message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a line of n colored squares in a row, numbered from 1 to n from left to right. The i-th square initially has the color c_i.
Let's say, that two squares i and j belong to the same connected component if c_i = c_j, and c_i = c_k for all k satisfying i < k < j. In other words, all squares on the segment from i to j should have the same color.
For example, the line [3, 3, 3] has 1 connected component, while the line [5, 2, 4, 4] has 3 connected components.
The game "flood fill" is played on the given line as follows:
* At the start of the game you pick any starting square (this is not counted as a turn).
* Then, in each game turn, change the color of the connected component containing the starting square to any other color.
Find the minimum number of turns needed for the entire line to be changed into a single color.
Input
The first line contains a single integer n (1 ≤ n ≤ 5000) — the number of squares.
The second line contains integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 5000) — the initial colors of the squares.
Output
Print a single integer — the minimum number of the turns needed.
Examples
Input
4
5 2 2 1
Output
2
Input
8
4 5 2 2 1 3 5 5
Output
4
Input
1
4
Output
0
Note
In the first example, a possible way to achieve an optimal answer is to pick square with index 2 as the starting square and then play as follows:
* [5, 2, 2, 1]
* [5, 5, 5, 1]
* [1, 1, 1, 1]
In the second example, a possible way to achieve an optimal answer is to pick square with index 5 as the starting square and then perform recoloring into colors 2, 3, 5, 4 in that order.
In the third example, the line already consists of one color only.
Submitted Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 2/13/19
"""
import collections
import time
import os
import sys
import bisect
import heapq
def solve(N, C):
A = []
for c in C:
if not A or A[-1] != c:
A.append(c)
colors = list(set(C))
colors.sort()
cmap = {c: i+1 for i, c in enumerate(colors)}
A = [cmap[a] for a in A]
N = len(A)
C = len(colors)
dp = [[[float('inf') for _ in range(2)] for _ in range(N+1)] for _ in range(N+1)]
# print(A)
for i in range(N):
dp[i][i][0] = 0
dp[i][i][1] = 0
for r in range(N):
for l in range(r, -1, -1):
for i in range(2):
c = A[l] if i == 0 else A[r]
if l:
dp[l-1][r][0] = min(dp[l-1][r][0], dp[l][r][i] + (1 if c != A[l-1] else 0))
if r+1 < N:
dp[l][r+1][1] = min(dp[l][r+1][1], dp[l][r][i] + (1 if c != A[r+1] else 0))
print(min(dp[0][N-1]))
def solve2(N, C):
A = []
for c in C:
if not A or A[-1] != c:
A.append(c)
B = A[::-1]
N = len(A)
dp = [[0 for _ in range(N+1)] for _ in range(N+1)]
for i in range(1, N+1):
for j in range(1, N+1):
dp[i][j] = dp[i-1][j-1] + 1 if A[i-1] == B[j-1] else max(dp[i-1][j], dp[i][j-1])
print(N-1-dp[N][N]//2)
N = int(input())
C = [int(x) for x in input().split()]
solve2(N, C)
``` | instruction | 0 | 52,261 | 7 | 104,522 |
Yes | output | 1 | 52,261 | 7 | 104,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a line of n colored squares in a row, numbered from 1 to n from left to right. The i-th square initially has the color c_i.
Let's say, that two squares i and j belong to the same connected component if c_i = c_j, and c_i = c_k for all k satisfying i < k < j. In other words, all squares on the segment from i to j should have the same color.
For example, the line [3, 3, 3] has 1 connected component, while the line [5, 2, 4, 4] has 3 connected components.
The game "flood fill" is played on the given line as follows:
* At the start of the game you pick any starting square (this is not counted as a turn).
* Then, in each game turn, change the color of the connected component containing the starting square to any other color.
Find the minimum number of turns needed for the entire line to be changed into a single color.
Input
The first line contains a single integer n (1 ≤ n ≤ 5000) — the number of squares.
The second line contains integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 5000) — the initial colors of the squares.
Output
Print a single integer — the minimum number of the turns needed.
Examples
Input
4
5 2 2 1
Output
2
Input
8
4 5 2 2 1 3 5 5
Output
4
Input
1
4
Output
0
Note
In the first example, a possible way to achieve an optimal answer is to pick square with index 2 as the starting square and then play as follows:
* [5, 2, 2, 1]
* [5, 5, 5, 1]
* [1, 1, 1, 1]
In the second example, a possible way to achieve an optimal answer is to pick square with index 5 as the starting square and then perform recoloring into colors 2, 3, 5, 4 in that order.
In the third example, the line already consists of one color only.
Submitted Solution:
```
def lcs(a, b):
c = [[0 for i in range(len(a)+1)] for i in range(len(b)+1)]
for i in range((len(b)+1)):
for j in range(len(b)+1):
if i == 0 or j == 0:
c[i][j] = 0
elif a[i-1] == b[j-1]:
c[i][j] = c[i-1][j-1] + 1
else:
c[i][j] = max(c[i-1][j], c[i][j-1])
return c
n = int(input())
a = list(map(int, input().split()))
b = [a[0]]
for i in range(1, n):
if a[i] != a[i-1]:
b.append(a[i])
a = b[:: -1]
c = lcs(a, b)
ans = 0
n = len(a)
for i in range(0, min(n+1,n//2+2)):
ans = max(ans, c[i][n-i])
print(n-ans-1)
``` | instruction | 0 | 52,262 | 7 | 104,524 |
No | output | 1 | 52,262 | 7 | 104,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a line of n colored squares in a row, numbered from 1 to n from left to right. The i-th square initially has the color c_i.
Let's say, that two squares i and j belong to the same connected component if c_i = c_j, and c_i = c_k for all k satisfying i < k < j. In other words, all squares on the segment from i to j should have the same color.
For example, the line [3, 3, 3] has 1 connected component, while the line [5, 2, 4, 4] has 3 connected components.
The game "flood fill" is played on the given line as follows:
* At the start of the game you pick any starting square (this is not counted as a turn).
* Then, in each game turn, change the color of the connected component containing the starting square to any other color.
Find the minimum number of turns needed for the entire line to be changed into a single color.
Input
The first line contains a single integer n (1 ≤ n ≤ 5000) — the number of squares.
The second line contains integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 5000) — the initial colors of the squares.
Output
Print a single integer — the minimum number of the turns needed.
Examples
Input
4
5 2 2 1
Output
2
Input
8
4 5 2 2 1 3 5 5
Output
4
Input
1
4
Output
0
Note
In the first example, a possible way to achieve an optimal answer is to pick square with index 2 as the starting square and then play as follows:
* [5, 2, 2, 1]
* [5, 5, 5, 1]
* [1, 1, 1, 1]
In the second example, a possible way to achieve an optimal answer is to pick square with index 5 as the starting square and then perform recoloring into colors 2, 3, 5, 4 in that order.
In the third example, the line already consists of one color only.
Submitted Solution:
```
n = int(input())
*a, = map(int, input().split())
b = [a[0]]
b = [a[i] for i in range(n) if a[i] != a[i - 1]]
dp = [[0 for j in range(n + 1)] for i in range(n + 1)]
for l in range(1, len(b)):
for i in range(len(b) - l):
j = i + l
dp[i][j] = min(dp[i][j - 1], dp[i + 1][j]) + 1
if b[i] == b[j]:
dp[i][j] = dp[i + 1][j - 1] + 1
print(dp[0][len(b) - 1])
``` | instruction | 0 | 52,263 | 7 | 104,526 |
No | output | 1 | 52,263 | 7 | 104,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a line of n colored squares in a row, numbered from 1 to n from left to right. The i-th square initially has the color c_i.
Let's say, that two squares i and j belong to the same connected component if c_i = c_j, and c_i = c_k for all k satisfying i < k < j. In other words, all squares on the segment from i to j should have the same color.
For example, the line [3, 3, 3] has 1 connected component, while the line [5, 2, 4, 4] has 3 connected components.
The game "flood fill" is played on the given line as follows:
* At the start of the game you pick any starting square (this is not counted as a turn).
* Then, in each game turn, change the color of the connected component containing the starting square to any other color.
Find the minimum number of turns needed for the entire line to be changed into a single color.
Input
The first line contains a single integer n (1 ≤ n ≤ 5000) — the number of squares.
The second line contains integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 5000) — the initial colors of the squares.
Output
Print a single integer — the minimum number of the turns needed.
Examples
Input
4
5 2 2 1
Output
2
Input
8
4 5 2 2 1 3 5 5
Output
4
Input
1
4
Output
0
Note
In the first example, a possible way to achieve an optimal answer is to pick square with index 2 as the starting square and then play as follows:
* [5, 2, 2, 1]
* [5, 5, 5, 1]
* [1, 1, 1, 1]
In the second example, a possible way to achieve an optimal answer is to pick square with index 5 as the starting square and then perform recoloring into colors 2, 3, 5, 4 in that order.
In the third example, the line already consists of one color only.
Submitted Solution:
```
from sys import stdin, stdout, setrecursionlimit
input = stdin.readline
# import string
# characters = string.ascii_lowercase
# digits = string.digits
# setrecursionlimit(int(1e6))
# dir = [-1,0,1,0,-1]
# moves = 'NESW'
inf = float('inf')
from functools import cmp_to_key
from collections import defaultdict as dd
from collections import Counter, deque
from heapq import *
import math
from math import floor, ceil, sqrt
def geti(): return map(int, input().strip().split())
def getl(): return list(map(int, input().strip().split()))
def getis(): return map(str, input().strip().split())
def getls(): return list(map(str, input().strip().split()))
def gets(): return input().strip()
def geta(): return int(input())
def print_s(s): stdout.write(s+'\n')
def solve():
n = geta()
a = set(getl())
print(len(a) - 1)
if __name__=='__main__':
solve()
``` | instruction | 0 | 52,264 | 7 | 104,528 |
No | output | 1 | 52,264 | 7 | 104,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a line of n colored squares in a row, numbered from 1 to n from left to right. The i-th square initially has the color c_i.
