, Div. 1 + Div. 2) | CF | 2,017 | 1 | 256 | Field expansion | In one of the games Arkady is fond of the game process happens on a rectangular field. In the game process Arkady can buy extensions for his field, each extension enlarges one of the field sizes in a particular number of times. Formally, there are n extensions, the i-th of them multiplies the width or the length (by Ar... | The first line contains five integers a, b, h, w and n (1 ≤ a, b, h, w, n ≤ 100 000) — the sizes of the rectangle needed to be placed, the initial sizes of the field and the number of available extensions.
The second line contains n integers a1, a2, ..., an (2 ≤ ai ≤ 100 000), where ai equals the integer a side multip... | Print the minimum number of extensions needed to reach Arkady's goal. If it is not possible to place the rectangle on the field with all extensions, print -1. If the rectangle can be placed on the initial field, print 0. | null | In the first example it is enough to use any of the extensions available. For example, we can enlarge h in 5 times using the second extension. Then h becomes equal 10 and it is now possible to place the rectangle on the field. | [{"input": "3 3 2 4 4\n2 5 4 10", "output": "1"}, {"input": "3 3 3 3 5\n2 3 5 4 2", "output": "0"}, {"input": "5 5 1 2 3\n2 2 3", "output": "-1"}, {"input": "3 4 1 1 3\n2 3 2", "output": "3"}] | 2,100 | ["brute force", "dp", "meet-in-the-middle"] | 103 | [{"input": "3 3 2 4 4\r\n2 5 4 10\r\n", "output": "1\r\n"}, {"input": "3 3 3 3 5\r\n2 3 5 4 2\r\n", "output": "0\r\n"}, {"input": "5 5 1 2 3\r\n2 2 3\r\n", "output": "-1\r\n"}, {"input": "3 4 1 1 3\r\n2 3 2\r\n", "output": "3\r\n"}, {"input": "572 540 6 2 12\r\n2 3 2 2 2 3 3 3 2 2 2 2\r\n", "output": "-1\r\n"}, {"input... | false | stdio | null | true |
515/B | 515 | B | PyPy 3-64 | TESTS | 40 | 140 | 1,331,200 | 141251359 | def solver():
n,m = map(int,input().split())
boys = [int(x) for x in input().split()[1:]]
girls = [int(x) for x in input().split()[1:]]
happy_boys = [False for i in range(n)]
happy_girls = [False for i in range(m)]
for e in boys:happy_boys[e] = True
for e in girls:happy_girls[e] =True
fo... | 56 | 61 | 4,505,600 | 133161183 | def gcd(a,b):
while b != 0:
r = a % b
a = b
b = r
return a
def lcm(a,b):
return a * b // gcd(a,b)
n, m = map(int, input().split())
boys = [False] * n
girls = [False] * m
happy_boys = list(map(int, input().split()))[1:]
happy_girls = list(map(int, input().split()))[1:]
for i in happy_boys:
boys[... | Codeforces Round 292 (Div. 2) | CF | 2,015 | 2 | 256 | Drazil and His Happy Friends | Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.
There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites $$(i \bmod n)$$-th... | The first line contains two integer n and m (1 ≤ n, m ≤ 100).
The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys.
The third line conatins integer g (0 ≤ ... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | null | By $$i \bmod k$$ we define the remainder of integer division of i by k.
In first sample case:
- On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day.
- On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so no... | [{"input": "2 3\n0\n1 0", "output": "Yes"}, {"input": "2 4\n1 0\n1 2", "output": "No"}, {"input": "2 3\n1 0\n1 1", "output": "Yes"}] | 1,300 | ["brute force", "dsu", "meet-in-the-middle", "number theory"] | 56 | [{"input": "2 3\r\n0\r\n1 0\r\n", "output": "Yes\r\n"}, {"input": "2 4\r\n1 0\r\n1 2\r\n", "output": "No\r\n"}, {"input": "2 3\r\n1 0\r\n1 1\r\n", "output": "Yes\r\n"}, {"input": "16 88\r\n6 5 14 2 0 12 7\r\n30 21 64 35 79 74 39 63 44 81 73 0 27 33 69 12 86 46 20 25 55 52 7 58 23 5 60 32 41 50 82\r\n", "output": "Yes\r... | false | stdio | null | true |
515/B | 515 | B | PyPy 3 | TESTS | 40 | 140 | 21,504,000 | 85997746 | from sys import stdin
###############################################################
def iinput(): return int(stdin.readline())
def sinput(): return input()
def minput(): return map(int, stdin.readline().split())
def linput(): return list(map(int, stdin.readline().split()))
############################################... | 56 | 62 | 0 | 9892582 | import sys
def readHappiness(count, stdin_happy):
line = stdin_happy.rstrip()
numbers = [int(nbr) for nbr in line.split(" ")]
numbers = numbers[1:]
happy_list = []
for n in range(count):
if n in numbers:
happy_list.append(1)
else:
happy_list.append(0)
r... | Codeforces Round 292 (Div. 2) | CF | 2,015 | 2 | 256 | Drazil and His Happy Friends | Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.
There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites $$(i \bmod n)$$-th... | The first line contains two integer n and m (1 ≤ n, m ≤ 100).
The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys.
The third line conatins integer g (0 ≤ ... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | null | By $$i \bmod k$$ we define the remainder of integer division of i by k.
In first sample case:
- On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day.
- On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so no... | [{"input": "2 3\n0\n1 0", "output": "Yes"}, {"input": "2 4\n1 0\n1 2", "output": "No"}, {"input": "2 3\n1 0\n1 1", "output": "Yes"}] | 1,300 | ["brute force", "dsu", "meet-in-the-middle", "number theory"] | 56 | [{"input": "2 3\r\n0\r\n1 0\r\n", "output": "Yes\r\n"}, {"input": "2 4\r\n1 0\r\n1 2\r\n", "output": "No\r\n"}, {"input": "2 3\r\n1 0\r\n1 1\r\n", "output": "Yes\r\n"}, {"input": "16 88\r\n6 5 14 2 0 12 7\r\n30 21 64 35 79 74 39 63 44 81 73 0 27 33 69 12 86 46 20 25 55 52 7 58 23 5 60 32 41 50 82\r\n", "output": "Yes\r... | false | stdio | null | true |
515/B | 515 | B | PyPy 3 | TESTS | 40 | 170 | 1,228,800 | 75318030 | n, m = [int(i) for i in input().split(" ")]
b = [0]*n
g = [0]*m
x = [int(i) for i in input().split(" ")][1:]
y = [int(i) for i in input().split(" ")][1:]
#print(x, y)
for i in x:
b[i] = 1
for i in y:
g[i] = 1
for i in range(n*m):
h = max(b[i%n], g[i%m])
b[i%n] = h
g[i%m] = h
if sum(b)+sum(g) == n+m... | 56 | 62 | 0 | 9896096 | import math
inp = input()
mas = [int(s) for s in inp.strip().split(' ')]
b = mas[0]
g = mas[1]
mas = [int(s) for s in input().strip().split(' ')]
hb = {}
if mas[0] != 0:
for i in mas[1::]:
hb[i] = True
mas = [int(s) for s in input().strip().split(' ')]
hg = {}
if mas[0] != 0:
for i in mas[1::]:
... | Codeforces Round 292 (Div. 2) | CF | 2,015 | 2 | 256 | Drazil and His Happy Friends | Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.
There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites $$(i \bmod n)$$-th... | The first line contains two integer n and m (1 ≤ n, m ≤ 100).
The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys.
The third line conatins integer g (0 ≤ ... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | null | By $$i \bmod k$$ we define the remainder of integer division of i by k.
In first sample case:
- On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day.
- On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so no... | [{"input": "2 3\n0\n1 0", "output": "Yes"}, {"input": "2 4\n1 0\n1 2", "output": "No"}, {"input": "2 3\n1 0\n1 1", "output": "Yes"}] | 1,300 | ["brute force", "dsu", "meet-in-the-middle", "number theory"] | 56 | [{"input": "2 3\r\n0\r\n1 0\r\n", "output": "Yes\r\n"}, {"input": "2 4\r\n1 0\r\n1 2\r\n", "output": "No\r\n"}, {"input": "2 3\r\n1 0\r\n1 1\r\n", "output": "Yes\r\n"}, {"input": "16 88\r\n6 5 14 2 0 12 7\r\n30 21 64 35 79 74 39 63 44 81 73 0 27 33 69 12 86 46 20 25 55 52 7 58 23 5 60 32 41 50 82\r\n", "output": "Yes\r... | false | stdio | null | true |
540/B | 540 | B | PyPy 3 | TESTS | 11 | 93 | 1,433,600 | 104390574 | from sys import stdin
n, k, p, x, y = map(int, stdin.buffer.readline().split())
a = list(map(int, stdin.buffer.readline().split()))
a.sort()
Sum = 0
id = 0
flag = False
for i in range(len(a)):
Sum += a[i]
if a[i] >= y and not flag:
flag = True
id = i
if Sum + n - k > x or id > (n // 2):
... | 78 | 62 | 4,710,400 | 10946495 | n, k, p, x, y = (int(x) for x in input().split())
alreadyGot = [int(x) for x in input().split()]
countToWrite = n - k
maxSumToWrite = x - sum(alreadyGot)
if maxSumToWrite < 0:
print(-1)
else:
newMarks = [1 for i in range(n-k)]
maxSumToWrite -= (n - k)
if maxSumToWrite < 0:
print(-1)
else:
newMarksIt = 0
whil... | Codeforces Round 301 (Div. 2) | CF | 2,015 | 2 | 256 | School Marks | Little Vova studies programming in an elite school. Vova and his classmates are supposed to write n progress tests, for each test they will get a mark from 1 to p. Vova is very smart and he can write every test for any mark, but he doesn't want to stand out from the crowd too much. If the sum of his marks for all tests... | The first line contains 5 space-separated integers: n, k, p, x and y (1 ≤ n ≤ 999, n is odd, 0 ≤ k < n, 1 ≤ p ≤ 1000, n ≤ x ≤ n·p, 1 ≤ y ≤ p). Here n is the number of tests that Vova is planned to write, k is the number of tests he has already written, p is the maximum possible mark for a test, x is the maximum total n... | If Vova cannot achieve the desired result, print "-1".
Otherwise, print n - k space-separated integers — the marks that Vova should get for the remaining tests. If there are multiple possible solutions, print any of them. | null | The median of sequence a1, ..., an where n is odd (in this problem n is always odd) is the element staying on (n + 1) / 2 position in the sorted list of ai.
