Codeforces 327A-Flipping Game（暴力枚举）

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 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

题意：翻牌游戏。

给出n张牌，每张牌仅仅有0和1两种状态。给出初始状态。对于翻牌操作这样规定：每次操作可将区间[i,j](1=<i<=j<=n)内牌的状态翻转（即0变1,1变0）。求一次翻转操作后，1的个数尽量多。

枚举区间+遍历区间推断，O（n^3）;

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <set>
#include <map>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
const int INF=0x3f3f3f3f;
#define LL long long
int a[110];
int main()
{
int n,num[2];
while(~scanf("%d",&n))
{
int ans=0;
for(int i=0;i<n;i++)
{
scanf("%d",a+i);
if(a[i]==1)
ans++;
}
int pos=ans;
if(pos==n)
{
printf("%d\n",n-1);
continue;
}
for(int i=0;i<n;i++)
for(int j=i;j<n;j++)
{
memset(num,0,sizeof(num));
for(int k=i;k<=j;k++)
num[a[k]]++;
if(num[0]>num[1])
ans=max(ans,pos+num[0]-num[1]);
}
printf("%d\n",ans);
}
return 0;
}