Hdoj 1019 Least Common Multiple

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本文介绍了一种求解多个正整数最小公倍数的方法，并提供了一个C++实现示例。通过输入一组正整数，程序能计算并输出这些数的最小公倍数。

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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
题意：求所有数的最小公倍数，不断去掉两个数，把他们的最下公倍数作为下一个数，再取，知道取完，就得到了答案。

///@zhangxiaoyu
///2015/8/13

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int gcd(int a,int b)
{
    if(b==0)return a;
    return gcd(b,a%b);
}

int main()
{
  int tt;
  int m,a,b;
  while(scanf("%d",&tt)!=EOF)
  {
      for(int i=0;i<tt;i++)
      {
          scanf("%d",&m);
          a=1;
          for(int j=0;j<m;j++)
          {
              scanf("%d",&b);
              a=a/gcd(a,b)*b;
          }
          printf("%d\n",a);
      }
  }
  return 0;
}