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

 1 #include<cstdio>
 2 #include<string.h>
 3 using namespace std;
 4 int hush[10005];
 5 int str[10005];
 6 int main()
 7 {
 8     int t,n;
 9     while(scanf("%d",&t)!=EOF)
10     {
11         while(t--)
12         {
13             scanf("%d",&n);
14             for(int i=1; i<=n; i++)
15             {
16                 scanf("%d",&str[i]);//把1——N保存到数组里
17                 hush[str[i]]=i;//记录每个数的位置
18             }
19             int flag=0;
20             for(int i=1; i<=n; i++)
21             {
22                  for(int j=i+1; j<=n; j++)
23                 {
24                    int temp=str[i]+str[j];//相加
25                    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
26                      continue;
27                    if(hush[temp/2]>i&&hush[temp/2]<j)
28                    {
29                        flag=1;
30                        break;
31                    }
32                 }
33                 if(flag==1) break;
34             }
35             printf("%c",flag==1?‘Y‘:‘N‘);
36             printf("\n");
37         }
38     }
39     return 0;
40 }

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