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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.
题意： 给定一个序列，找出一个子区间，中的1变为0, 0变为1，求最终这个序列中有几个1。
思路 ：暴力枚举区间，我做了小改动，将原序列的0变为1, 1变为-1，求区间和的最大值，然后加上区间外和区间内的1的个数。
代码：
#include<stdio.h>
int main()
{int a[10010],b[10010];int n;while(~scanf("%d",&n)){for(int i=0;i<n;i++)scanf("%d",&a[i]);for(int i=0;i<n;i++)if(a[i]==0)b[i]=1;elseb[i]=-1;int p=0,sum=0,max=-4324,z,y;for(int i=p;i<n;i++){sum+=b[i];if(sum>max){max=sum;z=p;y=i;}if(sum<0){sum=0;p=i+1;}}int s=0;for(int i=0;i<z;i++)if(a[i]==1)s++; for(int i=y+1;i<n;i++)if(a[i]==1)s++;int ss=0;for(int i=z;i<=y;i++)if(a[i]==1)ss++;printf("%d\n",s+max+ss); }return 0;
}
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