Let's say, that two squares i and j belong to the same connected component if c_i = c_j, and c_i = c_k for all k satisfying i < k < j. In other words, all squares on the segment from i to j should have the same color.
For example, the line [3, 3, 3] has 1 connected component, while the line [5, 2, 4, 4] has 3 connected components.
The game "flood fill" is played on the given line as follows:
* At the start of the game you pick any starting square (this is not counted as a turn).
* Then, in each game turn, change the color of the connected component containing the starting square to any other color.
Find the minimum number of turns needed for the entire line to be changed into a single color.
Input
The first line contains a single integer n (1 ≤ n ≤ 5000) — the number of squares.
The second line contains integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 5000) — the initial colors of the squares.
Output
Print a single integer — the minimum number of the turns needed.
Examples
Input
4
5 2 2 1
Output
2
Input
8
4 5 2 2 1 3 5 5
Output
4
Input
1
4
Output
0
Note
In the first example, a possible way to achieve an optimal answer is to pick square with index 2 as the starting square and then play as follows:
* [5, 2, 2, 1]
* [5, 5, 5, 1]
* [1, 1, 1, 1]
In the second example, a possible way to achieve an optimal answer is to pick square with index 5 as the starting square and then perform recoloring into colors 2, 3, 5, 4 in that order.
In the third example, the line already consists of one color only.
Submitted Solution:
```
def solution():
inp = input()
colors = input().replace(" ","")
lst = []
last = None
count = 1
for char in colors:
if last is None:
last = char
continue
if last != char:
lst.append([last] * count)
count = 1
last = char
else:
count += 1
lst.append([last] * count)
di = {}
for l in lst:
di[l[0]] = di.get(l[0],0) + 1
maks = max(di,key= lambda x: di[x])
count = 0
for l in lst:
if maks not in l:
count += 1
print(count)
solution()
``` | instruction | 0 | 52,265 | 7 | 104,530 |
No | output | 1 | 52,265 | 7 | 104,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ever since Kalevitch, a famous Berland abstractionist, heard of fractals, he made them the main topic of his canvases. Every morning the artist takes a piece of graph paper and starts with making a model of his future canvas. He takes a square as big as n × n squares and paints some of them black. Then he takes a clean square piece of paper and paints the fractal using the following algorithm:
Step 1. The paper is divided into n2 identical squares and some of them are painted black according to the model.
Step 2. Every square that remains white is divided into n2 smaller squares and some of them are painted black according to the model.
Every following step repeats step 2.
<image>
Unfortunately, this tiresome work demands too much time from the painting genius. Kalevitch has been dreaming of making the process automatic to move to making 3D or even 4D fractals.
Input
The first line contains integers n and k (2 ≤ n ≤ 3, 1 ≤ k ≤ 5), where k is the amount of steps of the algorithm. Each of the following n lines contains n symbols that determine the model. Symbol «.» stands for a white square, whereas «*» stands for a black one. It is guaranteed that the model has at least one white square.
Output
Output a matrix nk × nk which is what a picture should look like after k steps of the algorithm.
Examples
Input
2 3
.*
..
Output
.*******
..******
.*.*****
....****
.***.***
..**..**
.*.*.*.*
........
Input
3 2
.*.
***
.*.
Output
.*.***.*.
*********
.*.***.*.
*********
*********
*********
.*.***.*.
*********
.*.***.*.
Submitted Solution:
```
# n, k = map(int, input().split())
# ans = []
# for _ in range(n):
# input()
#ans.append(list(input()))
# for i in range(1, k):
# arr = [list(x) for x in ans]
# ans = [['*']*(n**(i+1)) for x in range(n**(i+1))]
# nx = n**i
# for r in range(nx):
# for c in range(nx):
# x = arr[r][c]
# for rx in range(n):
# for cx in range(n):
# ans[r*n+rx][c*n+cx] = '*' if x == '*' else arr[rx][cx]
# for row in ans:
# print(''.join(row))
``` | instruction | 0 | 52,548 | 7 | 105,096 |
No | output | 1 | 52,548 | 7 | 105,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ever since Kalevitch, a famous Berland abstractionist, heard of fractals, he made them the main topic of his canvases. Every morning the artist takes a piece of graph paper and starts with making a model of his future canvas. He takes a square as big as n × n squares and paints some of them black. Then he takes a clean square piece of paper and paints the fractal using the following algorithm:
Step 1. The paper is divided into n2 identical squares and some of them are painted black according to the model.
Step 2. Every square that remains white is divided into n2 smaller squares and some of them are painted black according to the model.
Every following step repeats step 2.
<image>
Unfortunately, this tiresome work demands too much time from the painting genius. Kalevitch has been dreaming of making the process automatic to move to making 3D or even 4D fractals.
Input
The first line contains integers n and k (2 ≤ n ≤ 3, 1 ≤ k ≤ 5), where k is the amount of steps of the algorithm. Each of the following n lines contains n symbols that determine the model. Symbol «.» stands for a white square, whereas «*» stands for a black one. It is guaranteed that the model has at least one white square.
Output
Output a matrix nk × nk which is what a picture should look like after k steps of the algorithm.
Examples
Input
2 3
.*
..
Output
.*******
..******
.*.*****
....****
.***.***
..**..**
.*.*.*.*
........
Input
3 2
.*.
***
.*.
Output
.*.***.*.
*********
.*.***.*.
*********
*********
*********
.*.***.*.
*********
.*.***.*.
Submitted Solution:
```
def pr(v):
n=len(v)
fw=open('output.txt','w')
for c in range(n-1):
s=''.join(v[c])
fw.write(s + '\n')
s=''.join(v[n-1])
fw.write(s)
def gbn(v,t):
n=len(v)
tn=len(t)
bl=list('*'*tn)
nv=[]
for c in range(n*tn):
nv.append([])
for x in range(n):
for z in range(n):
if v[z][x]=='.':
for h in range(tn):
nv[z*tn+h]+=t[h]
else:
for h in range(tn):
nv[z*tn+h]+=bl
return nv
def fact(v,k):
t=v.copy()
for c in range(k):
v=gbn(v,t)
return v
fr=open('input.txt','r')
ss=fr.readlines()
for c in range(len(ss)-1):
l=len(ss[c])
ss[c]=ss[c][:l-1]
n,k=map(int,ss[0].split(' '))
v=[]
for c in range(n):
v.append(list(ss[c+1]))
nv=fact(v,k-1)
pr(nv)
``` | instruction | 0 | 52,549 | 7 | 105,098 |
No | output | 1 | 52,549 | 7 | 105,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ever since Kalevitch, a famous Berland abstractionist, heard of fractals, he made them the main topic of his canvases. Every morning the artist takes a piece of graph paper and starts with making a model of his future canvas. He takes a square as big as n × n squares and paints some of them black. Then he takes a clean square piece of paper and paints the fractal using the following algorithm:
Step 1. The paper is divided into n2 identical squares and some of them are painted black according to the model.
Step 2. Every square that remains white is divided into n2 smaller squares and some of them are painted black according to the model.
Every following step repeats step 2.
<image>
Unfortunately, this tiresome work demands too much time from the painting genius. Kalevitch has been dreaming of making the process automatic to move to making 3D or even 4D fractals.
Input
The first line contains integers n and k (2 ≤ n ≤ 3, 1 ≤ k ≤ 5), where k is the amount of steps of the algorithm. Each of the following n lines contains n symbols that determine the model. Symbol «.» stands for a white square, whereas «*» stands for a black one. It is guaranteed that the model has at least one white square.
Output
Output a matrix nk × nk which is what a picture should look like after k steps of the algorithm.
Examples
Input
2 3
.*
..
Output
.*******
..******
.*.*****
....****
.***.***
..**..**
.*.*.*.*
........
Input
3 2
.*.
***
.*.
Output
.*.***.*.
*********
.*.***.*.
*********
*********
*********
.*.***.*.
*********
.*.***.*.
Submitted Solution:
```
# n, k = map(int, input().split())
ans = []
# for i in range(n):
# ans.append([c for c in input()])
# for i in range(1, k):
# arr = [list(x) for x in ans]
# ans = [['*']*(n**(i+1)) for x in range(n**(i+1))]
# nx = n**i
# for r in range(nx):
# for c in range(nx):
# x = arr[r][c]
# for rx in range(n):
# for cx in range(n):
# ans[r*n+rx][c*n+cx] = '*' if x == '*' else arr[rx][cx]
# for row in ans:
# print(''.join(row))
``` | instruction | 0 | 52,550 | 7 | 105,100 |
No | output | 1 | 52,550 | 7 | 105,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ever since Kalevitch, a famous Berland abstractionist, heard of fractals, he made them the main topic of his canvases. Every morning the artist takes a piece of graph paper and starts with making a model of his future canvas. He takes a square as big as n × n squares and paints some of them black. Then he takes a clean square piece of paper and paints the fractal using the following algorithm:
Step 1. The paper is divided into n2 identical squares and some of them are painted black according to the model.
Step 2. Every square that remains white is divided into n2 smaller squares and some of them are painted black according to the model.
Every following step repeats step 2.
<image>
Unfortunately, this tiresome work demands too much time from the painting genius. Kalevitch has been dreaming of making the process automatic to move to making 3D or even 4D fractals.
Input
The first line contains integers n and k (2 ≤ n ≤ 3, 1 ≤ k ≤ 5), where k is the amount of steps of the algorithm. Each of the following n lines contains n symbols that determine the model. Symbol «.» stands for a white square, whereas «*» stands for a black one. It is guaranteed that the model has at least one white square.
Output
Output a matrix nk × nk which is what a picture should look like after k steps of the algorithm.
Examples
Input
2 3
.*
..
Output
.*******
..******
.*.*****
....****
.***.***
..**..**
.*.*.*.*
........
Input
3 2
.*.
***
.*.
Output
.*.***.*.
*********
.*.***.*.
*********
*********
*********
.*.***.*.
*********
.*.***.*.