In the first sample the sum of marks equals 3 + 5 + 4 + 4 + 1 = 17, what doesn't exceed 18, that means that Vova won't be disturbed by his classmates. And the med... | [{"input": "5 3 5 18 4\n3 5 4", "output": "4 1"}, {"input": "5 3 5 16 4\n5 5 5", "output": "-1"}] | 1,700 | ["greedy", "implementation"] | 78 | [{"input": "5 3 5 18 4\r\n3 5 4\r\n", "output": "4 1\r\n"}, {"input": "5 3 5 16 4\r\n5 5 5\r\n", "output": "-1\r\n"}, {"input": "5 3 5 17 4\r\n5 5 5\r\n", "output": "1 1\r\n"}, {"input": "5 3 5 12 1\r\n5 5 1\r\n", "output": "-1\r\n"}, {"input": "5 3 5 13 1\r\n5 5 1\r\n", "output": "1 1\r\n"}, {"input": "7 4 5 26 5\r\n5... | false | stdio | import sys
def main(input_path, output_path, submission_path):
with open(input_path, 'r') as f_in:
n, k, p, x, y = map(int, f_in.readline().split())
existing = list(map(int, f_in.readline().split()))
with open(submission_path, 'r') as f_sub:
submission_line = f_sub.read().strip()
... | true |
45/I | 45 | I | Python 3 | TESTS | 35 | 92 | 0 | 185265675 | # LUOGU_RID: 97366388
input()
x=[int(i) for i in input().split()]
t=max(x)
if t<1:
print(" ".join([str(i)for i in x]))
elif t<2:
print(" ".join([str(t)]*x.count(t)))
else:
print(" ".join([str(i)for i in x if i>1])) | 65 | 248 | 0 | 67274149 | n=int(input())
lst=[*map(int,input().split())]
a,b=[],[]
for i,x in enumerate(lst):
if x>0:a.append(x)
if x<0:b.append(x)
b.sort()
if len(b)%2==0:a.extend(b)
else:a.extend(b[:-1])
if a==[]:print(max(lst))
else:print(*a) | School Team Contest 3 (Winter Computer School 2010/11) | ICPC | 2,010 | 2 | 256 | TCMCF+++ | Vasya has gotten interested in programming contests in TCMCF+++ rules. On the contest n problems were suggested and every problem had a cost — a certain integral number of points (perhaps, negative or even equal to zero). According to TCMCF+++ rules, only accepted problems can earn points and the overall number of poin... | The first line contains an integer n (1 ≤ n ≤ 100) — the number of the suggested problems. The next line contains n space-separated integers ci ( - 100 ≤ ci ≤ 100) — the cost of the i-th task. The tasks' costs may coinсide. | Print space-separated the costs of the problems that needed to be solved to get the maximal possible number of points. Do not forget, please, that it was necessary to solve at least one problem. If there are several solutions to that problem, print any of them. | null | null | [{"input": "5\n1 2 -3 3 3", "output": "3 1 2 3"}, {"input": "13\n100 100 100 100 100 100 100 100 100 100 100 100 100", "output": "100 100 100 100 100 100 100 100 100 100 100 100 100"}, {"input": "4\n-2 -2 -2 -2", "output": "-2 -2 -2 -2"}] | 1,400 | ["greedy"] | 65 | [{"input": "5\r\n1 2 -3 3 3\r\n", "output": "3 1 2 3 \r\n"}, {"input": "13\r\n100 100 100 100 100 100 100 100 100 100 100 100 100\r\n", "output": "100 100 100 100 100 100 100 100 100 100 100 100 100 \r\n"}, {"input": "4\r\n-2 -2 -2 -2\r\n", "output": "-2 -2 -2 -2 \r\n"}, {"input": "1\r\n1\r\n", "output": "1 \r\n"}, {"i... | false | stdio | import sys
from collections import Counter
def compute_max_product(ci):
if all(x == 0 for x in ci):
return 0
positives = [x for x in ci if x > 0]
negatives = [x for x in ci if x < 0]
zeros = [x for x in ci if x == 0]
non_zero = positives + negatives
if not non_zero:
return 0
... | true |
793/A | 793 | A | Python 3 | TESTS | 21 | 124 | 8,806,400 | 27098073 | x=input().split(" ")
N=int(x[0])
K=int(x[1])
mi=10**10
res=0
y=list(map(lambda a:int(a),input().split(" ")))
for i in range (len(y)):
if y[i]<mi:
mi=y[i]
temp=y[0]
y[0]=y[i]
y[i]=temp
for elem in y:
res+=(elem-mi)/K
print(int(res) if int(res)==res else -1) | 88 | 93 | 13,414,400 | 207367507 | n, k = map(int, input().split())
a = list(map(int, input().split()))
check = True
for i in range(1, n):
check = check and (a[i] % k == a[0] % k)
if not check:
print(-1)
else:
m = min(a)
ans = 0
for i in range(n):
ans += a[i] - m
print(ans // k) | Tinkoff Challenge - Elimination Round | CF | 2,017 | 1 | 256 | Oleg and shares | Oleg the bank client checks share prices every day. There are n share prices he is interested in. Today he observed that each second exactly one of these prices decreases by k rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg... | The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109) — the number of share prices, and the amount of rubles some price decreases each second.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the initial prices. | Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible. | null | Consider the first example.
Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds.
There... | [{"input": "3 3\n12 9 15", "output": "3"}, {"input": "2 2\n10 9", "output": "-1"}, {"input": "4 1\n1 1000000000 1000000000 1000000000", "output": "2999999997"}] | 900 | ["implementation", "math"] | 88 | [{"input": "3 3\r\n12 9 15\r\n", "output": "3"}, {"input": "2 2\r\n10 9\r\n", "output": "-1"}, {"input": "4 1\r\n1 1000000000 1000000000 1000000000\r\n", "output": "2999999997"}, {"input": "1 11\r\n123\r\n", "output": "0"}, {"input": "20 6\r\n38 86 86 50 98 62 32 2 14 62 98 50 2 50 32 38 62 62 8 14\r\n", "output": "151... | false | stdio | null | true |
793/A | 793 | A | Python 3 | TESTS | 22 | 77 | 8,704,000 | 192698191 | a,b=map(int,input().split())
c=[int(i) for i in input().split()]
v=sum(c)-a*min(c)
if a==4 and b==2:
print(-1)
elif v%b==0:
print(v//b)
else:
print(-1) | 88 | 93 | 13,414,400 | 222021703 | n, k = map(int, input().split())
arr = list(map(int, input().split()))
flag = True
for i in range(1, n):
flag = flag and (arr[0] % k == arr[i] % k)
if not flag:
print(-1)
else:
min_ = min(arr)
tot = 0
for ele in arr:
tot += ele - min_
print(tot // k) | Tinkoff Challenge - Elimination Round | CF | 2,017 | 1 | 256 | Oleg and shares | Oleg the bank client checks share prices every day. There are n share prices he is interested in. Today he observed that each second exactly one of these prices decreases by k rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg... | The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109) — the number of share prices, and the amount of rubles some price decreases each second.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the initial prices. | Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible. | null | Consider the first example.
Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds.
There... | [{"input": "3 3\n12 9 15", "output": "3"}, {"input": "2 2\n10 9", "output": "-1"}, {"input": "4 1\n1 1000000000 1000000000 1000000000", "output": "2999999997"}] | 900 | ["implementation", "math"] | 88 | [{"input": "3 3\r\n12 9 15\r\n", "output": "3"}, {"input": "2 2\r\n10 9\r\n", "output": "-1"}, {"input": "4 1\r\n1 1000000000 1000000000 1000000000\r\n", "output": "2999999997"}, {"input": "1 11\r\n123\r\n", "output": "0"}, {"input": "20 6\r\n38 86 86 50 98 62 32 2 14 62 98 50 2 50 32 38 62 62 8 14\r\n", "output": "151... | false | stdio | null | true |
900/A | 900 | A | Python 3 | TESTS | 21 | 187 | 0 | 182155044 | total_m = 0
total_p = 0
for i in range(int(input())):
x, y = input().split()
if int(x) < 0:
total_m += 1
else:
total_p +=1
if total_m > 0 and total_p == 0:
print('Yes')
elif total_m == 0 and total_p > 0:
print('Yes')
elif total_m > 1 and total_p == 1:
print('Yes')
elif total_p > ... | 71 | 124 | 22,323,200 | 213776904 | lst = [w.rstrip() for w in open(0).readlines()]
n = int(lst[0])
xs = [int(line.split()[0]) for line in lst[1:]]
pos = sum(1 for x in xs if x > 0)
neg = n - pos
print(["No","Yes"][min(pos, neg) < 2]) | Codeforces Round 450 (Div. 2) | CF | 2,017 | 1 | 256 | Find Extra One | You have n distinct points on a plane, none of them lie on OY axis. Check that there is a point after removal of which the remaining points are located on one side of the OY axis. | The first line contains a single positive integer n (2 ≤ n ≤ 105).
The following n lines contain coordinates of the points. The i-th of these lines contains two single integers xi and yi (|xi|, |yi| ≤ 109, xi ≠ 0). No two points coincide. | Print "Yes" if there is such a point, "No" — otherwise.
You can print every letter in any case (upper or lower). | null | In the first example the second point can be removed.
In the second example there is no suitable for the condition point.
In the third example any point can be removed. | [{"input": "3\n1 1\n-1 -1\n2 -1", "output": "Yes"}, {"input": "4\n1 1\n2 2\n-1 1\n-2 2", "output": "No"}, {"input": "3\n1 2\n2 1\n4 60", "output": "Yes"}] | 800 | ["geometry", "implementation"] | 71 | [{"input": "3\r\n1 1\r\n-1 -1\r\n2 -1\r\n", "output": "Yes"}, {"input": "4\r\n1 1\r\n2 2\r\n-1 1\r\n-2 2\r\n", "output": "No"}, {"input": "3\r\n1 2\r\n2 1\r\n4 60\r\n", "output": "Yes"}, {"input": "10\r\n1 1\r\n2 2\r\n3 3\r\n4 4\r\n5 5\r\n6 6\r\n7 7\r\n8 8\r\n9 9\r\n-1 -1\r\n", "output": "Yes"}, {"input": "2\r\n1000000... | false | stdio | null | true |
900/A | 900 | A | Python 3 | TESTS | 21 | 343 | 8,089,600 | 80979159 | def validarPuntos(puntos):
contadorR, contadorL = 0, 0
#calcular cantidad por cada eje
for i in range(len(puntos)):
if puntos[i][0] > 0:
#a la derecha
contadorR += 1
elif puntos[i][0] < 0:
#izquierda
contadorL += 1
#si hay co... | 71 | 139 | 0 | 153604583 | n = int(input())
c = 0
for _ in range(n):
s = input()
if s[0] == '-':
c += 1
print("NO" if c>1 and (n-c)>1 else "YES") | Codeforces Round 450 (Div. 2) | CF | 2,017 | 1 | 256 | Find Extra One | You have n distinct points on a plane, none of them lie on OY axis. Check that there is a point after removal of which the remaining points are located on one side of the OY axis. | The first line contains a single positive integer n (2 ≤ n ≤ 105).
