Greatest Greatest Common Divisor HD5207

Greatest Greatest Common Divisor

 

 Accepts: 271

 

 Submissions: 1138

 Time Limit: 4000/2000 MS (Java/Others)

 

 Memory Limit: 65536/65536 K (Java/Others)

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

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比赛时间内没做出来==

后来补的，从网上借鉴的方法，感觉非常巧妙，用数组直接存储，从大值开始扫，实现如下：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int t;
int n,temp;
int a[100000+10];
int main()
{
    scanf("%d",&t);
    for(int m=1;m<=t;m++)
    {
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        int maxi=0;
        for(int i=1;i<=n;i++)
        {
           scanf("%d",&temp);
           a[temp]++;
           maxi=max(maxi,temp);
        }
        int n=0,flag=0;
        for(int i=maxi;i>=1;i--)
        {
            n=0;
            for(int j=i;j<=maxi;j+=i)
            {
                if(a[j])
                    n+=a[j];
                if(n>1)
                {
                    printf("Case #%d: %d
",m,i);
                    flag=1;
                    break;
                }
            }
            if(flag)
                break;
        }
    }
}

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