codeforces 327A

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice 

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 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, ..., 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

51 0 0 1 0

Output

4

Input

41 0 0 1

Output

4

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.

题意：N张牌， 每张牌是0或1， 让你随意找区间，然后将这个区间里的数翻转(0变成1，1变成0)， 求翻转后这N张牌中1最多有几个？

【】暴力枚举【】

#include<stdio.h>int main(){    int s[101], t[101];    int n, ans, i, j, k, kk, max;    while(~scanf("%d", &n))    {        max = -1;        for(i=0; i<n; i++)            scanf("%d", &s[i]);        for(i=0; i<n; i++)        {            for(j=i; j<n; j++)            {                for(k=0; k<n; k++)                   t[k] = s[k];                for(kk=i; kk<=j; kk++)                {                    if(t[kk])                       t[kk] = 0;                    else t[kk] = 1;                }                ans = 0;               for(kk=0; kk<n; kk++)               {                   if(t[kk])                      ans++;               }               if(ans>max)                  max = ans;            }        }        printf("%d\n", max);    }    return 0;}