The following n lines contain coordinates of the points. The i-th of these lines contains two single integers xi and yi (|xi|, |yi| ≤ 109, xi ≠ 0). No two points coincide. | Print "Yes" if there is such a point, "No" — otherwise.
You can print every letter in any case (upper or lower). | null | In the first example the second point can be removed.
In the second example there is no suitable for the condition point.
In the third example any point can be removed. | [{"input": "3\n1 1\n-1 -1\n2 -1", "output": "Yes"}, {"input": "4\n1 1\n2 2\n-1 1\n-2 2", "output": "No"}, {"input": "3\n1 2\n2 1\n4 60", "output": "Yes"}] | 800 | ["geometry", "implementation"] | 71 | [{"input": "3\r\n1 1\r\n-1 -1\r\n2 -1\r\n", "output": "Yes"}, {"input": "4\r\n1 1\r\n2 2\r\n-1 1\r\n-2 2\r\n", "output": "No"}, {"input": "3\r\n1 2\r\n2 1\r\n4 60\r\n", "output": "Yes"}, {"input": "10\r\n1 1\r\n2 2\r\n3 3\r\n4 4\r\n5 5\r\n6 6\r\n7 7\r\n8 8\r\n9 9\r\n-1 -1\r\n", "output": "Yes"}, {"input": "2\r\n1000000... | false | stdio | null | true |
900/A | 900 | A | Python 3 | TESTS | 21 | 311 | 0 | 80624688 | n = int(input())
pos = 0
neg = 0
for _ in range(n):
x, y = map(int,input().split())
if x > 0:
pos+=1
else:
neg+=1
if (pos <= 1 and neg > 1) or (neg <=1 and pos > 1):
print('yes')
else:
print('no') | 71 | 140 | 0 | 138328075 | n = int(input())
c = 0
for i in range(n):
s = input()
if(s[0]== '-'):
c += 1
print("Yes" if c <= 1 or n - c <= 1 else "NO") | Codeforces Round 450 (Div. 2) | CF | 2,017 | 1 | 256 | Find Extra One | You have n distinct points on a plane, none of them lie on OY axis. Check that there is a point after removal of which the remaining points are located on one side of the OY axis. | The first line contains a single positive integer n (2 ≤ n ≤ 105).
The following n lines contain coordinates of the points. The i-th of these lines contains two single integers xi and yi (|xi|, |yi| ≤ 109, xi ≠ 0). No two points coincide. | Print "Yes" if there is such a point, "No" — otherwise.
You can print every letter in any case (upper or lower). | null | In the first example the second point can be removed.
In the second example there is no suitable for the condition point.
In the third example any point can be removed. | [{"input": "3\n1 1\n-1 -1\n2 -1", "output": "Yes"}, {"input": "4\n1 1\n2 2\n-1 1\n-2 2", "output": "No"}, {"input": "3\n1 2\n2 1\n4 60", "output": "Yes"}] | 800 | ["geometry", "implementation"] | 71 | [{"input": "3\r\n1 1\r\n-1 -1\r\n2 -1\r\n", "output": "Yes"}, {"input": "4\r\n1 1\r\n2 2\r\n-1 1\r\n-2 2\r\n", "output": "No"}, {"input": "3\r\n1 2\r\n2 1\r\n4 60\r\n", "output": "Yes"}, {"input": "10\r\n1 1\r\n2 2\r\n3 3\r\n4 4\r\n5 5\r\n6 6\r\n7 7\r\n8 8\r\n9 9\r\n-1 -1\r\n", "output": "Yes"}, {"input": "2\r\n1000000... | false | stdio | null | true |
900/A | 900 | A | Python 3 | TESTS | 21 | 311 | 0 | 89868083 | n = 0
p = 0
for i in range(int(input())):
x,y = list(map(int,input().split()))
if x < 0:
n += 1
else:
p += 1
if p>n and n in range(0,2):
print('Yes')
elif n>p and p in range(0,2):
print('Yes')
else:
print('No') | 71 | 140 | 0 | 216940003 | n = int(input())
s = 0
for i in range(n):
if int(input().split(' ')[0]) > 0:
s += 1
if not s or s == n or s == 1 or n - s == 1:
print('Yes')
else:
print('No') | Codeforces Round 450 (Div. 2) | CF | 2,017 | 1 | 256 | Find Extra One | You have n distinct points on a plane, none of them lie on OY axis. Check that there is a point after removal of which the remaining points are located on one side of the OY axis. | The first line contains a single positive integer n (2 ≤ n ≤ 105).
The following n lines contain coordinates of the points. The i-th of these lines contains two single integers xi and yi (|xi|, |yi| ≤ 109, xi ≠ 0). No two points coincide. | Print "Yes" if there is such a point, "No" — otherwise.
You can print every letter in any case (upper or lower). | null | In the first example the second point can be removed.
In the second example there is no suitable for the condition point.
In the third example any point can be removed. | [{"input": "3\n1 1\n-1 -1\n2 -1", "output": "Yes"}, {"input": "4\n1 1\n2 2\n-1 1\n-2 2", "output": "No"}, {"input": "3\n1 2\n2 1\n4 60", "output": "Yes"}] | 800 | ["geometry", "implementation"] | 71 | [{"input": "3\r\n1 1\r\n-1 -1\r\n2 -1\r\n", "output": "Yes"}, {"input": "4\r\n1 1\r\n2 2\r\n-1 1\r\n-2 2\r\n", "output": "No"}, {"input": "3\r\n1 2\r\n2 1\r\n4 60\r\n", "output": "Yes"}, {"input": "10\r\n1 1\r\n2 2\r\n3 3\r\n4 4\r\n5 5\r\n6 6\r\n7 7\r\n8 8\r\n9 9\r\n-1 -1\r\n", "output": "Yes"}, {"input": "2\r\n1000000... | false | stdio | null | true |
734/C | 734 | C | PyPy 3-64 | TESTS | 0 | 46 | 0 | 218779819 | n, m, k = map(int, input().split())
x, s = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
d = list(map(int, input().split()))
work1 = False
work2 = False
type1 = [[a[t], b[t]] for t in range(m)]
type2 = [[c[t], d[t]] for t in range... | 119 | 950 | 44,441,600 | 22239064 | import bisect
n, m, k = map(int, input().split())
x, s = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = list(map(int, input().split()))
d = list(map(int, input().split()))
z1 = list(zip(a, b))
z2 = list(zip(c, d))
result = n * x
for ai, bi in z1:
if bi > s:... | Codeforces Round 379 (Div. 2) | CF | 2,016 | 4 | 256 | Anton and Making Potions | Anton is playing a very interesting computer game, but now he is stuck at one of the levels. To pass to the next level he has to prepare n potions.
Anton has a special kettle, that can prepare one potions in x seconds. Also, he knows spells of two types that can faster the process of preparing potions.
1. Spells of t... | The first line of the input contains three integers n, m, k (1 ≤ n ≤ 2·109, 1 ≤ m, k ≤ 2·105) — the number of potions, Anton has to make, the number of spells of the first type and the number of spells of the second type.
The second line of the input contains two integers x and s (2 ≤ x ≤ 2·109, 1 ≤ s ≤ 2·109) — the i... | Print one integer — the minimum time one has to spent in order to prepare n potions. | null | In the first sample, the optimum answer is to use the second spell of the first type that costs 10 manapoints. Thus, the preparation time of each potion changes to 4 seconds. Also, Anton should use the second spell of the second type to instantly prepare 15 potions spending 80 manapoints. The total number of manapoints... | [{"input": "20 3 2\n10 99\n2 4 3\n20 10 40\n4 15\n10 80", "output": "20"}, {"input": "20 3 2\n10 99\n2 4 3\n200 100 400\n4 15\n100 800", "output": "200"}] | 1,600 | ["binary search", "dp", "greedy", "two pointers"] | 119 | [{"input": "20 3 2\r\n10 99\r\n2 4 3\r\n20 10 40\r\n4 15\r\n10 80\r\n", "output": "20\r\n"}, {"input": "20 3 2\r\n10 99\r\n2 4 3\r\n200 100 400\r\n4 15\r\n100 800\r\n", "output": "200\r\n"}, {"input": "10 3 3\r\n10 33\r\n1 7 6\r\n17 25 68\r\n2 9 10\r\n78 89 125\r\n", "output": "10\r\n"}, {"input": "94 1 1\r\n26 324\r\n... | false | stdio | null | true |
982/A | 982 | A | Python 3 | TESTS | 26 | 93 | 0 | 39745914 | n=int(input())
s=input()
if n==1:
if s=='1':
print('Yes')
else:
print('No')
elif n==2:
if s=='11':
print('No')
else:
print('Yes')
elif n==3:
if s=='001' or s=='100'or s=='111':
print('No')
else:
print('Yes')
elif s.startswith('001') or s.endswith('100'):
print('No')
else:
x,y=s.find('11'),s.find('0... | 55 | 46 | 0 | 198505201 | input()
s='0'+input()+'0'
print('No' if '000' in s or '11' in s else 'Yes') | Codeforces Round 484 (Div. 2) | CF | 2,018 | 1 | 256 | Row | You're given a row with $$$n$$$ chairs. We call a seating of people "maximal" if the two following conditions hold:
1. There are no neighbors adjacent to anyone seated.
2. It's impossible to seat one more person without violating the first rule.
The seating is given as a string consisting of zeros and ones ($$$0$$$ m... | The first line contains a single integer $$$n$$$ ($$$1 \leq n \leq 1000$$$) — the number of chairs.
The next line contains a string of $$$n$$$ characters, each of them is either zero or one, describing the seating. | Output "Yes" (without quotation marks) if the seating is "maximal". Otherwise print "No".
