HDU 1019 Least Common Multiple (最小公倍数)

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30285    Accepted Submission(s): 11455

Problem Description

The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.

 

Input

Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 ... nm where m is the number of integers in the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer.

 

Output

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

 

Sample Input

2

3

5 7 15

6

4 10296 936 1287 792 1

 

Sample Output

105

10296

 

Source

East Central North America 2003, Practice

 

Recommend

JGShining

 

求所给的数字的最小公倍数

#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int gcd(int a,int b)
{
    return b?gcd(b,a%b):a;
}
int lcm(int a,int b)
{
    return a/gcd(a,b)*b;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int kase,m;
    scanf("%d",&kase);
    while(kase--)
    {
        int ans,num;
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            int A;
            scanf("%d",&A);
            if(i==0){ans=A;continue;}
            num=A;
            ans=lcm(ans,num);
        }
        printf("%d
",ans);
    }
    return 0;
}