G - Least Common Multiple

最新推荐文章于 2019-03-01 11:16:09 发布

原创最新推荐文章于 2019-03-01 11:16:09 发布 · 274 阅读

0 ·

CC 4.0 BY-SA版权

程序设计专栏收录该内容

34 篇文章

订阅专栏

本文介绍了一种计算一组正整数最小公倍数的方法，并提供了一个C++实现示例。输入包括多组测试实例，每组包含一系列正整数，输出为对应的最小公倍数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

#include <cstring>
#include <stdio.h>
#include <algorithm>
#include <math.h>
#define MAXN 1e7+5
using namespace std;
typedef long long LL;
LL a,b;
LL gcd(LL a, LL b)
{
    return b == 0 ? a : gcd(b, a % b);
}

LL lcm(LL a, LL b)
{
    return a / gcd(a, b) * b; //先除后乘
}

void cmax(LL a,LL b)
{
    LL temp;
    if(a<b) temp=a,a=b,b=temp;
}

int main()
{
    int T;
	LL n;
	scanf("%d",&T);
    while(T--)
	{
		int n;
        scanf("%d",&n);
        scanf("%lld",&a);
        for(int i=1;i<n;i++){
            scanf("%lld",&b);
            cmax(a,b);
            a=lcm(a,b);
        }
        printf("%lld\n",a);
	}
	return 0;
}