Submitted Solution:
```
try:
n, k = map(int, input().split())
ans = []
for i in range(n):
ans.append([c for c in input()])
for i in range(1, k):
arr = [list(x) for x in ans]
ans = [['*']*(n**(i+1)) for x in range(n**(i+1))]
nx = n**i
for r in range(nx):
for c in range(nx):
x = arr[r][c]
for rx in range(n):
for cx in range(n):
ans[r*n+rx][c*n+cx] = '*' if x == '*' else arr[rx][cx]
for row in ans:
print(''.join(row))
except Exception as e:
print(e)
``` | instruction | 0 | 52,551 | 7 | 105,102 |
No | output | 1 | 52,551 | 7 | 105,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke loves colorful balls. He has a total of N×K balls, K in each of his favorite N colors. The colors are numbered 1 through N.
He will arrange all of the balls in a row from left to right, in arbitrary order. Then, for each of the N colors, he will paint the leftmost ball of that color into color 0, a color different from any of the N original colors.
After painting, how many sequences of the colors of the balls are possible? Find this number modulo 10^9+7.
Constraints
* 1≤N,K≤2,000
Input
The input is given from Standard Input in the following format:
N K
Output
Print the number of the possible sequences of the colors of the balls after painting, modulo 10^9+7.
Examples
Input
2 2
Output
4
Input
3 1
Output
1
Input
2 3
Output
14
Input
2000 2000
Output
546381702
Submitted Solution:
```
print(1)
``` | instruction | 0 | 52,934 | 7 | 105,868 |
No | output | 1 | 52,934 | 7 | 105,869 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke loves colorful balls. He has a total of N×K balls, K in each of his favorite N colors. The colors are numbered 1 through N.
He will arrange all of the balls in a row from left to right, in arbitrary order. Then, for each of the N colors, he will paint the leftmost ball of that color into color 0, a color different from any of the N original colors.
After painting, how many sequences of the colors of the balls are possible? Find this number modulo 10^9+7.
Constraints
* 1≤N,K≤2,000
Input
The input is given from Standard Input in the following format:
N K
Output
Print the number of the possible sequences of the colors of the balls after painting, modulo 10^9+7.
Examples
Input
2 2
Output
4
Input
3 1
Output
1
Input
2 3
Output
14
Input
2000 2000
Output
546381702
Submitted Solution:
```
def cmb(n, r, mod):#コンビネーションの高速計算
if ( r<0 or r>n ):
return 0
r = min(r, n-r)
return g1[n] * g2[r] * g2[n-r] % mod
mod = 10**9+7 #出力の制限
N = 4*10**6
g1 = [1, 1] # 元テーブル
g2 = [1, 1] #逆元テーブル
inverse = [0, 1] #逆元テーブル計算用テーブル
for i in range( 2, N + 1 ):
g1.append( ( g1[-1] * i ) % mod )
inverse.append( ( -inverse[mod % i] * (mod//i) ) % mod )
g2.append( (g2[-1] * inverse[-1]) % mod )
N,K=map(int,input().split())
if K==1:
print(1)
exit()
dp=[0 for i in range(N+1)]
dp[N]=1
for i in range(N,-1,-1):
for j in range(i-1,-1,-1):
n=(N-(j+1))*(K-1)+N-i
dp[j]+=(cmb(n+K-2,K-2,mod)*dp[j+1])%mod
dp[j]%=mod
print((dp[0]*g1[N])%mod)
``` | instruction | 0 | 52,935 | 7 | 105,870 |
No | output | 1 | 52,935 | 7 | 105,871 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke loves colorful balls. He has a total of N×K balls, K in each of his favorite N colors. The colors are numbered 1 through N.
He will arrange all of the balls in a row from left to right, in arbitrary order. Then, for each of the N colors, he will paint the leftmost ball of that color into color 0, a color different from any of the N original colors.
After painting, how many sequences of the colors of the balls are possible? Find this number modulo 10^9+7.
Constraints
* 1≤N,K≤2,000
Input
The input is given from Standard Input in the following format:
N K
Output
Print the number of the possible sequences of the colors of the balls after painting, modulo 10^9+7.
Examples
Input
2 2
Output
4
Input
3 1
Output
1
Input
2 3
Output
14
Input
2000 2000
Output
546381702
Submitted Solution:
```
def cmb(n, r, mod):#コンビネーションの高速計算
if ( r<0 or r>n ):
return 0
r = min(r, n-r)
return g1[n] * g2[r] * g2[n-r] % mod
mod = 10**9+7 #出力の制限
N = 4*10**6
g1 = [1, 1] # 元テーブル
g2 = [1, 1] #逆元テーブル
inverse = [0, 1] #逆元テーブル計算用テーブル
for i in range( 2, N + 1 ):
g1.append( ( g1[-1] * i ) % mod )
inverse.append( ( -inverse[mod % i] * (mod//i) ) % mod )
g2.append( (g2[-1] * inverse[-1]) % mod )
N,K=map(int,input().split())
if K==1:
print(1)
exit()
dp=[[0 for i in range(N+1)] for j in range(N+1)]
dp[N][N]=1
for i in range(N,-1,-1):
for j in range(N,-1,-1):
if i!=N:
dp[i][j]+=dp[i+1][j]
if j!=N and i>j:
n=(N-(j+1))*(K-1)+N-i
dp[i][j]+=cmb(n+K-2,K-2,mod)*dp[i][j+1]
dp[i][j]%=mod
print((dp[0][0]*g1[N])%mod)
``` | instruction | 0 | 52,936 | 7 | 105,872 |
No | output | 1 | 52,936 | 7 | 105,873 |
Provide a correct Python 3 solution for this coding contest problem.
Background
The site of Mr. A's house, which lives in a certain city, is surrounded by white walls. Feeling unsatisfied with the wall, Mr. A decided to invite the children in the neighborhood to paint the wall freely. Ask the children to choose their favorite section of the wall and paint it. So how was the wall painted?
Problem
There is a circular white wall as shown below, which consists of sections 0 to N-1.
Figure 1
M children specify the starting position a of this wall and the length L from the starting position, and paint counterclockwise from a to (a + L) mod N (however, a mod N is Represents the remainder when a is divided by N). Here, it is possible to paint over the section painted by another person, in which case it is regarded as one section painted with color. Output the colored sections in descending order. Also, output at the same time how many sections of that size are.
Constrains
Input meets the following conditions
* 2 ≤ N ≤ 100
* 1 ≤ M ≤ 50
* 0 ≤ ai <N
* 1 ≤ Li ≤ N
Input
The input is given in the following format.
N M
a0 L0
a1 L1
...
aM−1 LM−1
The first line is given two integers N, M, separated by blanks, which represent the length of the wall and the number of people who paint the wall, respectively. The following M line is given the start position ai and the length Li of the section to be painted by each person.
Output
The length of the colored section and the total number of the lengths are output in descending order of the colored section.
Examples
Input
5 3
0 1
2 1
3 1
Output
2 1
1 1
Input
4 2
0 2
1 2
Output
3 1
Input
10 3
2 1
4 1
9 2
Output
2 1
1 2 | instruction | 0 | 53,016 | 7 | 106,032 |
"Correct Solution:
```
N, M = map(int, input().split())
L = [0]*(N*2)
for i in range(M):
a, l = map(int, input().split())
for ll in range(a, a+l):
L[ll] = 1
for i in range(N, 2*N):
L[i-N] = max(L[i-N], L[i])
#print(L)
left = 0
i = 0
while L[i]==1:
left += 1
i += 1
if i==N:
print(N, 1)
exit()
A = []
s = 0
#p = 0
for i in range(i, N):
li = L[i]
if li==0:
if s!=0:
A.append(s)
s = 0
else:
s += 1
#print(A, left, s)
if s+left != 0:
A.append(s+left)
A.sort(reverse=True)
v = A[0]
n = 0
for a in A:
if a==v:
n+=1
else:
print(v, n)
n = 1
v = a
print(v, n)
``` | output | 1 | 53,016 | 7 | 106,033 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4. | instruction | 0 | 54,018 | 7 | 108,036 |
Tags: implementation, math, number theory
Correct Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools
from collections import deque,defaultdict,OrderedDict
import collections
def main():
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
#Solving Area Starts-->
zz=0
for _ in range(1):
n=ri()
s=rs()
c=0
ans=0
for i in s:
if i=='1':
c+=1
a=[]
b=[]
for i in range(n):
g,h=ria()
a.append(g)
b.append(h)
lolu=[0]*800
for i in range(n):
if s[i]=='1':
for k in range(b[i]):
lolu[k]+=1
# print(1,b[i]-1)
for k in range(1,150,2):
for j in range(k*a[i]+b[i],(k+1)*a[i]+b[i]):
lolu[j]+=1
# print(k*a[i]+b[i],(k+1)*a[i]+b[i]-1)
else:
for k in range(0,150,2):
for j in range(k*a[i]+b[i],(k+1)*a[i]+b[i]):
lolu[j]+=1
# print(k*a[i]+b[i],(k+1)*a[i]+b[i]-1)
# print("_______________________")
print(max(lolu))
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
``` | output | 1 | 54,018 | 7 | 108,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4. | instruction | 0 | 54,019 | 7 | 108,038 |
Tags: implementation, math, number theory
Correct Solution:
```
import sys
import copy
n = int(input())
res = 0
init_light = list(map(int, list(input())))
for i in range(n):
if init_light[i] == 1:
res += 1
toggle = list()
init_val = 0
for i in range(n):
toggle.append(list(map(int, input().split())) + [0])
init_val = max(toggle[i][1], init_val)
idx = 1
new_init = list()
if n == res:
print(n)
exit(0)
while idx <= 1000:
for i in range(n):
if toggle[i][1] <= idx:
if toggle[i][2] == 0:
init_light[i] += 1
init_light[i] %= 2
toggle[i][2] = toggle[i][0]-1
else:
toggle[i][2] -= 1
light_cnt = 0
for i in range(n):
if init_light[i] == 1:
light_cnt += 1
res = max(res, light_cnt)
idx += 1
print(res)
