Codeforces 327A. Flipping Game[ 最大子序列的和 ]

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.

Examples input Copy

5
1 0 0 1 0
      

output Copy

4
      

input Copy

4
1 0 0 1
      

output Copy

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.

题意：给定一串数字，只由01组成，对任意大小连续区间内进行翻转，求1的最大数量。

分析：这样子先看把0翻转我们就加1，将1翻转我们就加-1，那么我们只需要计算子序列和最大就可以了，再加上原先的1的和，就是最大的1的数量。。。

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn = 1e5+7;
int n, a[maxn], cunt = 0, dp[maxn];


int main()
{
    memset(dp, 0, sizeof(dp));
    scanf("%d",&n);
    for(int i = 0; i < n; i++)
    {
        int c; scanf("%d",&c);
        if(c == 1) a[i] = -1, cunt++;
        else a[i] = 1;
    }
    dp[0] = a[0];
    int maxc = a[0];
    for(int i = 1; i < n; i++)
        dp[i] = max(dp[i - 1] + a[i], a[i]), maxc = max(maxc, dp[i]);
    printf("%d\n", maxc + cunt);
    return 0;
}