hdu3833YY's new problem
这个题又是想难了，联想到poj2549 看了题解 自己写WA了好多次 ==其实之前犯过这个错误T^T 跳出循环时要注意 题中给的数得读入结束才行 长点心吧。。。
Problem Description
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.
P[i 1]-P[i 2]=P[i 2]-P[i 3], 1<=i 1<i 2<i 3<=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.
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 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
#include <iostream>
#include<cstdio>
#include<cstring>
//#include<algorithm>
#define mm 10003
using namespace std;
int t,n,s,flag;
int hsh[mm];
int main()
{
// freopen("cin.txt","r",stdin);
cin>>t;
while(t--)
{
cin>>n;
memset(hsh,0,sizeof(hsh));
flag=0;
for(int i=0;i<n;i++)
{
scanf("%d",&s);
hsh[s]=1;
if(flag==0)
{
for(int j=1;j<s&&j+s<=n;j++)
{
if(hsh[s+j]+hsh[s-j]==1)
{
flag=1;
break;
}
}
}
//if(flag) break;//就是这里 如果这么就跳出来了 会有数没读完
}
if(flag) cout<<"Y"<<endl;
else cout<<"N"<<endl;
}
return 0;
}