``` | output | 1 | 54,019 | 7 | 108,039 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4. | instruction | 0 | 54,020 | 7 | 108,040 |
Tags: implementation, math, number theory
Correct Solution:
```
#!/usr/bin/env python3
import sys
def rint():
return map(int, sys.stdin.readline().split())
#lines = stdin.readlines()
n = int(input())
ss = input()
a = []
b = []
for i in range(n):
tmpa, tmpb = rint()
a.append(tmpa)
b.append(tmpb)
rep = 5000
s = []
for i in range(n):
s.append(int(ss[i]))
ans = 0
for t in range(rep):
for i in range(n):
if t < b[i]:
continue
if (t - b[i])%a[i] == 0:
if s[i]:
s[i] = 0
else:
s[i] = 1
ans = max(ans, s.count(1))
print(ans)
``` | output | 1 | 54,020 | 7 | 108,041 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4. | instruction | 0 | 54,021 | 7 | 108,042 |
Tags: implementation, math, number theory
Correct Solution:
```
c = int(input())
c1 = list(input())
c2 = []
c3 = [c1.count('1')]
if c1 == [1] * c:
print(c)
else:
for i in range(c):
c2.append([int(i) for i in input().split()])
for i in range(1, 130):
t = 0
for j in range(len(c2)):
if i < c2[j][1]:
if c1[j] == '1':
t += 1
else:
if ((i - c2[j][1]) // c2[j][0] + 1) % 2 == 1:
if c1[j] == '0':
t += 1
else:
if c1[j] == '1':
t += 1
c3.append(t)
print(max(c3))
``` | output | 1 | 54,021 | 7 | 108,043 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4. | instruction | 0 | 54,022 | 7 | 108,044 |
Tags: implementation, math, number theory
Correct Solution:
```
n=int(input())
status=[None]*n
s=input()
for i in range(n):
status[i]=[0]*10000
a,b=map(int,input().split())
if(s[i]=='0'):
status[i][0]=0
else:
status[i][0]=1
for j in range(1,10000):
if j>=b and (j-b)%a==0:
status[i][j]=status[i][j-1]^1
else:
status[i][j]=status[i][j-1]
ans=0
for i in range(10000):
curr=0
for j in range(n):
curr+=status[j][i]
ans=max(ans,curr)
print(ans)
``` | output | 1 | 54,022 | 7 | 108,045 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4. | instruction | 0 | 54,023 | 7 | 108,046 |
Tags: implementation, math, number theory
Correct Solution:
```
a=int(input())
z=input()
d={}
for i in range(a):
b,c=map(int,input().split())
o=[0,1][z[i]=='1'];k=0;b,c=c,b
for y in range(b):
if o:d[y]=d.get(y,0)+1
while b<=14400:
if o==k:
for l in range(b,b+c):d[l]=d.get(l,0)+1
k^=1
b+=c
print(max(d.values()))
``` | output | 1 | 54,023 | 7 | 108,047 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4. | instruction | 0 | 54,024 | 7 | 108,048 |
Tags: implementation, math, number theory
Correct Solution:
```
n = int(input())
s = list(input())
lis = []
for i in range(n):
a, b = map(int, input().split())
lis.append((a, b))
m = s.count("1")
for i in range(1, 126):
for j in range(n):
if i >= lis[j][1] and (i-lis[j][1])%lis[j][0] == 0:
if s[j] == "0":
s[j] = "1"
else:
s[j] = "0"
m = max(m, s.count("1"))
print(m)
``` | output | 1 | 54,024 | 7 | 108,049 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4. | instruction | 0 | 54,025 | 7 | 108,050 |
Tags: implementation, math, number theory
Correct Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools
from collections import deque,defaultdict,OrderedDict
import collections
def main():
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
#Solving Area Starts-->
zz=0
for _ in range(1):
n=ri()
s=rs()
c=0
ans=0
for i in s:
if i=='1':
c+=1
a=[]
b=[]
for i in range(n):
g,h=ria()
a.append(g)
b.append(h)
lolu=[0]*1000000
for i in range(n):
if s[i]=='1':
for k in range(b[i]):
lolu[k]+=1
# print(1,b[i]-1)
for k in range(1,4000,2):
for j in range(k*a[i]+b[i],(k+1)*a[i]+b[i]):
lolu[j]+=1
# print(k*a[i]+b[i],(k+1)*a[i]+b[i]-1)
else:
for k in range(0,4000,2):
for j in range(k*a[i]+b[i],(k+1)*a[i]+b[i]):
lolu[j]+=1
# print(k*a[i]+b[i],(k+1)*a[i]+b[i]-1)
# print("_______________________")
print(max(lolu))
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
``` | output | 1 | 54,025 | 7 | 108,051 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4.
Submitted Solution:
```
R = lambda: map(int, input().split())
n = int(input())
s = input()
L = []
for i in range(n):
L.append(list(R()))
f = [0]*5001
for i in range(n):
c = int(s[i])
for j in range(L[i][1]):
f[j] += c
c ^= 1
for j in range(L[i][1],5001,L[i][0]):
for k in range(j,min(5001,j+L[i][0])):
f[k] += c
c ^= 1
#print(f)
#print(f)
print(max(f))
``` | instruction | 0 | 54,026 | 7 | 108,052 |
Yes | output | 1 | 54,026 | 7 | 108,053 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4.
Submitted Solution:
```
n = int(input())
s = list(input())
a = []
b = []
max_active = 0
prev = [0] * n
for i in range(n):
prev[i] = int(s[i])
x, y = map(int, input().split())
a.append(x)
b.append(y)
active = sum(prev)
max_active = max(active, max_active)
for i in range(1, 1000):
for j in range(n):
if i >= b[j]:
if (i - b[j]) % a[j] == 0:
if prev[j] == 1:
active -= 1
prev[j] = 0
else:
active += 1
prev[j] = 1
max_active = max(active, max_active)
print(max_active)
``` | instruction | 0 | 54,027 | 7 | 108,054 |
Yes | output | 1 | 54,027 | 7 | 108,055 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4.
Submitted Solution:
```
n = int(input())
s = [int(q) for q in input()]
a = [list(map(int, input().split())) for _ in range(n)]
d = s[::]
ans = sum(d)
for q in range(1179):
for q1 in range(len(a)):
if a[q1][1] <= q and (q-a[q1][1]) % a[q1][0] == 0:
d[q1] = 1-d[q1]
ans = max(ans, sum(d))
print(ans)
``` | instruction | 0 | 54,028 | 7 | 108,056 |
Yes | output | 1 | 54,028 | 7 | 108,057 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4.
Submitted Solution:
```
size = 500
t = [0] * size
n = int(input())
signs = []
for s in input():
signs.append(int(s))
for i in range(n):
a, b = map(int, input().split())
flag = True
if signs[i]:
for j in range(b):
t[j] += 1
start = b + a
end = start + a
else:
start = b
end = start + a
while flag:
for j in range(start, end):
if j >= size:
flag = False
break
t[j] += 1
start = end + a
end = start + a
print(max(t))
``` | instruction | 0 | 54,029 | 7 | 108,058 |
Yes | output | 1 | 54,029 | 7 | 108,059 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4.
Submitted Solution:
```
n=int(input())
s=input()
s=[int(i) for i in s]
lights=[[0 for j in range(1019)] for i in range(n)]
for i in range(n):
a,b=map(int,input().split())
prev=s[i]
lights[i][0]=s[i]
#print(s[i],'efiedfa')
for time in range(1,1004):
if time<b:
lights[i][time]=prev
elif time==b:
x=1
for x in range(1,1004-b):
lights[i][x+time-1]=1-prev
x+=1
if x%a==0:
prev=1-prev
break
maxi=0
for j in range(1002):
curr=0
for i in range(n):
if lights[i][j]:
curr+=1
maxi=max(maxi,curr)
print(maxi)
'''
for i in range(n):
print(lights[i][:20])'''
``` | instruction | 0 | 54,030 | 7 | 108,060 |
No | output | 1 | 54,030 | 7 | 108,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4.
Submitted Solution:
```
# Contest: Codeforces Round #584 - Dasha Code Championship - Elimination Round (rated, open for everyone, Div. 1 + Div. 2) (https://codeforces.com/contest/1209)
# Problem: B: Koala and Lights (https://codeforces.com/contest/1209/problem/B)
def rint():
return int(input())
def rints():
return list(map(int, input().split()))
n = rint()
ini = input()
c = [0 for _ in range(200)]
for i in range(n):
on = ini[i] == '1'
a, b = rints()
for j in range(200):
if j > b and (j - b) % a == 0:
on = not on
c[j] += on
print(max(c))
``` | instruction | 0 | 54,031 | 7 | 108,062 |
No | output | 1 | 54,031 | 7 | 108,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.
After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.
You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.
<image> Here is a graphic for the first example.
Input
The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.
The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".
The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.
Output
Print a single integer — the maximum number of lights that will ever be on at the same time.
Examples
Input
3
101
3 3
3 2
3 1
Output
2
Input
4
1111
3 4
5 2
3 1
3 2
Output
4
Input
6
011100
5 3
5 5
2 4
3 5
4 2
1 5
Output
6
Note
For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).
In the second example, all lights are initially on. So the answer is 4.
Submitted Solution:
```
n=int(input())
status=[None]*n
s=input()
A=[]
for i in range(n):
status[i]=[0]*400000
a,b=map(int,input().split())
A.append(a);
if(s[i]=='0'):
status[i][0]=0
else:
status[i][0]=1
for j in range(1,400000):
if (j-b)%a==0:
status[i][j]=status[i][j-1]^1
else:
status[i][j]=status[i][j-1]
ans=[0]*6
ANS=0
for i in range(100000):
curr=0
for j in range(n):
curr+=status[j][i]
ANS=max(ANS,curr)
for i in range(100000,400000):
curr=[0]*6
for j in range(n):
curr[A[j]]+=status[j][i]
for j in range(6):
ans[j]=max(ans[j],curr[j])
print(max(sum(ans),ANS))
``` | instruction | 0 | 54,033 | 7 | 108,066 |
No | output | 1 | 54,033 | 7 | 108,067 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There are $ N $ balls, each with its own color and value.
