Flipping Game（枚举）

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, ..., a_n. Each of those integers can be either 0 or 1. He's allowed to do exactly one move: he chooses two indicesi and j (1 ≤ i ≤ j ≤ n) and flips all valuesa_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 aren 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 test(s)

Input

51 0 0 1 0

Output

4

Input

41 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：有n张牌，只有0和1，问在[i，j]范围内翻转一次使1的数量最多。输出1最多的牌的数量

#include <stdio.h>#include <string.h>#include <stdlib.h>int main(){    int n,i,j,k,t;    int a[110];    int sum[2];    int cnt=0;    while(~scanf("%d",&n))    {        cnt=0;        for(i=0;i<n;i++)        {            scanf("%d",&a[i]);            if(a[i]==1)                cnt++;//记录开始时1的牌数        }        t=cnt;        if(cnt==n)        {            printf("%d\n",n-1);//如果全是1的话 你得翻一张牌 所以剩下的最大数为总数-1        }        else        {            for(i=0; i<n; i++)                for(j=i; j<n; j++)                {                    memset(sum,0,sizeof(sum));                    for(k=i; k<=j; k++)                        sum[a[k]]++;                    if(sum[0]>sum[1])                        {                            if(cnt<t+sum[0]-sum[1])                            {                                cnt=t+sum[0]-sum[1];                            }                        }                }            printf("%d\n",cnt);        }    }    return 0;}