Hdu 5207 Greatest Greatest Common Divisor（数论）

题目链接

Greatest Greatest Common Divisor

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 567    Accepted Submission(s): 257

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 ($)

题意：从n个数中，找出两个数之间的最大公约数的最大值。每个数<=100000,n<=100000。

题解：先用nlgn的方法预处理出1到10^5的数的约数。然后再统计出n个数中，每个数是多少个数的约数。最后的答案就是找所有约数中个数大于1的最大的数。

代码如下：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<vector>
#define nn 110000
typedef long long LL;
const int inf=0x3fffffff;
const LL inf64=(LL)inf*inf;
using namespace std;
vector<int>ve[nn];
int num[nn],n;
void init()
{
    int i,j;
    for(i=1;i<=100000;i++)
    {
        for(j=1;(LL)i*j<=100000;j++)
        {
            ve[i*j].push_back(i);
        }
    }
}
int main()
{
    init();
    int t,cas=1;
    int i,j,x;
    scanf("%d",&t);
    while(t--)
    {
        memset(num,0,sizeof(num));
        scanf("%d",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%d",&x);
            for(j=0;j<(int)ve[x].size();j++)
            {
                num[ve[x][j]]++;
            }
        }
        for(j=100000;j>=1;j--)
        {
            if(num[j]>1)
                break;
        }
        printf("Case #%d: %d\n",cas++,j);
    }
    return 0;
}