There are $ C $ types of ball colors from $ 1 $ to $ C $, and each color has an upper limit on the number of balls that can be selected.
Maximize the total value you get when choosing at most $ M $ balls in total.
Constraints
The input satisfies the following conditions.
* $ 1 \ leq M \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq C \ leq 10 ^ 5 $
* $ 0 \ leq l_i \ leq N $
* $ 1 \ leq c_i \ leq C $
* $ 1 \ leq w_i \ leq 1000 $
Input
The input is given in the following format.
$ N $ $ M $ $ C $
$ l_1 $ $ l_2 $ ... $ l_C $
$ c_1 $ $ w_1 $
$ c_2 $ $ w_2 $
...
$ c_N $ $ w_N $
All inputs are given as integers.
$ N $, $ M $, $ C $ are given on the first line, separated by blanks.
On the second line, the upper limit of the number of balls that can be selected for the color $ i $ ($ 1 \ leq i \ leq C $) is given, separated by blanks.
The color $ c_i $ of the ball $ i $ and the value $ w_i $ ($ 1 \ leq i \ leq N $) are given in the third and subsequent lines of $ N $, separated by blanks.
Output
Output the maximum value you can get on one line.
Examples
Input
3 3 2
1 1
1 1
1 100
2 10
Output
110
Input
3 3 3
1 0 1
1 1
2 100
3 1
Output
2
Input
22 7 26
11 14 15 3 11 7 16 17 1 4 2 19 4 14 16 16 3 13 17 12 7 11 2 20 12 22
6 10
1 3
13 1
16 5
4 1
20 7
18 4
26 6
9 1
12 2
21 1
21 7
18 1
14 5
24 5
6 1
3 1
2 5
21 2
7 6
10 9
15 7
Output
52 | instruction | 0 | 54,746 | 7 | 109,492 |
"Correct Solution:
```
N,M,C=map(int,input().split())
l=list(map(int, input().split()))
CW=[]
for i in range(N):
cw=list(map(int,input().split()))
CW.append(cw)
CW.sort(key=lambda x:x[1],reverse=True)
L=[0]*C
ans=0
count=0
for i in range(N):
if L[CW[i][0]-1]<l[CW[i][0]-1]:
ans+=CW[i][1]
L[CW[i][0]-1]+=1
count+=1
if count==M: break
print(ans)
``` | output | 1 | 54,746 | 7 | 109,493 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There are $ N $ balls, each with its own color and value.
There are $ C $ types of ball colors from $ 1 $ to $ C $, and each color has an upper limit on the number of balls that can be selected.
Maximize the total value you get when choosing at most $ M $ balls in total.
Constraints
The input satisfies the following conditions.
* $ 1 \ leq M \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq C \ leq 10 ^ 5 $
* $ 0 \ leq l_i \ leq N $
* $ 1 \ leq c_i \ leq C $
* $ 1 \ leq w_i \ leq 1000 $
Input
The input is given in the following format.
$ N $ $ M $ $ C $
$ l_1 $ $ l_2 $ ... $ l_C $
$ c_1 $ $ w_1 $
$ c_2 $ $ w_2 $
...
$ c_N $ $ w_N $
All inputs are given as integers.
$ N $, $ M $, $ C $ are given on the first line, separated by blanks.
On the second line, the upper limit of the number of balls that can be selected for the color $ i $ ($ 1 \ leq i \ leq C $) is given, separated by blanks.
The color $ c_i $ of the ball $ i $ and the value $ w_i $ ($ 1 \ leq i \ leq N $) are given in the third and subsequent lines of $ N $, separated by blanks.
Output
Output the maximum value you can get on one line.
Examples
Input
3 3 2
1 1
1 1
1 100
2 10
Output
110
Input
3 3 3
1 0 1
1 1
2 100
3 1
Output
2
Input
22 7 26
11 14 15 3 11 7 16 17 1 4 2 19 4 14 16 16 3 13 17 12 7 11 2 20 12 22
6 10
1 3
13 1
16 5
4 1
20 7
18 4
26 6
9 1
12 2
21 1
21 7
18 1
14 5
24 5
6 1
3 1
2 5
21 2
7 6
10 9
15 7
Output
52 | instruction | 0 | 54,747 | 7 | 109,494 |
"Correct Solution:
```
import heapq
n, m, c = map(int,input().split())
que = [[] for i in range(c + 1)]
can_buy = [0] + list(map(int,input().split()))
for i in range(n):
c, w = map(int, input().split())
heapq.heappush(que[c], w)
if len(que[c]) > can_buy[c]:
heapq.heappop(que[c])
line = [j for i in que for j in i]
line.sort()
ans = sum(line[-m:])
print(ans)
``` | output | 1 | 54,747 | 7 | 109,495 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There are $ N $ balls, each with its own color and value.
There are $ C $ types of ball colors from $ 1 $ to $ C $, and each color has an upper limit on the number of balls that can be selected.
Maximize the total value you get when choosing at most $ M $ balls in total.
Constraints
The input satisfies the following conditions.
* $ 1 \ leq M \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq C \ leq 10 ^ 5 $
* $ 0 \ leq l_i \ leq N $
* $ 1 \ leq c_i \ leq C $
* $ 1 \ leq w_i \ leq 1000 $
Input
The input is given in the following format.
$ N $ $ M $ $ C $
$ l_1 $ $ l_2 $ ... $ l_C $
$ c_1 $ $ w_1 $
$ c_2 $ $ w_2 $
...
$ c_N $ $ w_N $
All inputs are given as integers.
$ N $, $ M $, $ C $ are given on the first line, separated by blanks.
On the second line, the upper limit of the number of balls that can be selected for the color $ i $ ($ 1 \ leq i \ leq C $) is given, separated by blanks.
The color $ c_i $ of the ball $ i $ and the value $ w_i $ ($ 1 \ leq i \ leq N $) are given in the third and subsequent lines of $ N $, separated by blanks.
Output
Output the maximum value you can get on one line.
Examples
Input
3 3 2
1 1
1 1
1 100
2 10
Output
110
Input
3 3 3
1 0 1
1 1
2 100
3 1
Output
2
Input
22 7 26
11 14 15 3 11 7 16 17 1 4 2 19 4 14 16 16 3 13 17 12 7 11 2 20 12 22
6 10
1 3
13 1
16 5
4 1
20 7
18 4
26 6
9 1
12 2
21 1
21 7
18 1
14 5
24 5
6 1
3 1
2 5
21 2
7 6
10 9
15 7
Output
52 | instruction | 0 | 54,748 | 7 | 109,496 |
"Correct Solution:
```
N, M, C = map(int, input().split())
*L, = map(int, input().split())
B = []
for i in range(N):
c, w = map(int, input().split())
B.append((w, c))
B.sort(reverse=1)
ans = 0
for w, c in B:
if L[c-1] > 0:
ans += w
M -= 1
if M == 0:
break
L[c-1] -= 1
print(ans)
``` | output | 1 | 54,748 | 7 | 109,497 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There are $ N $ balls, each with its own color and value.
There are $ C $ types of ball colors from $ 1 $ to $ C $, and each color has an upper limit on the number of balls that can be selected.
Maximize the total value you get when choosing at most $ M $ balls in total.
Constraints
The input satisfies the following conditions.
* $ 1 \ leq M \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq C \ leq 10 ^ 5 $
* $ 0 \ leq l_i \ leq N $
* $ 1 \ leq c_i \ leq C $
* $ 1 \ leq w_i \ leq 1000 $
Input
The input is given in the following format.
$ N $ $ M $ $ C $
$ l_1 $ $ l_2 $ ... $ l_C $
$ c_1 $ $ w_1 $
$ c_2 $ $ w_2 $
...
$ c_N $ $ w_N $
All inputs are given as integers.
$ N $, $ M $, $ C $ are given on the first line, separated by blanks.
On the second line, the upper limit of the number of balls that can be selected for the color $ i $ ($ 1 \ leq i \ leq C $) is given, separated by blanks.
The color $ c_i $ of the ball $ i $ and the value $ w_i $ ($ 1 \ leq i \ leq N $) are given in the third and subsequent lines of $ N $, separated by blanks.
Output
Output the maximum value you can get on one line.
Examples
Input
3 3 2
1 1
1 1
1 100
2 10
Output
110
Input
3 3 3
1 0 1
1 1
2 100
3 1
Output
2
Input
22 7 26
11 14 15 3 11 7 16 17 1 4 2 19 4 14 16 16 3 13 17 12 7 11 2 20 12 22
6 10
1 3
13 1
16 5
4 1
20 7
18 4
26 6
9 1
12 2
21 1
21 7
18 1
14 5
24 5
6 1
3 1
2 5
21 2
7 6
10 9
15 7
Output
52 | instruction | 0 | 54,749 | 7 | 109,498 |
"Correct Solution:
```
n, m, c = map(int, input().split())
l = [0] + list(map(int, input().split()))
w = [tuple(map(int, input().split())) for _ in range(n)]
w.sort(key = lambda x:x[1], reverse = True)
v = 0
p = 0
for i in range(n):
if l[w[i][0]]:
l[w[i][0]] -= 1
v += w[i][1]
p += 1
if p == m:
break
print(v)
``` | output | 1 | 54,749 | 7 | 109,499 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There are $ N $ balls, each with its own color and value.
There are $ C $ types of ball colors from $ 1 $ to $ C $, and each color has an upper limit on the number of balls that can be selected.
Maximize the total value you get when choosing at most $ M $ balls in total.
Constraints
The input satisfies the following conditions.
* $ 1 \ leq M \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq C \ leq 10 ^ 5 $
* $ 0 \ leq l_i \ leq N $
* $ 1 \ leq c_i \ leq C $
* $ 1 \ leq w_i \ leq 1000 $
Input
The input is given in the following format.
