Least Common Multiple

最新推荐文章于 2021-12-10 15:59:44 发布

原创最新推荐文章于 2021-12-10 15:59:44 发布 · 135 阅读

0 ·

CC 4.0 BY-SA版权

HDU 专栏收录该内容

54 篇文章

订阅专栏

本文介绍了一种求解一组正整数最小公倍数（LCM）的算法，并提供了详细的C++代码实现。该算法首先接收一组正整数作为输入，然后通过逐次计算两数之间的最小公倍数来找出整组数的最小公倍数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题目大意：求给出每组数字的最小公倍数

#include<iostream>
using namespace std;
long long f(long long n,long long m){
    long long t;
    long long g;
    if(n<m){
        t=m;
        m=n;
        n=t;
    }
    g=n;
    while(n%m!=0){
        n+=g;
    }
    return n;
}
int main(){
    int n,m,i;
    long long t,p;
    cin>>n;
    while(n--){
        cin>>m;
        p=1;
        for(i=0;i<m;i++){
            cin>>t;
            p=f(p,t);
            
        }
        cout<<p<<endl;
    }
    return 0;
}