枚举

G - Flipping Game

Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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

He writes n integers a₁, a₂, ..., a_n. 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 a_k 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, ..., a_n. 

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 Input

Input

5
1 0 0 1 0

Output

4

Input

4
1 0 0 1

Output

4

Sample Output

Hint

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.

分析：n值不大，暴力破解就行，用三次循环，第一层从0~n移动 i，第二层移动 j,  第三层移动范围。

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
int main()
{
        int a[105],n,num[2];
        cin>>n;
        int ans=0;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
            if(a[i]==1)ans++;
        }
        int pos=ans;
        if(pos==n)
        printf("%d\n",n-1);
        else
        {
            for(int i=0;i<n;i++)
                for(int j=i;j<n;j++)
            {
                 memset(num,0,sizeof(num));
			   for(int k=i;k<=j;k++)
					num[a[k]]++;
				if(num[0]>num[1])
					ans=max(ans,pos+num[0]-num[1]);
            }
            printf("%d\n",ans);
        }
}