You are allowed to print letters in whatever case you'd like (uppercase or lowercase). | null | In sample case one the given seating is maximal.
In sample case two the person at chair three has a neighbour to the right.
In sample case three it is possible to seat yet another person into chair three. | [{"input": "3\n101", "output": "Yes"}, {"input": "4\n1011", "output": "No"}, {"input": "5\n10001", "output": "No"}] | 1,200 | ["brute force", "constructive algorithms"] | 55 | [{"input": "3\r\n101\r\n", "output": "Yes\r\n"}, {"input": "4\r\n1011\r\n", "output": "No\r\n"}, {"input": "5\r\n10001\r\n", "output": "No\r\n"}, {"input": "1\r\n0\r\n", "output": "No\r\n"}, {"input": "1\r\n1\r\n", "output": "Yes\r\n"}, {"input": "100\r\n01010010101010010100100101010010101001010010010010100101010100101... | false | stdio | null | true |
900/A | 900 | A | Python 3 | TESTS | 21 | 374 | 0 | 40556543 | n=int(input())
pos=0
neg=0
for k in range(n):
a,b=map(int,input().split())
if(a<0):
neg+=1
else:
pos+=1
if(pos>1):
if(neg==0 or neg==1):
print("Yes")
else:
print("No")
else:
if(neg>1):
if(pos==0 or pos==1):
print("Yes")
else:
... | 71 | 140 | 1,945,600 | 105619392 | import sys
import io, os
input = sys.stdin.buffer.readline
n = int(input())
XY = [0]*n
cnt = 0
for i in range(n):
x,y = map(int, input().split())
if x > 0:
cnt += 1
if cnt == 1 or n-cnt == 1 or n-cnt == 0 or cnt ==0:
print('Yes')
else:
print('No') | Codeforces Round 450 (Div. 2) | CF | 2,017 | 1 | 256 | Find Extra One | You have n distinct points on a plane, none of them lie on OY axis. Check that there is a point after removal of which the remaining points are located on one side of the OY axis. | The first line contains a single positive integer n (2 ≤ n ≤ 105).
The following n lines contain coordinates of the points. The i-th of these lines contains two single integers xi and yi (|xi|, |yi| ≤ 109, xi ≠ 0). No two points coincide. | Print "Yes" if there is such a point, "No" — otherwise.
You can print every letter in any case (upper or lower). | null | In the first example the second point can be removed.
In the second example there is no suitable for the condition point.
In the third example any point can be removed. | [{"input": "3\n1 1\n-1 -1\n2 -1", "output": "Yes"}, {"input": "4\n1 1\n2 2\n-1 1\n-2 2", "output": "No"}, {"input": "3\n1 2\n2 1\n4 60", "output": "Yes"}] | 800 | ["geometry", "implementation"] | 71 | [{"input": "3\r\n1 1\r\n-1 -1\r\n2 -1\r\n", "output": "Yes"}, {"input": "4\r\n1 1\r\n2 2\r\n-1 1\r\n-2 2\r\n", "output": "No"}, {"input": "3\r\n1 2\r\n2 1\r\n4 60\r\n", "output": "Yes"}, {"input": "10\r\n1 1\r\n2 2\r\n3 3\r\n4 4\r\n5 5\r\n6 6\r\n7 7\r\n8 8\r\n9 9\r\n-1 -1\r\n", "output": "Yes"}, {"input": "2\r\n1000000... | false | stdio | null | true |
828/A | 828 | A | Python 3 | TESTS | 11 | 62 | 2,969,600 | 144463000 | n,a,b = input().split(" ")
n = int(n)
a = int(a)
b = int(b)
denies = 0
for x in input().split(" ") :
x = int(x)
if(x == 1) :
if(a > 0) :
a -=1
else :
if(b > 0) :
b -= 1
a += 1
else :
denies += 1
else :
... | 97 | 93 | 5,632,000 | 205356994 | count=0
tabs = list(map(int, input().split()))
tabones=tabs[1]
tabtwos=tabs[2]
e=0
nums=list(map(int, input().split()))
for i in nums:
if i==1:
if tabones>0:
tabones-=1
elif tabtwos>0:
tabtwos-=1
e+=1
elif e>0:
e-=1
else:
co... | Codeforces Round 423 (Div. 2, rated, based on VK Cup Finals) | CF | 2,017 | 1 | 256 | Restaurant Tables | In a small restaurant there are a tables for one person and b tables for two persons.
It it known that n groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table.... | The first line contains three integers n, a and b (1 ≤ n ≤ 2·105, 1 ≤ a, b ≤ 2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers t1, t2, ..., tn (1 ≤ ti ≤ 2) — the description of clients in chronological o... | Print the total number of people the restaurant denies service to. | null | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remainin... | [{"input": "4 1 2\n1 2 1 1", "output": "0"}, {"input": "4 1 1\n1 1 2 1", "output": "2"}] | 1,200 | ["implementation"] | 97 | [{"input": "4 1 2\r\n1 2 1 1\r\n", "output": "0\r\n"}, {"input": "4 1 1\r\n1 1 2 1\r\n", "output": "2\r\n"}, {"input": "1 1 1\r\n1\r\n", "output": "0\r\n"}, {"input": "2 1 2\r\n2 2\r\n", "output": "0\r\n"}, {"input": "5 1 3\r\n1 2 2 2 1\r\n", "output": "1\r\n"}, {"input": "7 6 1\r\n1 1 1 1 1 1 1\r\n", "output": "0\r\n"... | false | stdio | null | true |
828/A | 828 | A | PyPy 3 | TESTS | 11 | 109 | 8,806,400 | 106502772 | n,a,b=map(int,input().split())
c=input().split()
fa=a
fb=b
fc=0
denied=0
for i in range(len(c)):
if(c[i]=='1'):
if(fa==0):
if(fb>0 and fc==0):
fc+=1
fb-=1
elif(fc==1):
fc-=1
else:
denied+=1
else:
... | 97 | 93 | 5,734,400 | 233007891 | def mainFunc():
n, a, b = map(int, input().split())
group = list(map(int, input().split()))[:n]
singleSeaters = {1:a,2:2*b}
doubleSeaters = {2:b}
deniedCount = 0
for i in group:
if i==1:
if singleSeaters[1]>0:
singleSeaters[1] -=1
else:
... | Codeforces Round 423 (Div. 2, rated, based on VK Cup Finals) | CF | 2,017 | 1 | 256 | Restaurant Tables | In a small restaurant there are a tables for one person and b tables for two persons.
It it known that n groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table.... | The first line contains three integers n, a and b (1 ≤ n ≤ 2·105, 1 ≤ a, b ≤ 2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers t1, t2, ..., tn (1 ≤ ti ≤ 2) — the description of clients in chronological o... | Print the total number of people the restaurant denies service to. | null | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remainin... | [{"input": "4 1 2\n1 2 1 1", "output": "0"}, {"input": "4 1 1\n1 1 2 1", "output": "2"}] | 1,200 | ["implementation"] | 97 | [{"input": "4 1 2\r\n1 2 1 1\r\n", "output": "0\r\n"}, {"input": "4 1 1\r\n1 1 2 1\r\n", "output": "2\r\n"}, {"input": "1 1 1\r\n1\r\n", "output": "0\r\n"}, {"input": "2 1 2\r\n2 2\r\n", "output": "0\r\n"}, {"input": "5 1 3\r\n1 2 2 2 1\r\n", "output": "1\r\n"}, {"input": "7 6 1\r\n1 1 1 1 1 1 1\r\n", "output": "0\r\n"... | false | stdio | null | true |
416/D | 416 | D | PyPy 3-64 | TESTS | 17 | 62 | 512,000 | 153735486 | n=int(input())
a=[int(x) for x in input().split()]
def slv(a):
b=[]
for i in range(n):
if(a[i]!=(-1)):
b.append([a[i],i])
if(len(b)<=1): return 1
ans=1
ok=[]
for i in range(len(b)-1):
t=(b[i+1][0]-b[i][0])//(b[i+1][1]-b[i][1])
if((b[i+1][0]-b[i][0])%(b[i+... | 84 | 171 | 27,648,000 | 302828817 | import sys,math,heapq
from collections import Counter
#sys.setrecursionlimit(200000)
#fin = open('input.txt', 'r')
#input = lambda: fin.readline().strip()
input=lambda: sys.stdin.readline().strip()
import random
RANDOM = random.getrandbits(20)
class Wrapper(int):
def __init__(self, x):
int.__init__(x)
... | Codeforces Round 241 (Div. 2) | CF | 2,014 | 1 | 256 | Population Size | Polycarpus develops an interesting theory about the interrelation of arithmetic progressions with just everything in the world. His current idea is that the population of the capital of Berland changes over time like an arithmetic progression. Well, or like multiple arithmetic progressions.
Polycarpus believes that if... | The first line of the input contains integer n (1 ≤ n ≤ 2·105) — the number of elements in the sequence. The second line contains integer values a1, a2, ..., an separated by a space (1 ≤ ai ≤ 109 or ai = - 1). | Print the minimum number of arithmetic progressions that you need to write one after another to get sequence a. The positions marked as -1 in a can be represented by any positive integers. | null | null | [{"input": "9\n8 6 4 2 1 4 7 10 2", "output": "3"}, {"input": "9\n-1 6 -1 2 -1 4 7 -1 2", "output": "3"}, {"input": "5\n-1 -1 -1 -1 -1", "output": "1"}, {"input": "7\n-1 -1 4 5 1 2 3", "output": "2"}] | 2,400 | ["greedy", "implementation", "math"] | 84 | [{"input": "9\r\n8 6 4 2 1 4 7 10 2\r\n", "output": "3\r\n"}, {"input": "9\r\n-1 6 -1 2 -1 4 7 -1 2\r\n", "output": "3\r\n"}, {"input": "5\r\n-1 -1 -1 -1 -1\r\n", "output": "1\r\n"}, {"input": "7\r\n-1 -1 4 5 1 2 3\r\n", "output": "2\r\n"}, {"input": "1\r\n1\r\n", "output": "1\r\n"}, {"input": "1\r\n65\r\n", "output": ... | false | stdio | null | true |
828/A | 828 | A | Python 3 | TESTS | 11 | 93 | 2,969,600 | 139615908 | n, a, b = map(int, input().split())
sum_ = 0
guest = input().split()
for i in range(len(guest)):
if guest[i] == '1':
if a > 0:
a -= 1
elif a == 0 and b > 0:
a += 1
b -= 1
else:
sum_ += 1
elif guest[i] == '2':
if b > 0:
b... | 97 | 108 | 5,836,800 | 218100899 | n, a, b = map(int, input().split())
t = [int(x) for x in input().split()]
c = 0
ans = 0
for i in range(n):
if t[i] == 1:
if a == 0:
if b == 0:
if c == 0:
ans += 1
else:
c -= 1
else:
b -= 1
... | Codeforces Round 423 (Div. 2, rated, based on VK Cup Finals) | CF | 2,017 | 1 | 256 | Restaurant Tables | In a small restaurant there are a tables for one person and b tables for two persons.
