Codeforces Round #191 (Div. 2)-A. Flipping Game

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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 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.

题目很简单：但是本人还是很菜，做了很久

这道题，我的思路是看了别人代码综合写出的：对于给出的每个数如果它是0，那么就加1如果它是1，那么就减1并且把1的个数记录起来看减了多少次，并且每次加1或者减1后要与上一次的和相比较大小，这样就得出翻转后最大数字和（如果我说的不太明确那么就直接看代码吧）

AC代码：

#include<iostream>
#include<cstring>
const int MAX=101;
int s[MAX];
using namespace std;
int main()
{
int n,m,i,j,q,p;
cin>>n;
p=0;
j=0;
m=0;
q=-1;
for(i=1;i<=n;i++)
{
cin>>s[i];
if(s[i]==1)//减1
{
p=max(p-1,-1);
m++;
}
else if(s[i]==0)//加1
{
p=max(p+1,1);
}
q=max(q,p);//与上一次比较
}
cout<<q+m<<endl;//输出后加上被减去的1就是最大和
return 0;
}