（CodeForces - 327A）Flipping Game

（CodeForces - 327A）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 x means 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.

Examples

input

5
1 0 0 1 0

output

4

input

4
1 0 0 1

output

4

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.

题目大意：有n个0，1序列，在区间[i,j]进行一次翻转，会使得0变成1，1变成0，问翻转一次使得这n个序列中1的个数最多是多少。

思路：因为n最大100，所以直接暴力枚举需要翻的区间0( n3 )，即可。

#include
#include
#include
using namespace std;

const int maxn=105;
int a[maxn];
int sum[2];

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        int cnt=0;
        for(int i=0;iscanf("%d",a+i);
            if(a[i]) cnt++;
        }
        if(cnt==n)//如果全是1得翻一张牌，不能不翻 
        {
            printf("%d\n",n-1);
            continue;
        }
        int ans=cnt;
        for(int i=0;ifor(int j=i;jmemset(sum,0,sizeof(sum));
                for(int k=i;k<=j;k++) sum[a[k]]++;
                if(sum[0]>sum[1])
                    ans=max(ans,cnt+sum[0]-sum[1]); 
            }
        printf("%d\n",ans);
    }
    return 0;
}