Flipping Game



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 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 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, ..., 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.

一些01，改变一个区间，然后把这个区间内1变为0,0变为1，求变化后1的最大个数（必须修改一个数）。一个简单的动规，只需要判断这个数字能最大改变多少1的个数。

if(a[i]==0)dp[i]=max(0,dp[i-1]+1);

if(a[i]==1)dp[i]=max(0,dp[i-1]-1);

然后注意如果全是1的话，这个通项出来的是n，然而题目要求至少改变一个数，所以这个情况就是n-1。特判。

#include<cstdio>#include<cstring>#include<iostream>using namespace std;int dp[105], a[105];int max(int a, int b){return a > b ? a : b;}int main(){int n, i, j, m, ans, pre;cin >> n;pre = 0;for (i = 1; i <= n; i++){cin >> a[i];if (a[i] == 1)pre++;}if (a[1] == 0)dp[1] = 1;else dp[1] = 0;for (i = 2; i <= n; i++){if (a[i] == 1)dp[i] = max(dp[i - 1] - 1, 0);else if (a[i] == 0)dp[i] = max(0, dp[i - 1] + 1);}ans = dp[1];for (i = 2; i <= n; i++)ans = max(ans, dp[i]);if (ans == 0)cout << pre - 1 << endl;elsecout << pre + ans << endl;return 0;}