G - YY's new problem(HUSH算法，目前还不懂什么是HUSH算法）

 

Time Limit:4000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 3833

Description

Given a permutation P of 1 to N, YY wants to know whether there exists such three elements P[i₁], P[i₂], P[i₃] that
P[i₁]-P[i₂]=P[i₂]-P[i₃], 1<=i₁<i₂<i₃<=N.

 

Input

The 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.

 

Output

For each test case, just output 'Y' if such i₁, i₂, i₃ can be found, else 'N'.

 

Sample Input

2 3 1 3 2 4 3 2 4 1

 

Sample Output

N Y

 

#include<cstdio>
#include<string.h>
using namespace std;
int hush[10005];
int str[10005];
int main()
{
    int t,n;
    while(scanf("%d",&t)!=EOF)
    {
        while(t--)
        {
            scanf("%d",&n);
            for(int i=1; i<=n; i++)
            {
                scanf("%d",&str[i]);//把1——N保存到数组里
                hush[str[i]]=i;//记录每个数的位置
            }
            int flag=0;
            for(int i=1; i<=n; i++)
            {
                 for(int j=i+1; j<=n; j++)
                {
                   int temp=str[i]+str[j];//相加
                   if(temp%2)//优化，必须是偶数才行，尽量别写成temp%2==1,直接写temp%2
                     continue;
                   if(hush[temp/2]>i&&hush[temp/2]<j)
                   {
                       flag=1;
                       break;
                   }
                }
                if(flag==1) break;
            }
            printf("%c",flag==1?'Y':'N');
            printf("\n");
        }
    }
    return 0;
}