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 · 405 阅读

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文章标签：

#最大公约数 #最小公倍数 #数学问题

本文介绍了一个用于计算一组正整数的最小公倍数（LCM）的算法实现。该算法通过逐个处理输入的整数并利用辗转相除法计算最大公约数（GCD），进而求得最小公倍数。程序可以处理多个实例，并确保结果在32位整数范围内。

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
2 3 5
3 4 6 12

样例输出

15
12

提示

#include <cstdio>
int gcd(int a,int b){
    return !b?a:gcd(b,a%b);
}

int lcm(int a,int b){
    return a/gcd(a,b)*b;
}

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int m;
        scanf("%d",&m);
        int ans=1;
        for(int i=1;i<=m;i++){
            int x;
            scanf("%d",&x);
            ans=lcm(ans,x);
        }
        printf("%d\n",ans);
    }
    return 0;
}