hdu Least Common Multiple

Least Common Multiple

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 31   Accepted Submission(s) : 14

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

Problem 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

Source

East Central North America 2003, Practice

此题大意：求所给数的最小公倍数。

思路：先两两相求，最后得结果。我是先求两个数的最大公约数再用两数之积去除最大公约数。求最大公约有多种方法，我用的辗转相除法，即两数相除，小的做除数大的为被除数求余数，若余数不为零，让原来的除数做被除数，余数做除数相除求余数，一直做下去知道余数为零，此时的得除数便是两个数的最大公约数。
代码：

#include <iostream>
using namespace std;
//定义一个求最大公约数的函数
//辗转相除求最大公约数
int gcd(long long int a,long long int b)
{
    long long int r,t;
    if(a<b)//保证a为最大的数
    {
        t=a;
        a=b;
        b=t;
    }
    while((r=(a%b))!=0)//直至余数为零
    {
        a=b;
        b=r;
    }
    return b;//返回除数数
}
int main()
{
    long long int i,j,s,m,n,k;
    cin>>n;
    for(i=0;i<n;i++)
    {
        cin>>m;
        cin>>k;
        s=k;
        //两个两个的求
        for(j=1;j<m;j++)
        {
           cin>>k;
           s=(s*k)/gcd(s,k);//最小公倍数为两个数之积除以最大公约数
        }
        cout<<s<<endl;
    }
    return 0;
}