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[];
 int str[];
 int main()
 {
     int t,n;
     while(scanf("%d",&t)!=EOF)
     {
         while(t--)
         {
             scanf("%d",&n);
             for(int i=; 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=;
             for(int i=; i<=n; i++)
             {
                  for(int j=i+; j<=n; j++)
                 {
                    int temp=str[i]+str[j];//相加
                    if(temp%)//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/]>i&&hush[temp/]<j)
                    {
                        flag=;
                        break;
                    }
                 }
                 if(flag==) break;
             }
             printf("%c",flag==?'Y':'N');
             printf("\n");
         }
     }
     return ;
 }