hdu 5210 Greatest Greatest Common Divisor

最新推荐文章于 2019-09-21 21:29:55 发布

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Greatest Greatest Common Divisor

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1014 Accepted Submission(s): 435

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 #

ans,

x is the case number, starting from

ans is the maximum value of greatest common divisor.

Sample Input

Sample Output

Case #1: 2
Case #2: 3

题目大意：简单
题目分析：
先将10^5以内的所有数的因数筛出来，统计每个因数出现的次数，然后倒序遍历，知道cnt[i]>1时，得到最大的最大公约数

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#define MAX 100007

using namespace std;

typedef long long LL;

int t,n;
int a[MAX];
int cnt[MAX];

int main ( )
{
    scanf ( "%d" , &t );
    int cc = 1;
    while ( t-- )
    {
        scanf ( "%d" , &n );
        for ( int i = 0 ; i < n ; i++ )
            scanf ( "%d" , &a[i] );

        memset ( cnt  ,  0 , sizeof ( cnt ) );

        for ( int i = 0 ; i < n ; i++ )
            for ( int j = 1 ; j*j <= a[i] ; j++ )
                if ( a[i]%j == 0 ) 
                {
                    cnt[j]++;
                    if ( j != a[i]/j ) cnt[a[i]/j]++;
                }
        for ( int i = MAX-1 ; i >= 1 ; i-- )
            if ( cnt[i] >= 2 )
            {
                printf ( "Case #%d: %d\n" , cc++ , i );
                break;
            }
    }
}