(CodeForces - 327A)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
5
1 0 0 1 0
output
4
input
4
1 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.
题目大意:有n个0,1序列,在区间[i,j]进行一次翻转,会使得0变成1,1变成0,问翻转一次使得这n个序列中1的个数最多是多少。
思路:因为n最大100,所以直接暴力枚举需要翻的区间0(n3),即可。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=105;
int a[maxn];
int sum[2];
int main()
{
int n;
while(~scanf("%d",&n))
{
int cnt=0;
for(int i=0;i<n;i++)
{
scanf("%d",a+i);
if(a[i]) cnt++;
}
if(cnt==n)//如果全是1得翻一张牌,不能不翻
{
printf("%d\n",n-1);
continue;
}
int ans=cnt;
for(int i=0;i<n;i++)
for(int j=i;j<n;j++)
{
memset(sum,0,sizeof(sum));
for(int k=i;k<=j;k++) sum[a[k]]++;
if(sum[0]>sum[1])
ans=max(ans,cnt+sum[0]-sum[1]);
}
printf("%d\n",ans);
}
return 0;
}