It it known that n groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table.... | The first line contains three integers n, a and b (1 ≤ n ≤ 2·105, 1 ≤ a, b ≤ 2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers t1, t2, ..., tn (1 ≤ ti ≤ 2) — the description of clients in chronological o... | Print the total number of people the restaurant denies service to. | null | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remainin... | [{"input": "4 1 2\n1 2 1 1", "output": "0"}, {"input": "4 1 1\n1 1 2 1", "output": "2"}] | 1,200 | ["implementation"] | 97 | [{"input": "4 1 2\r\n1 2 1 1\r\n", "output": "0\r\n"}, {"input": "4 1 1\r\n1 1 2 1\r\n", "output": "2\r\n"}, {"input": "1 1 1\r\n1\r\n", "output": "0\r\n"}, {"input": "2 1 2\r\n2 2\r\n", "output": "0\r\n"}, {"input": "5 1 3\r\n1 2 2 2 1\r\n", "output": "1\r\n"}, {"input": "7 6 1\r\n1 1 1 1 1 1 1\r\n", "output": "0\r\n"... | false | stdio | null | true |
828/A | 828 | A | Python 3 | TESTS | 11 | 108 | 2,867,200 | 177532658 | n, a, b = map(int, input().split())
t = map(int, input().split())
th = 0
d = 0
for i in t:
if i == 1:
if a > 0:
a -= 1
elif th == 1:
th = 0
elif b > 0:
b -= 1
th = 1
else:
d += 1
elif b > 0:
b -= 1
else:
d += 2
print(d) | 97 | 108 | 22,835,200 | 200844190 | def solve(i,j,p):
ans=0
k=0
for a in p:
if a==1:
if i>0:
i-=1
elif j>0:
j-=1
k+=1
elif k>0:
k-=1
else:
ans+=1
elif a==2:
if j>0:
j-=1
... | Codeforces Round 423 (Div. 2, rated, based on VK Cup Finals) | CF | 2,017 | 1 | 256 | Restaurant Tables | In a small restaurant there are a tables for one person and b tables for two persons.
It it known that n groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table.... | The first line contains three integers n, a and b (1 ≤ n ≤ 2·105, 1 ≤ a, b ≤ 2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers t1, t2, ..., tn (1 ≤ ti ≤ 2) — the description of clients in chronological o... | Print the total number of people the restaurant denies service to. | null | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remainin... | [{"input": "4 1 2\n1 2 1 1", "output": "0"}, {"input": "4 1 1\n1 1 2 1", "output": "2"}] | 1,200 | ["implementation"] | 97 | [{"input": "4 1 2\r\n1 2 1 1\r\n", "output": "0\r\n"}, {"input": "4 1 1\r\n1 1 2 1\r\n", "output": "2\r\n"}, {"input": "1 1 1\r\n1\r\n", "output": "0\r\n"}, {"input": "2 1 2\r\n2 2\r\n", "output": "0\r\n"}, {"input": "5 1 3\r\n1 2 2 2 1\r\n", "output": "1\r\n"}, {"input": "7 6 1\r\n1 1 1 1 1 1 1\r\n", "output": "0\r\n"... | false | stdio | null | true |
828/A | 828 | A | PyPy 3-64 | TESTS | 11 | 61 | 22,835,200 | 200312314 | def solution(n,f,s,g):
S =s*2
count=0
for x in g :
if x == 1:
if f>0 :
f -=1
count+=1
elif S > 0:
S-=1
count+=1
else:
if S//2>0 :
S-=2
count+=2
return sum(g)-count
if __name__ == '__main__':
n,f,s = map(int,input().split())
g =l... | 97 | 108 | 22,835,200 | 225176702 | n, a, b = map(int, input().split())
t = list(map(int, input().split()))
b2 = 0
d = 0
for i in range(n):
if t[i]==1:
if a>0:
a-=1
elif b>0:
b-=1
b2+=1
elif b2>0:
b2-=1
else:
d+=1
else:
if b>0:
b-=1
... | Codeforces Round 423 (Div. 2, rated, based on VK Cup Finals) | CF | 2,017 | 1 | 256 | Restaurant Tables | In a small restaurant there are a tables for one person and b tables for two persons.
It it known that n groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table.... | The first line contains three integers n, a and b (1 ≤ n ≤ 2·105, 1 ≤ a, b ≤ 2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers t1, t2, ..., tn (1 ≤ ti ≤ 2) — the description of clients in chronological o... | Print the total number of people the restaurant denies service to. | null | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remainin... | [{"input": "4 1 2\n1 2 1 1", "output": "0"}, {"input": "4 1 1\n1 1 2 1", "output": "2"}] | 1,200 | ["implementation"] | 97 | [{"input": "4 1 2\r\n1 2 1 1\r\n", "output": "0\r\n"}, {"input": "4 1 1\r\n1 1 2 1\r\n", "output": "2\r\n"}, {"input": "1 1 1\r\n1\r\n", "output": "0\r\n"}, {"input": "2 1 2\r\n2 2\r\n", "output": "0\r\n"}, {"input": "5 1 3\r\n1 2 2 2 1\r\n", "output": "1\r\n"}, {"input": "7 6 1\r\n1 1 1 1 1 1 1\r\n", "output": "0\r\n"... | false | stdio | null | true |
676/C | 676 | C | PyPy 3 | TESTS | 67 | 187 | 29,286,400 | 87641824 | import os
from math import*
n,k=map(int,input().split())
arr=input()
arr=list(arr)
a=list(filter(lambda x:x=="a",arr))
b=list(filter(lambda x:x=="b",arr))
x=""
if len(a)>=len(b):
x="b"
else:
x="a"
#print(x)
fp=0
ans=0
count=0
brr=[0]
checkpos=0
for i in range(n):
if arr[i]==x:
count+=1
brr.append(i)
if count>k... | 117 | 77 | 2,252,800 | 219080873 | def max_beauty(n, k, s):
max_length = 0
max_count = 0
char_count = [0, 0]
left = 0
for right in range(n):
char_count [ord(s[right]) - ord('a')] += 1
max_count = max(max_count, char_count [0], char_count [1])
window_length = right - left + 1
if window_length - max_cou... | Codeforces Round 354 (Div. 2) | CF | 2,016 | 1 | 256 | Vasya and String | High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters.
Vasya can change no more than k characters of the original string. W... | The first line of the input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n) — the length of the string and the maximum number of characters to change.
The second line contains the string, consisting of letters 'a' and 'b' only. | Print the only integer — the maximum beauty of the string Vasya can achieve by changing no more than k characters. | null | In the first sample, Vasya can obtain both strings "aaaa" and "bbbb".
In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". | [{"input": "4 2\nabba", "output": "4"}, {"input": "8 1\naabaabaa", "output": "5"}] | 1,500 | ["binary search", "dp", "strings", "two pointers"] | 117 | [{"input": "4 2\r\nabba\r\n", "output": "4\r\n"}, {"input": "8 1\r\naabaabaa\r\n", "output": "5\r\n"}, {"input": "1 0\r\na\r\n", "output": "1\r\n"}, {"input": "1 1\r\nb\r\n", "output": "1\r\n"}, {"input": "1 0\r\nb\r\n", "output": "1\r\n"}, {"input": "1 1\r\na\r\n", "output": "1\r\n"}, {"input": "10 10\r\nbbbbbbbbbb\r\... | false | stdio | null | true |
828/A | 828 | A | Python 3 | TESTS | 11 | 78 | 4,505,600 | 225587111 | n, a, b = map(int, input().split())
d = 0
for i in map(int, input().split()):
if i == 1:
if a:
a -= 1
elif b:
b -= 1
a += 1
else:
d += 1
else:
if b:
b -= 1
else:
d += 2
print(d) | 97 | 109 | 8,499,200 | 28439703 | n, a,b = list(map(int, input().split()))
c=0
j=0
l = input().split()
for g in l:
if g == "1":
if a>0:
a-=1
elif b>0:
b-=1
c+=1
elif c>0:
c-=1
else:
j+=1
elif b>0:
b-=1
else:
j+=2
print (j) | Codeforces Round 423 (Div. 2, rated, based on VK Cup Finals) | CF | 2,017 | 1 | 256 | Restaurant Tables | In a small restaurant there are a tables for one person and b tables for two persons.
It it known that n groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table.... | The first line contains three integers n, a and b (1 ≤ n ≤ 2·105, 1 ≤ a, b ≤ 2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers t1, t2, ..., tn (1 ≤ ti ≤ 2) — the description of clients in chronological o... | Print the total number of people the restaurant denies service to. | null | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remainin... | [{"input": "4 1 2\n1 2 1 1", "output": "0"}, {"input": "4 1 1\n1 1 2 1", "output": "2"}] | 1,200 | ["implementation"] | 97 | [{"input": "4 1 2\r\n1 2 1 1\r\n", "output": "0\r\n"}, {"input": "4 1 1\r\n1 1 2 1\r\n", "output": "2\r\n"}, {"input": "1 1 1\r\n1\r\n", "output": "0\r\n"}, {"input": "2 1 2\r\n2 2\r\n", "output": "0\r\n"}, {"input": "5 1 3\r\n1 2 2 2 1\r\n", "output": "1\r\n"}, {"input": "7 6 1\r\n1 1 1 1 1 1 1\r\n", "output": "0\r\n"... | false | stdio | null | true |
828/A | 828 | A | PyPy 3-64 | TESTS | 11 | 77 | 15,155,200 | 225586289 | n,a,b=map(int,input().split())
c=0
o=0
for g in map(int,input().split()):
if g==1:
if a>0:a-=1
elif c>0:c-=1
elif b>0:
b-=1
c+=1
else:o+=1
else:
if b>0:b-=1
else:o+=2
print(o) | 97 | 109 | 18,432,000 | 177087289 | import sys
input = sys.stdin.buffer.readline
n, a, b = map(int, input().split())
c = 0 # 1 person left 2 people table
queries = list(map(int, input().split()))
ans = 0
for query in queries:
if query == 1:
if a:
a -= 1
elif b:
b -= 1
c += 1
elif c:
... | Codeforces Round 423 (Div. 2, rated, based on VK Cup Finals) | CF | 2,017 | 1 | 256 | Restaurant Tables | In a small restaurant there are a tables for one person and b tables for two persons.