$ N $ $ M $ $ C $
$ l_1 $ $ l_2 $ ... $ l_C $
$ c_1 $ $ w_1 $
$ c_2 $ $ w_2 $
...
$ c_N $ $ w_N $
All inputs are given as integers.
$ N $, $ M $, $ C $ are given on the first line, separated by blanks.
On the second line, the upper limit of the number of balls that can be selected for the color $ i $ ($ 1 \ leq i \ leq C $) is given, separated by blanks.
The color $ c_i $ of the ball $ i $ and the value $ w_i $ ($ 1 \ leq i \ leq N $) are given in the third and subsequent lines of $ N $, separated by blanks.
Output
Output the maximum value you can get on one line.
Examples
Input
3 3 2
1 1
1 1
1 100
2 10
Output
110
Input
3 3 3
1 0 1
1 1
2 100
3 1
Output
2
Input
22 7 26
11 14 15 3 11 7 16 17 1 4 2 19 4 14 16 16 3 13 17 12 7 11 2 20 12 22
6 10
1 3
13 1
16 5
4 1
20 7
18 4
26 6
9 1
12 2
21 1
21 7
18 1
14 5
24 5
6 1
3 1
2 5
21 2
7 6
10 9
15 7
Output
52 | instruction | 0 | 54,750 | 7 | 109,500 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
def inpl(): return list(map(int, input().split()))
N, M, C = inpl()
L = {i+1:v for i, v in enumerate(inpl())}
balls = []
for _ in range(N):
c, w = inpl()
balls.append([w, c])
balls = sorted(balls, reverse=True)
i = 0
count = 0
ans = 0
for i in range(N):
w, c = balls[i]
if count == M:
break
if L[c] == 0:
continue
else:
count += 1
ans += w
L[c] -= 1
print(ans)
``` | output | 1 | 54,750 | 7 | 109,501 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There are $ N $ balls, each with its own color and value.
There are $ C $ types of ball colors from $ 1 $ to $ C $, and each color has an upper limit on the number of balls that can be selected.
Maximize the total value you get when choosing at most $ M $ balls in total.
Constraints
The input satisfies the following conditions.
* $ 1 \ leq M \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq C \ leq 10 ^ 5 $
* $ 0 \ leq l_i \ leq N $
* $ 1 \ leq c_i \ leq C $
* $ 1 \ leq w_i \ leq 1000 $
Input
The input is given in the following format.
$ N $ $ M $ $ C $
$ l_1 $ $ l_2 $ ... $ l_C $
$ c_1 $ $ w_1 $
$ c_2 $ $ w_2 $
...
$ c_N $ $ w_N $
All inputs are given as integers.
$ N $, $ M $, $ C $ are given on the first line, separated by blanks.
On the second line, the upper limit of the number of balls that can be selected for the color $ i $ ($ 1 \ leq i \ leq C $) is given, separated by blanks.
The color $ c_i $ of the ball $ i $ and the value $ w_i $ ($ 1 \ leq i \ leq N $) are given in the third and subsequent lines of $ N $, separated by blanks.
Output
Output the maximum value you can get on one line.
Examples
Input
3 3 2
1 1
1 1
1 100
2 10
Output
110
Input
3 3 3
1 0 1
1 1
2 100
3 1
Output
2
Input
22 7 26
11 14 15 3 11 7 16 17 1 4 2 19 4 14 16 16 3 13 17 12 7 11 2 20 12 22
6 10
1 3
13 1
16 5
4 1
20 7
18 4
26 6
9 1
12 2
21 1
21 7
18 1
14 5
24 5
6 1
3 1
2 5
21 2
7 6
10 9
15 7
Output
52 | instruction | 0 | 54,751 | 7 | 109,502 |
"Correct Solution:
```
N,M,C = map(int,input().split())
colors = [int(l) for l in input().split()]
ball = []
#print(len(colors),colors)
for _ in range(0,N):
c,w = map(int,input().split())
ball.append([c,w])
ball.sort(key= lambda x:x[1]) #価値でソートした
ball.reverse()
#print(ball)
take = 0 #ボール取った
cost = 0 #合計金額
i = 0 #リストのナンバリング
while (take <= M - 1) and (i <= N - 1):
if colors[ball[i][0] - 1] == 0:
i += 1
#print(i)
continue
else:
cost = cost + ball[i][1]
colors[ball[i][0] - 1] = colors[ball[i][0] - 1] -1
take += 1
i += 1
#print(i)
print(cost)
``` | output | 1 | 54,751 | 7 | 109,503 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
There are $ N $ balls, each with its own color and value.
There are $ C $ types of ball colors from $ 1 $ to $ C $, and each color has an upper limit on the number of balls that can be selected.
Maximize the total value you get when choosing at most $ M $ balls in total.
Constraints
The input satisfies the following conditions.
* $ 1 \ leq M \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq C \ leq 10 ^ 5 $
* $ 0 \ leq l_i \ leq N $
* $ 1 \ leq c_i \ leq C $
* $ 1 \ leq w_i \ leq 1000 $
Input
The input is given in the following format.
$ N $ $ M $ $ C $
$ l_1 $ $ l_2 $ ... $ l_C $
$ c_1 $ $ w_1 $
$ c_2 $ $ w_2 $
...
$ c_N $ $ w_N $
All inputs are given as integers.
$ N $, $ M $, $ C $ are given on the first line, separated by blanks.
On the second line, the upper limit of the number of balls that can be selected for the color $ i $ ($ 1 \ leq i \ leq C $) is given, separated by blanks.
The color $ c_i $ of the ball $ i $ and the value $ w_i $ ($ 1 \ leq i \ leq N $) are given in the third and subsequent lines of $ N $, separated by blanks.
Output
Output the maximum value you can get on one line.
Examples
Input
3 3 2
1 1
1 1
1 100
2 10
Output
110
Input
3 3 3
1 0 1
1 1
2 100
3 1
Output
2
Input
22 7 26
11 14 15 3 11 7 16 17 1 4 2 19 4 14 16 16 3 13 17 12 7 11 2 20 12 22
6 10
1 3
13 1
16 5
4 1
20 7
18 4
26 6
9 1
12 2
21 1
21 7
18 1
14 5
24 5
6 1
3 1
2 5
21 2
7 6
10 9
15 7
Output
52 | instruction | 0 | 54,752 | 7 | 109,504 |
"Correct Solution:
```
N, M, C = map(int, input().split())
*L, = map(int, input().split())
BALL = [list(map(int, input().split())) for i in range(N)]
BALL.sort(key=(lambda x: x[1]), reverse=1)
ans = 0
for c, w in BALL:
if L[c-1]:
ans += w
L[c-1] -= 1
M -= 1
if M == 0:
break
print(ans)
``` | output | 1 | 54,752 | 7 | 109,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
There are $ N $ balls, each with its own color and value.
There are $ C $ types of ball colors from $ 1 $ to $ C $, and each color has an upper limit on the number of balls that can be selected.
Maximize the total value you get when choosing at most $ M $ balls in total.
Constraints
The input satisfies the following conditions.
* $ 1 \ leq M \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq C \ leq 10 ^ 5 $
* $ 0 \ leq l_i \ leq N $
* $ 1 \ leq c_i \ leq C $
* $ 1 \ leq w_i \ leq 1000 $
Input
The input is given in the following format.
$ N $ $ M $ $ C $
$ l_1 $ $ l_2 $ ... $ l_C $
$ c_1 $ $ w_1 $
$ c_2 $ $ w_2 $
...
$ c_N $ $ w_N $
All inputs are given as integers.
$ N $, $ M $, $ C $ are given on the first line, separated by blanks.
On the second line, the upper limit of the number of balls that can be selected for the color $ i $ ($ 1 \ leq i \ leq C $) is given, separated by blanks.
The color $ c_i $ of the ball $ i $ and the value $ w_i $ ($ 1 \ leq i \ leq N $) are given in the third and subsequent lines of $ N $, separated by blanks.
Output
Output the maximum value you can get on one line.
Examples
Input
3 3 2
1 1
1 1
1 100
2 10
Output
110
Input
3 3 3
1 0 1
1 1
2 100
3 1
Output
2
Input
22 7 26
11 14 15 3 11 7 16 17 1 4 2 19 4 14 16 16 3 13 17 12 7 11 2 20 12 22
6 10
1 3
13 1
16 5
4 1
20 7
18 4
26 6
9 1
12 2
21 1
21 7
18 1
14 5
24 5
6 1
3 1
2 5
21 2
7 6
10 9
15 7
Output
52
Submitted Solution:
```
n, m, c = map(int, input().split())
l = (0,) + tuple(map(int, input().split()))
a = [0] * (c + 1)
w = [tuple(map(int, input().split())) for _ in range(n)]
w.sort(key = lambda x:x[1], reverse = True)
v = 0
for i in range(n):
if a[w[i][0]] < l[w[i][0]]:
v += w[i][1]
a[w[i][0]] += 1
if sum(a) == m:
break
print(v)
``` | instruction | 0 | 54,753 | 7 | 109,506 |
No | output | 1 | 54,753 | 7 | 109,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
There are $ N $ balls, each with its own color and value.
There are $ C $ types of ball colors from $ 1 $ to $ C $, and each color has an upper limit on the number of balls that can be selected.
Maximize the total value you get when choosing at most $ M $ balls in total.
Constraints
The input satisfies the following conditions.
* $ 1 \ leq M \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq C \ leq 10 ^ 5 $
* $ 0 \ leq l_i \ leq N $
* $ 1 \ leq c_i \ leq C $
* $ 1 \ leq w_i \ leq 1000 $
Input
The input is given in the following format.
$ N $ $ M $ $ C $
$ l_1 $ $ l_2 $ ... $ l_C $
$ c_1 $ $ w_1 $
$ c_2 $ $ w_2 $
...
$ c_N $ $ w_N $
All inputs are given as integers.
