Kadane Algorithm

Kadane Algorithm

　　求一个数组里连续子序列的和最大。

for(int i=1;i<=n;i++)
    {
        nmax+=a[i];
        maxx=max(maxx,nmax);
        nmax=max(nmax,0);
    }

A. Flipping Game

Description

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 akfor 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

正确解法：

一直在想O（n）怎么写

题目是说 把一个子序列翻转，求最大1的个数。

我们把1的收益为 -1,0的收益为 1

求一个序列的最大收益，也就是求这个序列的最大子序列和

最大收益加上原本1的个数就是答案了。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 3000000+10;
const ll inf = 1e17;
int n;
int a[150],yi,maxx=-999999,nmax;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if(a[i]==1)
        {
            yi++;
            a[i]=-1;
        }
        else
            a[i]=1;
    }
    for(int i=1;i<=n;i++)
    {
        nmax+=a[i];
        maxx=max(maxx,nmax);
        nmax=max(nmax,0);
    }
    printf("%d
",maxx+yi);

    return 0;
}