It it known that n groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table.... | The first line contains three integers n, a and b (1 ≤ n ≤ 2·105, 1 ≤ a, b ≤ 2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers t1, t2, ..., tn (1 ≤ ti ≤ 2) — the description of clients in chronological o... | Print the total number of people the restaurant denies service to. | null | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remainin... | [{"input": "4 1 2\n1 2 1 1", "output": "0"}, {"input": "4 1 1\n1 1 2 1", "output": "2"}] | 1,200 | ["implementation"] | 97 | [{"input": "4 1 2\r\n1 2 1 1\r\n", "output": "0\r\n"}, {"input": "4 1 1\r\n1 1 2 1\r\n", "output": "2\r\n"}, {"input": "1 1 1\r\n1\r\n", "output": "0\r\n"}, {"input": "2 1 2\r\n2 2\r\n", "output": "0\r\n"}, {"input": "5 1 3\r\n1 2 2 2 1\r\n", "output": "1\r\n"}, {"input": "7 6 1\r\n1 1 1 1 1 1 1\r\n", "output": "0\r\n"... | false | stdio | null | true |
828/A | 828 | A | Python 3 | TESTS | 11 | 78 | 3,481,600 | 174338080 | n, a, b = map(int, input().split())
l = list(map(int, input().split()))
ans = 0
c = 0
for i in l:
if i == 1:
if a != 0:
a -= i;ans += 1
elif c != 0:
ans += i;c -= 1
elif b != 0:
ans += i;b -= 1;c += 1
else:
if b != 0:
ans += i;b -= ... | 97 | 124 | 3,788,800 | 116314423 | def checkSat(a, b, l):
OneT = a;
TwoT = b;
ParTwoT = 0;
Satisfied = 0;
for i in range(len(l)):
if(l[i]==1):
if(OneT!=0):
OneT -= 1
Satisfied +=1
else:
if(TwoT!=0):
TwoT -= 1
ParTwo... | Codeforces Round 423 (Div. 2, rated, based on VK Cup Finals) | CF | 2,017 | 1 | 256 | Restaurant Tables | In a small restaurant there are a tables for one person and b tables for two persons.
It it known that n groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table.... | The first line contains three integers n, a and b (1 ≤ n ≤ 2·105, 1 ≤ a, b ≤ 2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers t1, t2, ..., tn (1 ≤ ti ≤ 2) — the description of clients in chronological o... | Print the total number of people the restaurant denies service to. | null | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remainin... | [{"input": "4 1 2\n1 2 1 1", "output": "0"}, {"input": "4 1 1\n1 1 2 1", "output": "2"}] | 1,200 | ["implementation"] | 97 | [{"input": "4 1 2\r\n1 2 1 1\r\n", "output": "0\r\n"}, {"input": "4 1 1\r\n1 1 2 1\r\n", "output": "2\r\n"}, {"input": "1 1 1\r\n1\r\n", "output": "0\r\n"}, {"input": "2 1 2\r\n2 2\r\n", "output": "0\r\n"}, {"input": "5 1 3\r\n1 2 2 2 1\r\n", "output": "1\r\n"}, {"input": "7 6 1\r\n1 1 1 1 1 1 1\r\n", "output": "0\r\n"... | false | stdio | null | true |
828/A | 828 | A | Python 3 | TESTS | 11 | 78 | 3,993,600 | 141470577 | def main():
n, a, b = map(int, input().split())
lst = list(map(int, input().split()))
c, ans = 0, 0
for i in lst:
if i == 1:
if a > 0:
a -= 1
elif c > 0: # c = Table for 2 occupied by 1
c -= 1
elif b > 0:
b -... | 97 | 124 | 8,499,200 | 28431129 | n, a, b = list(map(int,input().split()))
l = input().split()
o = 0
c = 0
for i in l:
if i == '2':
if b > 0:
b -= 1
else:
o += 2
if i == '1':
if a > 0:
a -= 1
elif b > 0:
b -= 1
c += 1
elif c > 0:
c -=... | Codeforces Round 423 (Div. 2, rated, based on VK Cup Finals) | CF | 2,017 | 1 | 256 | Restaurant Tables | In a small restaurant there are a tables for one person and b tables for two persons.
It it known that n groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table.... | The first line contains three integers n, a and b (1 ≤ n ≤ 2·105, 1 ≤ a, b ≤ 2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers t1, t2, ..., tn (1 ≤ ti ≤ 2) — the description of clients in chronological o... | Print the total number of people the restaurant denies service to. | null | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remainin... | [{"input": "4 1 2\n1 2 1 1", "output": "0"}, {"input": "4 1 1\n1 1 2 1", "output": "2"}] | 1,200 | ["implementation"] | 97 | [{"input": "4 1 2\r\n1 2 1 1\r\n", "output": "0\r\n"}, {"input": "4 1 1\r\n1 1 2 1\r\n", "output": "2\r\n"}, {"input": "1 1 1\r\n1\r\n", "output": "0\r\n"}, {"input": "2 1 2\r\n2 2\r\n", "output": "0\r\n"}, {"input": "5 1 3\r\n1 2 2 2 1\r\n", "output": "1\r\n"}, {"input": "7 6 1\r\n1 1 1 1 1 1 1\r\n", "output": "0\r\n"... | false | stdio | null | true |
828/A | 828 | A | PyPy 3-64 | TESTS | 11 | 92 | 22,835,200 | 167448617 | #!/usr/bin/env python3
import math
def test_case(casen):
n, a, b = map(int, input().split())
t = list(map(int, input().split()))
ans = 0
twoone = 0
for i in range(n):
if t[i] == 1 and a > 0:
a -= 1
continue
if t[i] == 1 and twoone > 0:
twoone ... | 97 | 124 | 8,499,200 | 28436370 | n, a, b = (input().split())
n = int(n)
a = int(a)
b = int(b)
potatos = 0
counter = 0
customers = input().split()
for i in customers:
if(i == '1'):
if(a > 0):
a -= 1
elif(b > 0):
b -= 1
potatos += 1
elif(potatos > 0):
potatos -= 1
else:... | Codeforces Round 423 (Div. 2, rated, based on VK Cup Finals) | CF | 2,017 | 1 | 256 | Restaurant Tables | In a small restaurant there are a tables for one person and b tables for two persons.
It it known that n groups of people come today, each consisting of one or two people.
If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table.... | The first line contains three integers n, a and b (1 ≤ n ≤ 2·105, 1 ≤ a, b ≤ 2·105) — the number of groups coming to the restaurant, the number of one-seater and the number of two-seater tables.
The second line contains a sequence of integers t1, t2, ..., tn (1 ≤ ti ≤ 2) — the description of clients in chronological o... | Print the total number of people the restaurant denies service to. | null | In the first example the first group consists of one person, it is seated at a vacant one-seater table. The next group occupies a whole two-seater table. The third group consists of one person, it occupies one place at the remaining two-seater table. The fourth group consists of one person, he is seated at the remainin... | [{"input": "4 1 2\n1 2 1 1", "output": "0"}, {"input": "4 1 1\n1 1 2 1", "output": "2"}] | 1,200 | ["implementation"] | 97 | [{"input": "4 1 2\r\n1 2 1 1\r\n", "output": "0\r\n"}, {"input": "4 1 1\r\n1 1 2 1\r\n", "output": "2\r\n"}, {"input": "1 1 1\r\n1\r\n", "output": "0\r\n"}, {"input": "2 1 2\r\n2 2\r\n", "output": "0\r\n"}, {"input": "5 1 3\r\n1 2 2 2 1\r\n", "output": "1\r\n"}, {"input": "7 6 1\r\n1 1 1 1 1 1 1\r\n", "output": "0\r\n"... | false | stdio | null | true |
623/A | 623 | A | Python 3 | TESTS | 4 | 62 | 409,600 | 104469006 | from collections import defaultdict
n, m = map(int, input("").split())
edge_list =defaultdict(list)
for i in range (m):
start, end = map(int, input("").split())
edge_list[start].append(end)
edge_list[end].append(start)
def dfs(vertice, visited, connected_component):
if vertice in visited:
retur... | 106 | 732 | 8,908,800 | 227484209 | n, m = [int(i) for i in input().split(' ')]
adj = [[1] * n for _ in range(n)]
for i in range(n):
adj[i][i] = 0
for _ in range(m):
u, v = [int(i) - 1 for i in input().split(' ')]
adj[u][v], adj[v][u] = 0, 0
b = set([i for i in range(n) if not any(adj[i])])
a = None
for i in range(n):
if i in b:
c... | AIM Tech Round (Div. 1) | CF | 2,016 | 2 | 256 | Graph and String | One day student Vasya was sitting on a lecture and mentioned a string s1s2... sn, consisting of letters "a", "b" and "c" that was written on his desk. As the lecture was boring, Vasya decided to complete the picture by composing a graph G with the following properties:
- G has exactly n vertices, numbered from 1 to n.... | The first line of the input contains two integers n and m $$( 1 \leq n \leq 500, 0 \leq m \leq \frac { n ( n - 1 ) } { 2 } )$$ — the number of vertices and edges in the graph found by Petya, respectively.
Each of the next m lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the edges of the graph G. It ... | In the first line print "Yes" (without the quotes), if the string s Petya is interested in really exists and "No" (without the quotes) otherwise.
If the string s exists, then print it on the second line of the output. The length of s must be exactly n, it must consist of only letters "a", "b" and "c" only, and the gra... | null | In the first sample you are given a graph made of two vertices with an edge between them. So, these vertices can correspond to both the same and adjacent letters. Any of the following strings "aa", "ab", "ba", "bb", "bc", "cb", "cc" meets the graph's conditions.