$ N $, $ M $, $ C $ are given on the first line, separated by blanks.
On the second line, the upper limit of the number of balls that can be selected for the color $ i $ ($ 1 \ leq i \ leq C $) is given, separated by blanks.
The color $ c_i $ of the ball $ i $ and the value $ w_i $ ($ 1 \ leq i \ leq N $) are given in the third and subsequent lines of $ N $, separated by blanks.
Output
Output the maximum value you can get on one line.
Examples
Input
3 3 2
1 1
1 1
1 100
2 10
Output
110
Input
3 3 3
1 0 1
1 1
2 100
3 1
Output
2
Input
22 7 26
11 14 15 3 11 7 16 17 1 4 2 19 4 14 16 16 3 13 17 12 7 11 2 20 12 22
6 10
1 3
13 1
16 5
4 1
20 7
18 4
26 6
9 1
12 2
21 1
21 7
18 1
14 5
24 5
6 1
3 1
2 5
21 2
7 6
10 9
15 7
Output
52
Submitted Solution:
```
n, m, c = map(int, input().split())
l = (0,) + tuple(map(int, input().split()))
a = [0] * (c + 1)
w = [tuple(map(int, input().split())) for _ in range(n)]
w.sort(key = lambda x:x[1], reverse = True)
v = 0
p = 0
for i in range(n):
if l[w[i][0]]:
l[w[i][0]] -= 1
v += w[i][1]
p += 1
if p == m:
break
print(v)
``` | instruction | 0 | 54,754 | 7 | 109,508 |
No | output | 1 | 54,754 | 7 | 109,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Problem
There are $ N $ balls, each with its own color and value.
There are $ C $ types of ball colors from $ 1 $ to $ C $, and each color has an upper limit on the number of balls that can be selected.
Maximize the total value you get when choosing at most $ M $ balls in total.
Constraints
The input satisfies the following conditions.
* $ 1 \ leq M \ leq N \ leq 10 ^ 5 $
* $ 1 \ leq C \ leq 10 ^ 5 $
* $ 0 \ leq l_i \ leq N $
* $ 1 \ leq c_i \ leq C $
* $ 1 \ leq w_i \ leq 1000 $
Input
The input is given in the following format.
$ N $ $ M $ $ C $
$ l_1 $ $ l_2 $ ... $ l_C $
$ c_1 $ $ w_1 $
$ c_2 $ $ w_2 $
...
$ c_N $ $ w_N $
All inputs are given as integers.
$ N $, $ M $, $ C $ are given on the first line, separated by blanks.
On the second line, the upper limit of the number of balls that can be selected for the color $ i $ ($ 1 \ leq i \ leq C $) is given, separated by blanks.
The color $ c_i $ of the ball $ i $ and the value $ w_i $ ($ 1 \ leq i \ leq N $) are given in the third and subsequent lines of $ N $, separated by blanks.
Output
Output the maximum value you can get on one line.
Examples
Input
3 3 2
1 1
1 1
1 100
2 10
Output
110
Input
3 3 3
1 0 1
1 1
2 100
3 1
Output
2
Input
22 7 26
11 14 15 3 11 7 16 17 1 4 2 19 4 14 16 16 3 13 17 12 7 11 2 20 12 22
6 10
1 3
13 1
16 5
4 1
20 7
18 4
26 6
9 1
12 2
21 1
21 7
18 1
14 5
24 5
6 1
3 1
2 5
21 2
7 6
10 9
15 7
Output
52
Submitted Solution:
```
n, m, c = map(int, input().split())
l = (0,) + tuple(map(int, input().split()))
a = [0] * (c + 1)
w = [tuple(map(int, input().split())) for _ in range(n)]
w.sort(key = lambda x:x[1], reverse = True)
v = 0
for i in range(m):
if a[w[i][0]] < l[w[i][0]]:
v += w[i][1]
a[w[i][0]] += 1
if sum(a) == m:
break
print(v)
``` | instruction | 0 | 54,755 | 7 | 109,510 |
No | output | 1 | 54,755 | 7 | 109,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3 | instruction | 0 | 55,650 | 7 | 111,300 |
Tags: brute force
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
from collections import defaultdict
from itertools import permutations
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
n, q = map(int, input().split())
zab = [0] * 5001
a = []
for i in range(q):
l, r = map(int, input().split())
a.append([l, r])
for j in range(l, r + 1):
zab[j] += 1
ans = 0
for i in range(q):
fl = 0
d = zab.copy()
for j in range(a[i][0], a[i][1] + 1):
d[j] -= 1
for j in range(5001):
if d[j] > 0:
fl += 1
b = [0] * 5001
for j in range(1, n + 1):
b[j] = b[j - 1]
if d[j] == 1:
b[j] += 1
for j in range(i + 1, q):
ans = max(ans, fl - b[a[j][1]] + b[a[j][0] - 1])
print(ans)
``` | output | 1 | 55,650 | 7 | 111,301 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3 | instruction | 0 | 55,651 | 7 | 111,302 |
Tags: brute force
Correct Solution:
```
n,q=map(int,input().split())
sec=[list(map(int,input().split())) for _ in range(q)]
sec=sorted(sec,key=lambda x:(x[0],x[1]))
fence=[0]*(n+1)
for i in sec:
x,y=i[0],i[1]
x-=1;y-=1
fence[x]+=1
fence[y+1]-=1
for i in range(1,n+1):
fence[i]+=fence[i-1]
zeroes=[0]*(n);ones=[0]*(n);twos=[0]*(n)
zeroes[0]=1 if fence[0]==0 else 0
ones[0]=1 if fence[0]==1 else 0
twos[0]=1 if fence[0]==2 else 0
for i in range(1,n):
if fence[i]==0:
zeroes[i]+=zeroes[i-1]+1
else:
zeroes[i]=zeroes[i-1]
for i in range(1,n):
if fence[i]==1:
ones[i]+=ones[i-1]+1
else:
ones[i]=ones[i-1]
for i in range(1,n):
if fence[i]==2:
twos[i]+=twos[i-1]+1
else:
twos[i]=twos[i-1]
np=0
for i in range(q):
x1,y1=sec[i][0],sec[i][1]
x1-=1;y1-=1
co1=co2=ct=0
for j in range(i+1,q):
x2,y2=sec[j][0],sec[j][1]
x2-=1;y2-=1
co1=ones[y1]-(0 if x1==0 else ones[x1-1])
co2=ones[y2]-(0 if x2==0 else ones[x2-1])
if x2<=y1:
ct=twos[min(y1,y2)]-(0 if x2==0 else twos[x2-1])
else:
ct=0
np=max(np,n-(co1+co2+ct+zeroes[-1]))
#print(i,j,np,co1,co2,ct,zeroes[-1],x2,y1)
print(np)
``` | output | 1 | 55,651 | 7 | 111,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3 | instruction | 0 | 55,652 | 7 | 111,304 |
Tags: brute force
Correct Solution:
```
import collections
n , q = map(int , input().split())
sections = [0]*n
p = []
for _ in range(q):
l , r = map(int , input().split())
p.append((l,r))
for j in range(l,r+1):
sections[j-1]+=1
aux = n-collections.Counter(sections)[0]
number1 = [0]*n
number2 = [0]*n
for i in range(n):
if(sections[i]==1):
for j in range(i,n):
number1[j]+=1
elif(sections[i]==2):
for j in range(i,n):
number2[j]+=1
ans = -float('inf')
for i in range(len(p)):
for j in range(len(p)):
if(j>i):
a, b = p[i]
c, d = p[j]
if(a>c):
a , c = c , a
b , d = d , b
aux1 = number1[b-1]-number1[a-1]+1*(sections[a-1]==1)
aux2 = number1[d-1]-number1[c-1]+1*(sections[c-1]==1)
aux3 = abs(number2[c-1]-number2[min(b,d)-1])+1*(sections[c-1]==2)
if(b<c): aux3 = 0
ans = max(ans , aux-(aux1+aux2+aux3))
print(ans)
``` | output | 1 | 55,652 | 7 | 111,305 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3 | instruction | 0 | 55,653 | 7 | 111,306 |
Tags: brute force
Correct Solution:
```
n,q=map(int,input().split())
arr=[[] for i in range(n)]
for i in range(q):
l,r=map(int,input().split())
for j in range(l-1,r):
arr[j].append(i)
count=[[0 for i in range(q)] for i in range(q)]
total=0
for i in arr:
if len(i) != 0:
total += 1
if len(i) == 1:
for j in range(q):
if j == i[0]:
continue
count[i[0]][j] += 1
count[j][i[0]] += 1
elif len(i) == 2:
count[i[0]][i[1]] += 1
count[i[1]][i[0]] += 1
maxi=-1
for i in range(q):
for j in range(q):
if i!=j:
maxi=max(maxi,total-count[i][j])
print(maxi)
``` | output | 1 | 55,653 | 7 | 111,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3 | instruction | 0 | 55,654 | 7 | 111,308 |
Tags: brute force
Correct Solution:
```
n, q = map(int, input().split())
painters = []
sections = [0] * (n + 1)
for i in range(q):
l, r = map(int, input().split())
l -= 1
r -= 1
painters.append([l, r])
sections[l] += 1
sections[r + 1] -= 1
cnt1 = [0] * (n + 1)
cnt2 = [0] * (n + 1)
p = 0
total = 0
for i in range(n):
p += sections[i]
if p == 1:
cnt1[i + 1] = cnt1[i] + 1
else:
cnt1[i + 1] = cnt1[i]
if p == 2:
cnt2[i + 1] = cnt2[i] + 1
else:
cnt2[i + 1] = cnt2[i]
if p > 0:
total += 1
ans = 0
for i in range(q - 1):
for j in range(i + 1, q):
[l1, r1] = painters[i]
[l2, r2] = painters[j]
l = max(l1, l2)
r = min(r1, r2)
if l <= r:
t = total - (cnt2[r + 1] - cnt2[l]) - (cnt1[max(r1, r2) + 1] - cnt1[min(l1, l2)])
ans = max(ans, t)
else:
t = total - (cnt1[r1 + 1] - cnt1[l1]) - (cnt1[r2 + 1] - cnt1[l2])
