Least Common Multiple

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

解题思路：

题意是给出n个数，求出这些数的最小公倍数，输出，先求前两个数的最小公倍数，再用这个数和第三个数求最小公倍数，依此类推，直到求出来。

代码如下:

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
//long long b[100000000];
int g(int a,int bb)
{
    if(a%bb==0)
        return bb;
    else
        return g(bb,a%bb);
}
int main()
{
    int a,i,j,k,q,c,t,d,b;
    scanf("%d",&a);
    while(a--) 
    {
        scanf("%d",&c);
        scanf("%d",&b);
        for(i=0;i<c-1;i++)
        {
            scanf("%d",&d);
            t=g(b,d);
            b=b*(d/t);
        }
        printf("%d\n",b);
    }
    return 0;
}