杭电1019Least Common Multiple

地址：http://acm.hdu.edu.cn/showproblem.php?pid=1019

题目：

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

 

思路：最小公倍数（lcm），最大公约数（gcd）。

　　求gcd（a，b)，用辗转相除法。

　　代码：

long gcd(long a,long b)
{

    long r=;
    while (r>)
    {
        r=a%b;
        a=b;
        b=r;
    }
    return a;}
lcm(a,b）=a*b/gcd(a,b);
代码：
　　

long lcm(long a,long b)
{
    return a/gcd(a,b)*b;
}
ac代码：

 #include <iostream>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <queue>
 #include <stack>
 #include <map>
 #include <vector>

 #define PI acos((double)-1)
 #define E exp(double(1))
 using namespace std;

 long gcd(long a,long b)
 {

     long r=;
     while (r>)
     {
         r=a%b;
         a=b;
         b=r;
     }
     return a;
 //    if(b)
 //        while( (a%=b)  && (b %= a));
 //    return a+b;
 }

 long lcm(long a,long b)
 {
     return a/gcd(a,b)*b;
 }
 int main (void)
 {
     int t,n;
     cin>>t;
     while(t--)
     {
         long temp,m;
         cin>>n;
         scanf("%ld",&temp);
         for(int i = ;i<n;i++)
         {
             scanf("%ld",&m);
             temp = lcm(temp,m);
         }
         cout<<temp<<endl;

     }
     return ;
 }