hdu3833（暴力）
YY's new problem
Time Limit: 12000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 5384 Accepted Submission(s): 1508
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
暴力要有方法。。。。。
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
typedef long long ll;
/*const int M=200000;
int ip;
int head[M],cd[M],rd[M];
struct data
{
int v,next;
}tu[M];
void init(int n)
{
memset(head,-1,sizeof(head));
memset(rd,0,sizeof(rd));
memset(cd,0,sizeof(cd));
ip=0;
}
void add(int u,int v)
{
tu[ip].v=v,tu[ip].next=head[u],head[u]=ip++;
}
*/
int zb[10010],a[10010];
int main()
{
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
for(int i=0; i<n; i++)
{scanf("%d",&a[i]);zb[a[i]]=i;}
int flag=0;
for(int i=0; i<n; i++)
{
for(int j=i+1; j<n; j++)
{
if((a[i]+a[j])%2==0)
{
if(zb[(a[i]+a[j])/2]>i&&zb[(a[i]+a[j])/2]<j)
{
flag=1;
break;
}
}
}
if(flag)break;
}
if(flag)printf("Y\n");
else printf("N\n");
}
return 0;
}