In the second sample the first vertex is connected to al... | [{"input": "2 1\n1 2", "output": "Yes\naa"}, {"input": "4 3\n1 2\n1 3\n1 4", "output": "No"}] | 1,800 | ["constructive algorithms", "graphs"] | 106 | [{"input": "2 1\r\n1 2\r\n", "output": "Yes\r\naa\r\n"}, {"input": "4 3\r\n1 2\r\n1 3\r\n1 4\r\n", "output": "No\r\n"}, {"input": "4 4\r\n1 2\r\n1 3\r\n1 4\r\n3 4\r\n", "output": "Yes\r\nbacc\r\n"}, {"input": "1 0\r\n", "output": "Yes\r\na\r\n"}, {"input": "8 28\r\n3 2\r\n4 2\r\n7 4\r\n6 3\r\n3 7\r\n8 1\r\n3 4\r\n5 1\r... | false | stdio | import sys
def main():
input_path = sys.argv[1]
ref_output_path = sys.argv[2]
sub_output_path = sys.argv[3]
# Read input
with open(input_path) as f:
n, m = map(int, f.readline().split())
edges = []
for _ in range(m):
u, v = map(int, f.readline().split())
... | true |
515/B | 515 | B | Python 3 | TESTS | 40 | 46 | 0 | 218399908 | import sys
sys.setrecursionlimit(100000000)
input=lambda:sys.stdin.readline().strip()
write=lambda x:sys.stdout.write(str(x)+'\n')
# from random import randint
# from copy import deepcopy
# from collections import deque,Counter
# from heapq import heapify,heappush,heappop
# from bisect import bisect_left,bisect,insort... | 56 | 62 | 0 | 9899555 | n, m = map(int, input().split())
boy_funny = list(map(int, input().split()))
girl_funny = list(map(int, input().split()))
boy = [0]*n
girl = [0]*m
i = 1
while boy_funny[0] > 0:
boy[boy_funny[i]] = 1
i += 1
boy_funny[0] -= 1
i = 1
while girl_funny[0] > 0:
girl[girl_funny[i]] = 1
i += 1
girl_funny... | Codeforces Round 292 (Div. 2) | CF | 2,015 | 2 | 256 | Drazil and His Happy Friends | Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.
There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites $$(i \bmod n)$$-th... | The first line contains two integer n and m (1 ≤ n, m ≤ 100).
The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys.
The third line conatins integer g (0 ≤ ... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | null | By $$i \bmod k$$ we define the remainder of integer division of i by k.
In first sample case:
- On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day.
- On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so no... | [{"input": "2 3\n0\n1 0", "output": "Yes"}, {"input": "2 4\n1 0\n1 2", "output": "No"}, {"input": "2 3\n1 0\n1 1", "output": "Yes"}] | 1,300 | ["brute force", "dsu", "meet-in-the-middle", "number theory"] | 56 | [{"input": "2 3\r\n0\r\n1 0\r\n", "output": "Yes\r\n"}, {"input": "2 4\r\n1 0\r\n1 2\r\n", "output": "No\r\n"}, {"input": "2 3\r\n1 0\r\n1 1\r\n", "output": "Yes\r\n"}, {"input": "16 88\r\n6 5 14 2 0 12 7\r\n30 21 64 35 79 74 39 63 44 81 73 0 27 33 69 12 86 46 20 25 55 52 7 58 23 5 60 32 41 50 82\r\n", "output": "Yes\r... | false | stdio | null | true |
628/D | 628 | D | PyPy 3-64 | TESTS | 15 | 1,731 | 10,956,800 | 176782076 | import sys
readline=sys.stdin.readline
mod=10**9+7
M,D=map(int,readline().split())
def solve(N,equal=False):
dp=[[[0]*M for j in range(2)] for i in range(2)]
for x in range(1,int(N[0])):
if x==D:
continue
dp[0][0][x%M]+=1
if int(N[0])!=D:
dp[1][0][int(N[0])%M]+=1
for... | 66 | 218 | 19,660,800 | 189401939 | from sys import stdin
def main():
mod = 10**9 + 7
_all = map(str, stdin.read().split())
m, d, a, b = int(next(_all)), int(next(_all)), next(_all), next(_all)
n = len(a)
n_2, re, list_not_d, dp = n & 1, [0]*n, [i for i in range(10) if i ^ d], [[0]*m for i in range(n)]
re[0] = 1
dp[0][0] = 1... | Educational Codeforces Round 8 | ICPC | 2,016 | 2 | 256 | Magic Numbers | Consider the decimal presentation of an integer. Let's call a number d-magic if digit d appears in decimal presentation of the number on even positions and nowhere else.
For example, the numbers 1727374, 17, 1 are 7-magic but 77, 7, 123, 34, 71 are not 7-magic. On the other hand the number 7 is 0-magic, 123 is 2-magic... | The first line contains two integers m, d (1 ≤ m ≤ 2000, 0 ≤ d ≤ 9) — the parameters from the problem statement.
The second line contains positive integer a in decimal presentation (without leading zeroes).
The third line contains positive integer b in decimal presentation (without leading zeroes).
It is guaranteed ... | Print the only integer a — the remainder after dividing by 109 + 7 of the number of d-magic numbers in segment [a, b] that are multiple of m. | null | The numbers from the answer of the first example are 16, 26, 36, 46, 56, 76, 86 and 96.
The numbers from the answer of the second example are 2, 4, 6 and 8.
The numbers from the answer of the third example are 1767, 2717, 5757, 6707, 8797 and 9747. | [{"input": "2 6\n10\n99", "output": "8"}, {"input": "2 0\n1\n9", "output": "4"}, {"input": "19 7\n1000\n9999", "output": "6"}] | 2,200 | ["dp"] | 66 | [{"input": "2 6\r\n10\r\n99\r\n", "output": "8\r\n"}, {"input": "2 0\r\n1\r\n9\r\n", "output": "4\r\n"}, {"input": "19 7\r\n1000\r\n9999\r\n", "output": "6\r\n"}, {"input": "9 4\r\n33\r\n52\r\n", "output": "0\r\n"}, {"input": "10 8\r\n18\r\n59\r\n", "output": "0\r\n"}, {"input": "43 3\r\n587\r\n850\r\n", "output": "1\r... | false | stdio | null | true |
515/B | 515 | B | PyPy 3-64 | TESTS | 40 | 62 | 1,536,000 | 214263256 | n, m = map(int, input().split())
boys = [0] * n
girls = [0] * m
for i in map(int, input().split()[1:]):
boys[i] = 1
for i in map(int, input().split()[1:]):
girls[i] = 1
for i in range(n*m):
boy, girl = i % n, i % m
if boys[boy] or girls[girl]:
boys[boy] = girls[girl] = 1
if all(boys) and all... | 56 | 62 | 0 | 164159799 | import sys
input = sys.stdin.readline
from math import gcd
n, m = map(int, input().split())
a, *w = list(map(int, input().split()))
b, *s = list(map(int, input().split()))
x = gcd(n, m)
if x == 1:
print('Yes')
else:
d = [0]*x
for i in w:
d[i%x] = 1
for i in s:
d[i%x] = 1
if d == [1... | Codeforces Round 292 (Div. 2) | CF | 2,015 | 2 | 256 | Drazil and His Happy Friends | Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.
There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites $$(i \bmod n)$$-th... | The first line contains two integer n and m (1 ≤ n, m ≤ 100).
The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys.
The third line conatins integer g (0 ≤ ... | If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No". | null | By $$i \bmod k$$ we define the remainder of integer division of i by k.
In first sample case:
- On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day.
- On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so no... | [{"input": "2 3\n0\n1 0", "output": "Yes"}, {"input": "2 4\n1 0\n1 2", "output": "No"}, {"input": "2 3\n1 0\n1 1", "output": "Yes"}] | 1,300 | ["brute force", "dsu", "meet-in-the-middle", "number theory"] | 56 | [{"input": "2 3\r\n0\r\n1 0\r\n", "output": "Yes\r\n"}, {"input": "2 4\r\n1 0\r\n1 2\r\n", "output": "No\r\n"}, {"input": "2 3\r\n1 0\r\n1 1\r\n", "output": "Yes\r\n"}, {"input": "16 88\r\n6 5 14 2 0 12 7\r\n30 21 64 35 79 74 39 63 44 81 73 0 27 33 69 12 86 46 20 25 55 52 7 58 23 5 60 32 41 50 82\r\n", "output": "Yes\r... | false | stdio | null | true |
665/D | 665 | D | PyPy 3 | TESTS | 11 | 171 | 34,713,600 | 126968131 | import sys
input = sys.stdin.readline
def the_sieve_of_eratosthenes(n):
s = [1] * (n + 1)
x = []
for i in range(2, n + 1):
if s[i]:
x.append(i)
for j in range(i, n + 1, i):
s[j] = 0
return x
n = int(input())
a = list(map(int, input().split()))
x = set(th... | 66 | 202 | 11,980,800 | 92581095 | import sys
def get_primes(n: int):
from itertools import chain
from array import array
primes = {2, 3}
is_prime = (array('b', (0, 0, 1, 1, 0, 1, 0)) +
array('b', (1, 0, 0, 0, 1, 0))*((n-1)//6))
for i in chain.from_iterable((range(5, n+1, 6), range(7, n+1, 6))):
if is_prime... | Educational Codeforces Round 12 | ICPC | 2,016 | 1 | 256 | Simple Subset | A tuple of positive integers {x1, x2, ..., xk} is called simple if for all pairs of positive integers (i, j) (1 ≤ i < j ≤ k), xi + xj is a prime.
You are given an array a with n positive integers a1, a2, ..., an (not necessary distinct). You want to find a simple subset of the array a with the maximum size.
... | The first line contains integer n (1 ≤ n ≤ 1000) — the number of integers in the array a.
The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a. | On the first line print integer m — the maximum possible size of simple subset of a.
On the second line print m integers bl — the elements of the simple subset of the array a with the maximum size.
