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.
Examples
Input
5 1 0 0 1 0
Output
4
Input
4 1 0 0 1
Output
题意：求把某个区间翻转（即1变0，0变1）之后，所有数加起来的和最大，即求最大和
坑点：必须反转一次
#include"stdio.h"
#include"string.h"
#include"iostream"
#include"algorithm"
using namespace std;
int dp[101][101],b[101];
int main()
{
int i,j,k=0,n,m,s=0;
cin>>n;
for(i=1; i<=n; i++)
{
cin>>b[i];
s+=b[i];
}
if(s!=n)
{
for(i=1; i<=n; i++)
{
dp[i][i]=dp[i-1][i-1];
for(j=i; j<=n; j++)
if(b[j]==0)
dp[i][j]=dp[i][j-1]+1;
else
dp[i][j]=dp[i][j-1]-1;
}
for(i=1; i<=n; i++)
for(j=1; j<=n; j++)
k=max(dp[i][j],k);
s=s+k;
}
else
s=s-1;
printf("%d",s);
return 0;
}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.