hdu 5207 求数组中任取两数的gcd最大值（灵活题）

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
   
   
   
   

http://acm.hdu.edu.cn/showproblem.php?pid=5207

做法很巧妙。记录每个数的所有因子，排序，再找整个数组里出现次数>=2的最大因子

#include<iostream>  
#include<algorithm>  
#include<string>  
#include<map>  
#include<vector>  
#include<cmath>  
#include<string.h>  
#include<stdlib.h>  
#include<cstdio>  
#define ll long long  
using namespace std;
int x[100001];
int main(){
    int t;
    cin>>t;
    int cnt=0;
    while(t--){
        memset(x,0,sizeof(x));
        int n,a;
        cin>>n;
        int max=-1;
        for(int i=0;i<n;++i){
            cin>>a;
            if(a>max)
                max=a;
            int g=sqrt((double)a);
            for(int i=1;i<=g;++i){
                if(a%i==0){
                    x[i]++;
                    x[a/i]++;
                }
            }
            if(g*g==a)
                x[g]--;
        }
        printf("Case #%d: ", ++cnt);
        for(int i = max; i >= 1; i--){
            if(x[i]>=2){
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}