九度题目1439：Least Common Multiple

最新推荐文章于 2019-02-25 20:53:09 发布

原创最新推荐文章于 2019-02-25 20:53:09 发布 · 263 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#最小公倍数 #LCM

ACM 同时被 2 个专栏收录

46 篇文章

订阅专栏

ACM-数学问题

15 篇文章

订阅专栏

本文介绍九度OJ题目1439：LeastCommonMultiple的解决思路及源代码实现，通过求最小公倍数（LCM）来解决多组整数集合的问题，并提供完整的C++代码示例。

九度题目1439：Least Common Multiple

原题OJ链接：http://ac.jobdu.com/problem.php?pid=1439

题目描述：

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.

输入：

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.

输出：

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

样例输入：

2
3 5 7 15
6 4 10296 936 1287 792 1

样例输出：

105
10296

源代码：

#include<iostream>
using namespace std;
long long gcd(long long a,long long b){
    if(b==0) return a;
    else return gcd(b,a%b);
}

int main(){
    int ins;
    cin>>ins; 
    for(int i=0;i<ins;i++){
        int m;
        long long a,b,c;
        cin>>m>>a;
        for(int j=0;j<m-1;j++){
            cin>>b;
            a=a*b/gcd(a,b);
        }
        cout<<a<<endl;
    }
    return 0;
}