Rumah  >  Artikel  >  hujung hadapan web  >  Codeforces Round #191 (Div. 2)-A. Flipping Game_html/css_WEB-ITnose

Codeforces Round #191 (Div. 2)-A. Flipping Game_html/css_WEB-ITnose

WBOY
WBOYasal
2016-06-24 11:55:091807semak imbas

Flipping Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Iahub got bored, so he invented a game to be played on paper.

He writes n integers a1,?a2,?...,?an. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indices i and j (1?≤?i?≤?j?≤?n) and flips all values ak for which their positions are in range [i,?j] (that is i?≤?k?≤?j). Flip the value of xmeans to apply operation x?=?1 - x.

The goal of the game is that after exactly one move to obtain the maximum number of ones. Write a program to solve the little game of Iahub.

Input

The first line of the input contains an integer n (1?≤?n?≤?100). In the second line of the input there are n integers: a1,?a2,?...,?an. It is guaranteed that each of those n values is either 0 or 1.

Output

Print an integer ? the maximal number of 1s that can be obtained after exactly one move.

Sample test(s)

input

51 0 0 1 0

output

input

41 0 0 1

output

Note

In the first case, flip the segment from 2 to 5 (i?=?2,?j?=?5). That flip changes the sequence, it becomes: [1 1 1 0 1]. So, it contains four ones. There is no way to make the whole sequence equal to [1 1 1 1 1].

In the second case, flipping only the second and the third element (i?=?2,?j?=?3) will turn all numbers into 1.







解题思路:题意是讲有一个0,1序列,现允许你对任意一个子序列取反,问操作后得到的序列,最多能有多少个1。

数据不大,直接暴力。直接两层循环枚举取反序列的起点和终点,统计每种情况下1的个数,保存一个最大值即可。






AC代码:

#include <iostream>#include <cstdio>using namespace std;int a[105];int main(){//  freopen("in.txt","r",stdin);    int n;    while(cin>>n){        for(int i =0; i<n i cin>>a[i];        int ans = 0;        for(int i=0; i<n i for j="i;" int sum="0;" k="0;" if>=i && k  <br>  <br>  <p></p> </n></n></cstdio></iostream>
Kenyataan:
Kandungan artikel ini disumbangkan secara sukarela oleh netizen, dan hak cipta adalah milik pengarang asal. Laman web ini tidak memikul tanggungjawab undang-undang yang sepadan. Jika anda menemui sebarang kandungan yang disyaki plagiarisme atau pelanggaran, sila hubungi admin@php.cn