Codeforces Round #191 (Div. 2)---A. 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, ..., 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 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, ..., 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.

解题思路：题意是讲有一个0,1序列，现同意你对随意一个子序列取反，问操作后得到的序列，最多能有多少个1。

数据不大，直接暴力。直接两层循环枚举取反序列的起点和终点，统计每种情况下1的个数，保存一个最大值就可以。

AC代码：

#include <iostream>
#include <cstdio>
using namespace std;

int a[105];

int main(){
//  freopen("in.txt","r",stdin);
int n;
while(cin>>n){
for(int i =0; i<n; i++)
cin>>a[i];
int ans = 0;
for(int i=0; i<n; i++){
for(int j=i; j<n; j++){
int sum = 0;
for(int k=0; k<n; k++){
if(k>=i && k<=j) sum += a[k]^1;          //用异或^取反
else sum += a[k];
}
if(ans < sum)  ans = sum;
}
}
cout<<ans<<endl;
}
return 0;
}