HDOJ 5207 Greatest Greatest Common Divisor 暴力枚举

暴力枚举公约数，有两个数是这个公约数倍数的就可以

Greatest Greatest Common Divisor

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 395    Accepted Submission(s): 171

Problem Description

Pick two numbers  ai,aj(i≠j)  from a sequence to maximize the value of their greatest common divisor.

 

Input

Multiple test cases. In the first line there is an integer  T , indicating the number of test cases. For each test cases, the first line contains an integer  n , the size of the sequence. Next line contains  n  numbers, from  a1  to  an .  1≤T≤100,2≤n≤105,1≤ai≤105 . The case for  n≥104  is no more than  10 .

 

Output

For each test case, output one line. The output format is Case # x :  ans ,  x  is the case number, starting from  1 ,  ans  is the maximum value of greatest common divisor.

 

Sample Input

   
   
   
   

    
    
    
    2
4
1 2 3 4
3
3 6 9
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case #1: 2
Case #2: 3
   
   
   
   

 

Source

BestCoder Round #38 ($)

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年04月18日 星期六 19时04分39秒
File Name     :B.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int maxn=100100;

int n;
int a[maxn];
int vis[maxn];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	int T_T,cas=1;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d",&n);
		int mx=0,ans=-1;
		memset(vis,0,sizeof(vis));
		for(int i=0;i<n;i++) 
		{
			scanf("%d",a+i);
			mx=max(mx,a[i]);
			vis[a[i]]++;
		}
		for(int i=mx;i>=1;i--)
		{
			if(ans>0) break;
			int temp=0;
			for(int j=i;j<=mx;j+=i) temp+=vis[j];
			if(temp>=2) ans=i;
		}
		printf("Case #%d: %d\n",cas++,ans);
	}
    
    return 0;
}