L - 辗转相除法（第二季水）

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

 

 

如下为一般方法写的代码

#include<iostream>
using namespace std;
int s[500];
__int64 x;
int cmp ( const void *a , const void *b ) 
{  
    return *(int *)b - *(int *)a; 
}
void f(int s[],int n)
{
    int p=1;
    bool flag;
    while(1){
        flag=true;
        x=s[0]*p;
        for(int i=1;i<n;i++){
            if(x%s[i]!=0){
                flag=false;
                break;
            }
        }
        if(flag==true)break;
        p++;
    }
    cout<<x<<endl;
}
int main()
{
    int n;
    cin>>n;
    while(n--){
        int a;
        cin>>a;
        for(int i=0;i<a;i++)cin>>s[i];
        qsort(s,a,sizeof(s[0]),cmp);
        f(s,a);
    }
    return 0;
}

如下为运用辗转相除法写的代码

 

#include<iostream>
using namespace std;
int s[500];
__int64 x,y;
int g(int a,int b)      
{
    int i,t;
    y=a*b;
    if(a<b){
        t=a;
        a=b;
        b=t;
    }
    while(b){
        t=b;
        b=a%b;
        a=t;
    }
    return y/a;    
}
void f(int s[],int n)
{
    x=s[0];
    for(int i=1;i<n;i++){
        x=g(s[i],x);
    }
    cout<<x<<endl;
}
int main()
{
    int n;
    cin>>n;
    while(n--){
        int a;
        cin>>a;
        for(int i=0;i<a;i++)cin>>s[i];
        f(s,a);
    }
    //system("pause");
    return 0;
}