ans = max(ans, t)
print(ans)
``` | output | 1 | 55,654 | 7 | 111,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3 | instruction | 0 | 55,655 | 7 | 111,310 |
Tags: brute force
Correct Solution:
```
import sys
input=sys.stdin.readline
n,q=map(int,input().split())
ar=[]
for i in range(q):
ar.append(list(map(int,input().split())))
li=[0]*(n+1)
for i in range(q):
for j in range(ar[i][0],ar[i][1]+1):
li[j]+=1
dic={}
matrix=[]
for i in range(q+1):
tem=[]
for j in range(q+1):
tem.append(0)
matrix.append(tem.copy())
for i in range(1,n+1):
mem=[]
if(li[i]==1):
for j in range(q):
if(ar[j][0]<=i and ar[j][1]>=i):
mem.append(j)
break
elif(li[i]==2):
for j in range(q):
if(ar[j][0]<=i and ar[j][1]>=i):
mem.append(j)
break
for j in range(q):
if(ar[j][0]<=i and ar[j][1]>=i and mem[0]!=j):
mem.append(j)
break
if(li[i]==1):
matrix[mem[0]][0]+=1
elif(li[i]==2):
matrix[mem[0]][mem[1]]+=1
su=0
for i in range(1,n+1):
if(li[i]!=0):
su+=1
ans=0
for i in range(q):
for j in range(i+1,q):
s1=matrix[i][0]
s2=matrix[j][0]
s3=matrix[i][j]
ans=max(ans,su-s1-s2-s3)
print(ans)
``` | output | 1 | 55,655 | 7 | 111,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3 | instruction | 0 | 55,656 | 7 | 111,312 |
Tags: brute force
Correct Solution:
```
import sys,os,io
from sys import stdin
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop
from bisect import bisect_left , bisect_right
import math
alphabets = list('abcdefghijklmnopqrstuvwxyz')
def isPrime(x):
for i in range(2,x):
if i*i>x:
break
if (x%i==0):
return False
return True
def ncr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return (num * pow(den,
p - 2, p)) % p
def primeFactors(n):
l = []
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(int(i))
n = n / i
if n > 2:
l.append(n)
return list(set(l))
def power(x, y, p) :
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def si():
return input()
def prefix_sum(arr):
r = [0] * (len(arr)+1)
for i, el in enumerate(arr):
r[i+1] = r[i] + el
return r
def divideCeil(n,x):
if (n%x==0):
return n//x
return n//x+1
def ii():
return int(input())
def li():
return list(map(int,input().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
def power_set(L):
cardinality=len(L)
n=2 ** cardinality
powerset = []
for i in range(n):
a=bin(i)[2:]
subset=[]
for j in range(len(a)):
if a[-j-1]=='1':
subset.append(L[j])
powerset.append(subset)
powerset_orderred=[]
for k in range(cardinality+1):
for w in powerset:
if len(w)==k:
powerset_orderred.append(w)
return powerset_orderred
def fastPlrintNextLines(a):
# 12
# 3
# 1
#like this
#a is list of strings
print('\n'.join(map(str,a)))
def sortByFirstAndSecond(A):
A = sorted(A,key = lambda x:x[0])
A = sorted(A,key = lambda x:x[1])
return list(A)
#__________________________TEMPLATE__________________OVER_______________________________________________________
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w")
# else:
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def solve():
n,q = li()
a = [[] for i in range(n)]
for i in range(q):
l,r = li()
for j in range(l-1,r):
a[j].append(i)
count = [[0 for i in range(q)] for j in range(q)]
ind = -1
total = 0
for i in a:
if len(i)!=0:
total+=1
if len(i)==1:
for j in range(q):
if j==i[0]:
continue
count[i[0]][j]+=1
count[j][i[0]]+=1
elif len(i)==2:
count[i[0]][i[1]]+=1
count[i[1]][i[0]]+=1
ans = -1
for i in range(q):
for j in range(i+1,q):
# if i!=j:
ans = max(ans,total-count[i][j])
print(ans)
t = 1
# t = int(input())
for _ in range(t):
solve()
``` | output | 1 | 55,656 | 7 | 111,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3 | instruction | 0 | 55,657 | 7 | 111,314 |
Tags: brute force
Correct Solution:
```
from operator import itemgetter
n,q=map(int,input().split())
cnt=0
ans=[0]*(n)
arr=[0]*q
for i in range(q):
arr[i]=list(map(int,input().split()))
for j in range(arr[i][0]-1,arr[i][1],1):
ans[j]+=1
if ans[j]==1:
cnt+=1
cnt1=[0]*(n+1)
cnt2=[0]*(n+1)
# print("ans",*ans)
for i in range(n):
cnt1[i+1]=cnt1[i]
cnt2[i+1]=cnt2[i]
if ans[i]==1:
cnt1[i+1]+=1
if ans[i]==2:
cnt2[i+1]+=1
# print(cnt2)
mac=0
for i in range(q):
for j in range(i+1,q,1):
delete=cnt1[arr[i][1]]-cnt1[arr[i][0]-1]+cnt1[arr[j][1]]-cnt1[arr[j][0]-1]
if arr[j][0]>arr[i][1] or arr[j][1]<arr[i][0]:
pass
elif arr[j][0]<=arr[i][1]:
# print("****",cnt2[min(arr[i][1],arr[j][1])],cnt2[max(arr[j][0]-1,arr[i][0]-1)])
delete+=cnt2[min(arr[i][1],arr[j][1])]-cnt2[max(arr[j][0]-1,arr[i][0]-1)]
# print(i,j,delete)
if cnt-delete>mac:
mac=cnt-delete
print(mac)
``` | output | 1 | 55,657 | 7 | 111,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3
Submitted Solution:
```
n, q = map(int, input().split())
a = []
ar = [0 for i in range(n + 1)]
for i in range(q):
l, r = map(int, input().split())
l -= 1
r -= 1
a.append((l, r))
ar[l] += 1
ar[r + 1] += -1
plus = 0
for i in range(n):
plus += ar[i]
ar[i] = plus
ans = 0
for i in range(q):
for j in range(a[i][0], a[i][1] + 1):
ar[j] -= 1
pref = [0]
count = 0
for pos in range(n):
if ar[pos] > 0:
count += 1
value = 0
if ar[pos] == 1:
value = 1
pref.append(value + pref[-1])
for pos in range(q):
if pos != i:
ans = max(ans, count - (pref[a[pos][1] + 1] - pref[a[pos][0]]))
for j in range(a[i][0], a[i][1] + 1):
ar[j] += 1
print(ans)
``` | instruction | 0 | 55,658 | 7 | 111,316 |
Yes | output | 1 | 55,658 | 7 | 111,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3
Submitted Solution:
```
def solve(N, Q, intervals):
# dp[i]: most painted sections at j-th section
dp = [0] * (N + 1)
# leftmost idx that cover i-th section
L = [i + 1 for i in range(N + 1)]
for l, r in intervals:
for i in range(l, r + 1, 1):
L[i] = min(L[i], l)
for _ in range(Q - 2):
for i in range(N, 0, -1):
l = L[i]
dp[i] = max(
dp[i],
dp[l - 1] + i - l + 1
)
for i in range(1, N + 1, 1):
dp[i] = max(dp[i], dp[i - 1])
return dp[N]
N, Q = map(int, input().split())
intervals = []
for _ in range(Q):
l, r = map(int, input().split())
intervals.append((l, r))
print(solve(N, Q, intervals))
``` | instruction | 0 | 55,659 | 7 | 111,318 |
Yes | output | 1 | 55,659 | 7 | 111,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3
Submitted Solution:
```
n, q = map(int, input().split())
a = []
for i in range(q):
l, r = map(int, input().split())
l -= 1
r -= 1
a.append([l, r])
ct = [0] * (n + 1)
for i in a:
ct[i[0]] += 1
ct[i[1] + 1] -= 1
ones, twos = [0] * n, [0] * n
s = 0
for i in range(n):
if i > 0:
ct[i] += ct[i - 1]
ones[i] += ones[i - 1]
twos[i] += twos[i - 1]
if ct[i] == 1:
ones[i] += 1
elif ct[i] == 2:
twos[i] += 1
if ct[i] != 0:
s += 1
ones.append(0)
twos.append(0)
ans = 0
for i in range(q):
for j in range(i + 1, q):
rem = 0;
rem += ones[a[i][1]] - ones[a[i][0] - 1]
rem += ones[a[j][1]] - ones[a[j][0] - 1]
l, r = max(a[i][0], a[j][0]), min(a[i][1], a[j][1])
if r >= l:
rem += twos[r] - twos[l - 1]
ans = max(ans, s - rem)
print(ans)
``` | instruction | 0 | 55,660 | 7 | 111,320 |
Yes | output | 1 | 55,660 | 7 | 111,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
from copy import copy
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
MOD = 10 ** 9 + 7
N,Q=MAP()
imos=[0]*(N+2)
Pts=[None]*Q
for i in range(Q):
l,r=MAP()
Pts[i]=(l,r)
imos[l]+=1
imos[r+1]-=1
for i in range(N+1):
imos[i+1]+=imos[i]
mx=0
for i in range(Q):
cp=copy(imos)
l,r=Pts[i]
for j in range(l, r+1):
cp[j]-=1
sm=0
cnt1=[0]*(N+2)
for j in range(1, N+1):
if cp[j]>0:
sm+=1
if cp[j]==1:
cnt1[j]+=1
cnt1[j+1]+=cnt1[j]
for j in range(i+1, Q):
l2,r2=Pts[j]
mx=max(mx, sm-(cnt1[r2]-cnt1[l2-1]))
print(mx)
``` | instruction | 0 | 55,661 | 7 | 111,322 |
Yes | output | 1 | 55,661 | 7 | 111,323 |
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