Flipping Game 翻转游戏

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
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.

可以说是水题，一股脑遍历就行了 花了我近半个小时，一直超时，最后才发现memset这个函数是罪魁祸首 ⊙﹏⊙b汗........
代码很容易懂

代码如下：

#include <stdio.h>

#include <string.h>

int main()

{

 int n;

 while(scanf("%d",&n)==1)

 {

  int a[100],b[100],p=0;

  int count[10001]={0};

  int i,j,l,m,q;

 // memset(count,0,sizeof(count));

  for(l=0;l<n;l++)

   scanf("%d",&a[l]);

    for(i=0;i<=l;i++)  

 {  

  for(j=l;j>i;j--)

   {

    for(l=0;l<n;l++)  

    b[l]=a[l];  

   for(m=i;m<=j;m++)

     if(a[m]==1)     

  b[m]=0;  

    else if(a[m]==0)   

    b[m]=1;  

   for(q=0;q<n;q++)   

   if(b[q]==1)     

  count[p]++;     

 p++;  

  }  

 }

   int max,tmp;

   max=count[0];

   for(i=1;i<p;i++)  

   if(count[i]>max)

     max=count[i];

 

   printf("%d\n",max);

 }

 return 0;

}