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

Flipping Game

time limit per test

memory limit per test

input

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(int j=i; j<n; j++){                int sum = 0;                for(int k=0; k<n; k++){                    if(k>=i && k<=j) sum += a[k]^1;          //用异或^取反                    else sum += a[k];                }                if(ans < sum)  ans = sum;            }        }        cout<<ans<<endl;    }    return 0;}