A. 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, ..., 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.

Sample test(s)

input

5
1 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.

解题说明：这道题判断如何选择区间进行01变换让数列中的1个数最多，可以用暴力做法来做，每选择一个区间求出一个值，最后找到一个最大值。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include<set>
#include <algorithm>
using namespace std;

int main()
{
int n,a[100],i,j,k,c=0,max=0,b[100];
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<n;i++)
{
for(j=i;j<n;j++)
{
for(k=0;k<n;k++)
{
b[k]=a[k];
}
for(k=i;k<=j;k++)
{
b[k]=1-b[k];
}
c=0;
for(k=0;k<n;k++)
{
if(b[k]==1)
{
c++;
}
}
if(c>max)
{
max=c;
}
}
}
printf("%d",max);
return 0;
}