hdu 3833 YY's new problem（hash表）

YY's new problem

Time Limit: 12000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3501    Accepted Submission(s): 984

Problem 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

 

231 3 243 2 4 1

 

 

Sample Output

 

NY

 

思路：1到n每个元素只会出现一次，引入hash[]来记录该数是否已经出现，出现为1，否则为0 ；

读入一个数t ，从1到t-1依次判断是否有hash[t-i]+hash[t+i]==1即以t为中项，

对于t-i,t+i是否仅出现过一个，由于是按顺序读入的，即可保证t-i和t+i在原序列中一定是在t的两边

 

#include"stdio.h"#include"string.h"#define N 10005int main(){int T,t,n,i,m,flag,hash[N];scanf("%d",&T);while(T--){scanf("%d",&n);memset(hash,0,sizeof(hash));m=n;flag=0;while(m--){scanf("%d",&t);hash[t]=1;for(i=1;!flag&&i<t&&t+i<=n;i++) {if(hash[t-i]+hash[t+i]==1) //这两个元素有且只有一个出现flag=1;}}if(flag)printf("Y\n");elseprintf("N\n");}return 0;}