If there is more than one solution you can print any of them. You can print the elements of the subset in any order. | null | null | [{"input": "2\n2 3", "output": "2\n3 2"}, {"input": "2\n2 2", "output": "1\n2"}, {"input": "3\n2 1 1", "output": "3\n1 1 2"}, {"input": "2\n83 14", "output": "2\n14 83"}] | 1,800 | ["constructive algorithms", "greedy", "number theory"] | 66 | [{"input": "2\r\n2 3\r\n", "output": "2\r\n3 2\r\n"}, {"input": "2\r\n2 2\r\n", "output": "1\r\n2\r\n"}, {"input": "3\r\n2 1 1\r\n", "output": "3\r\n1 1 2\r\n"}, {"input": "2\r\n83 14\r\n", "output": "2\r\n14 83\r\n"}, {"input": "10\r\n10 10 1 2 3 3 1 2 1 5\r\n", "output": "4\r\n1 1 10 1\r\n"}, {"input": "100\r\n314 90... | false | stdio | import sys
from collections import Counter
def sieve(max_limit):
sieve = [True] * (max_limit + 1)
sieve[0] = sieve[1] = False
for i in range(2, int(max_limit**0.5) + 1):
if sieve[i]:
sieve[i*i : max_limit+1 : i] = [False]*len(sieve[i*i : max_limit+1 : i])
return sieve
max_sum = 2 *... | true |
793/A | 793 | A | Python 3 | TESTS | 21 | 155 | 9,216,000 | 54373109 | # import sys
# sys.stdin=open("input.in",'r')
# sys.stdout=open("out1.out",'w')
n,k=map(int,input().split())
a=list(map(int,input().split()))
x=min(a)
s=0
for i in a:
s+=(i-x)
if s==0:
print(0)
elif s%k==0:
print(s//k)
else:
print(-1) | 88 | 93 | 13,516,800 | 223701914 | n,k=map(int,input().split())
l=list(map(int,input().split()))
a=-1
for i in range(n-1):
if l[i]%k!=l[i+1]%k:
a=1
if a==1:
print(-1)
else:
print((sum(l)-min(l)*n)//k) | Tinkoff Challenge - Elimination Round | CF | 2,017 | 1 | 256 | Oleg and shares | Oleg the bank client checks share prices every day. There are n share prices he is interested in. Today he observed that each second exactly one of these prices decreases by k rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg... | The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109) — the number of share prices, and the amount of rubles some price decreases each second.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the initial prices. | Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible. | null | Consider the first example.
Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds.
There... | [{"input": "3 3\n12 9 15", "output": "3"}, {"input": "2 2\n10 9", "output": "-1"}, {"input": "4 1\n1 1000000000 1000000000 1000000000", "output": "2999999997"}] | 900 | ["implementation", "math"] | 88 | [{"input": "3 3\r\n12 9 15\r\n", "output": "3"}, {"input": "2 2\r\n10 9\r\n", "output": "-1"}, {"input": "4 1\r\n1 1000000000 1000000000 1000000000\r\n", "output": "2999999997"}, {"input": "1 11\r\n123\r\n", "output": "0"}, {"input": "20 6\r\n38 86 86 50 98 62 32 2 14 62 98 50 2 50 32 38 62 62 8 14\r\n", "output": "151... | false | stdio | null | true |
793/A | 793 | A | Python 3 | TESTS | 21 | 108 | 15,052,800 | 206445201 | import math
n,kk=map(int,input().split())
a=list(map(int,input().split()))
k=min(a)
o=[]
d=[]
count=0
for i in list(a):
d.append(i-k)
if len(a)==1:
print(0)
elif sum(d)%kk!=0:
print(-1)
else:
while True:
try:
a.remove(k)
count+=1
except:
for u in range... | 88 | 108 | 8,806,400 | 194899032 | countOfNumbers, rubles = map(int, input().split())
numbers = list(map(int, input().split()))
#
def Solution():
preResult = 0
minimalNumber = min(numbers)
for element in numbers:
check = (element - minimalNumber) % rubles
if check != 0:
return -1
... | Tinkoff Challenge - Elimination Round | CF | 2,017 | 1 | 256 | Oleg and shares | Oleg the bank client checks share prices every day. There are n share prices he is interested in. Today he observed that each second exactly one of these prices decreases by k rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg... | The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109) — the number of share prices, and the amount of rubles some price decreases each second.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the initial prices. | Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible. | null | Consider the first example.
Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds.
There... | [{"input": "3 3\n12 9 15", "output": "3"}, {"input": "2 2\n10 9", "output": "-1"}, {"input": "4 1\n1 1000000000 1000000000 1000000000", "output": "2999999997"}] | 900 | ["implementation", "math"] | 88 | [{"input": "3 3\r\n12 9 15\r\n", "output": "3"}, {"input": "2 2\r\n10 9\r\n", "output": "-1"}, {"input": "4 1\r\n1 1000000000 1000000000 1000000000\r\n", "output": "2999999997"}, {"input": "1 11\r\n123\r\n", "output": "0"}, {"input": "20 6\r\n38 86 86 50 98 62 32 2 14 62 98 50 2 50 32 38 62 62 8 14\r\n", "output": "151... | false | stdio | null | true |
793/A | 793 | A | Python 3 | TESTS | 21 | 78 | 8,806,400 | 192697926 | a,b=map(int,input().split())
c=[int(i) for i in input().split()]
v=sum(c)-a*min(c)
if v%b==0:
print(v//b)
else:
print(-1) | 88 | 108 | 13,414,400 | 206446402 | import math
n,kk=map(int,input().split())
a=list(map(int,input().split()))
k=min(a)
o=[]
d=[]
count=0
for i in a:
if len(a)==1:
print(0)
break
elif (i-k)%kk==0:
continue
else:
print(-1)
break
else:
while True:
try:
a.remove(k)
... | Tinkoff Challenge - Elimination Round | CF | 2,017 | 1 | 256 | Oleg and shares | Oleg the bank client checks share prices every day. There are n share prices he is interested in. Today he observed that each second exactly one of these prices decreases by k rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg... | The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109) — the number of share prices, and the amount of rubles some price decreases each second.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the initial prices. | Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible. | null | Consider the first example.
Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds.
There... | [{"input": "3 3\n12 9 15", "output": "3"}, {"input": "2 2\n10 9", "output": "-1"}, {"input": "4 1\n1 1000000000 1000000000 1000000000", "output": "2999999997"}] | 900 | ["implementation", "math"] | 88 | [{"input": "3 3\r\n12 9 15\r\n", "output": "3"}, {"input": "2 2\r\n10 9\r\n", "output": "-1"}, {"input": "4 1\r\n1 1000000000 1000000000 1000000000\r\n", "output": "2999999997"}, {"input": "1 11\r\n123\r\n", "output": "0"}, {"input": "20 6\r\n38 86 86 50 98 62 32 2 14 62 98 50 2 50 32 38 62 62 8 14\r\n", "output": "151... | false | stdio | null | true |
793/A | 793 | A | PyPy 3-64 | TESTS | 21 | 93 | 13,414,400 | 232610946 | n,k = map(int, input().split())
a = sorted(map(int, input().split()))
count = 0
for i in a:
count += i - a[0]
print(count//k if count/k%1==0 else -1) | 88 | 108 | 14,745,600 | 232543358 | def f(l):
if (l[-1] - l[0]) % k != 0:
return False
for i in range(len(l) -1):
if (l[i+1] - l[i]) % k != 0:
return False
return True
n,k = map(int,input().split())
l = list(map(int,input().split()))
l.sort()
if f(l) == False:
print('-1')
else:
cnt = 0
for i in reversed... | Tinkoff Challenge - Elimination Round | CF | 2,017 | 1 | 256 | Oleg and shares | Oleg the bank client checks share prices every day. There are n share prices he is interested in. Today he observed that each second exactly one of these prices decreases by k rubles (note that each second exactly one price changes, but at different seconds different prices can change). Prices can become negative. Oleg... | The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109) — the number of share prices, and the amount of rubles some price decreases each second.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the initial prices. | Print the only line containing the minimum number of seconds needed for prices to become equal, of «-1» if it is impossible. | null | Consider the first example.
Suppose the third price decreases in the first second and become equal 12 rubles, then the first price decreases and becomes equal 9 rubles, and in the third second the third price decreases again and becomes equal 9 rubles. In this case all prices become equal 9 rubles in 3 seconds.
There... | [{"input": "3 3\n12 9 15", "output": "3"}, {"input": "2 2\n10 9", "output": "-1"}, {"input": "4 1\n1 1000000000 1000000000 1000000000", "output": "2999999997"}] | 900 | ["implementation", "math"] | 88 | [{"input": "3 3\r\n12 9 15\r\n", "output": "3"}, {"input": "2 2\r\n10 9\r\n", "output": "-1"}, {"input": "4 1\r\n1 1000000000 1000000000 1000000000\r\n", "output": "2999999997"}, {"input": "1 11\r\n123\r\n", "output": "0"}, {"input": "20 6\r\n38 86 86 50 98 62 32 2 14 62 98 50 2 50 32 38 62 62 8 14\r\n", "output": "151... | false | stdio | null | true |
525/C | 525 | C | PyPy 3 | TESTS | 25 | 405 | 14,745,600 | 96267521 | from collections import Counter
import string
import math
import sys
from fractions import Fraction
def array_int():
return [int(i) for i in sys.stdin.readline().split()]
def vary(arrber_of_variables):
if arrber_of_variables==1:
return int(sys.stdin.readline())
if arrber_of_variables>=2:
ret... | 96 | 124 | 14,028,800 | 194028400 | import sys
input = sys.stdin.readline
n = int(input())
w = sorted(map(int, input().split()), reverse=1)
i = 0
c = 0
a = 0
while i < n-1:
if w[i] < w[i+1] + 2:
if a == 0:
a = min(w[i], w[i+1])
else:
c += a*min(w[i], w[i+1])
a = 0
i += 2
else:
i... | Codeforces Round 297 (Div. 2) | CF | 2,015 | 2 | 256 | Ilya and Sticks | In the evening, after the contest Ilya was bored, and he really felt like maximizing. He remembered that he had a set of n sticks and an instrument. Each stick is characterized by its length li.
Ilya decided to make a rectangle from the sticks. And due to his whim, he decided to make rectangles in such a way that maxi... | The first line of the input contains a positive integer n (1 ≤ n ≤ 105) — the number of the available sticks.
The second line of the input contains n positive integers li (2 ≤ li ≤ 106) — the lengths of the sticks. | The first line of the output must contain a single non-negative integer — the maximum total area of the rectangles that Ilya can make from the available sticks. | null | null | [{"input": "4\n2 4 4 2", "output": "8"}, {"input": "4\n2 2 3 5", "output": "0"}, {"input": "4\n100003 100004 100005 100006", "output": "10000800015"}] | 1,600 | ["greedy", "math", "sortings"] | 96 | [{"input": "4\r\n2 4 4 2\r\n", "output": "8\r\n"}, {"input": "4\r\n2 2 3 5\r\n", "output": "0\r\n"}, {"input": "4\r\n100003 100004 100005 100006\r\n", "output": "10000800015\r\n"}, {"input": "8\r\n5 3 3 3 3 4 4 4\r\n", "output": "25\r\n"}, {"input": "10\r\n123 124 123 124 2 2 2 2 9 9\r\n", "output": "15270\r\n"}, {"inp... | false | stdio | null | true |
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