HDU3833 YY's new problem 卡时间第一题

Given a permutation P of 1 to N, YY wants to know whether there exists such three elements P[i ₁], P[i ₂], P[i ₃] that 
P[i ₁]-P[i ₂]=P[i ₂]-P[i ₃], 1<=i ₁<i ₂<i ₃<=N.

InputThe first line is T(T<=60), representing the total test cases. 
Each test case comes two lines, the former one is N, 3<=N<=10000, the latter is a permutation of 1 to N.OutputFor each test case, just output 'Y' if such i ₁, i ₂, i ₃ can be found, else 'N'.Sample Input

2
3
1 3 2
4
3 2 4 1

Sample Output

N
Y

不稍微优化一下很容易超时：

# include<iostream>
# include<cstdio>
# include<string.h>
# include<algorithm>
using namespace std;
int vis[10010],a[10010];
int main()
{
    int i,j,n,T;
    scanf("%d",&T);
    while(T--){
        memset(vis,0,sizeof(vis));
        bool flag=false;
        scanf("%d",&n);
        for(i=1;i<=n;i++)scanf("%d",&a[i]);
        for(i=1;i<=n&&!flag;i++){
             vis[a[i]]=1;
             for(j=1;j<a[i];j++) {
                    if(j+a[i]<=n&&vis[a[i]-j]+vis[a[i]+j]==1){
                          flag=true;
                          break;
                    }
             }
        }
        if(flag) printf("Y\n");
        else printf("N\n");
    }
    return 0;
}