Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34436    Accepted Submission(s): 12968

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

23 5 7 156 4 10296 936 1287 792 1

 

Sample Output

10510296

 

Source

East Central North America 2003, Practice

 

思路：这个题好像以前做过，但是忘了思路了，百度了一下，现在知道了。求所有数的最小公倍数，就是先两个求最小公倍数，然后所得的公倍数再与下一个数求最小公倍数。。两个数乘积除以最大公约数等于最小公倍数。

知道方法了，那么写代码就比较容易了。。我注释掉的代码是我的方法，求最大公约数，但是提交说超时了。。百度一下，可以用欧几里得定理。这样就ac了。。

节省时间，记住这个方法求最大公约数。

int gys(int a,int b)  //欧几里得求最大公约数

{

  if(b==0)

      return a;

  return gys(b,a%b);

}

#include<iostream>using namespace std;int gys(int a,int b){  /* int i,result;               //此方法超时   for(i=1;i<=a;i++)   {       if(a%i==0 && b%i==0)          result=i;   }   result=a*b/result;   return result;*/    if(b==0) return a;          //欧几里得求最大公约数    return gys(b,a%b);}int main(){    int T,n,i,a[1000];    cin>>T;    while(T--)    {       cin>>n;       for(i=0;i<n;i++)           cin>>a[i];       for(i=0;i<n-1;i++)  //x x x x x       {           a[i+1]=a[i]/gys(a[i],a[i+1])*a[i+1];       }       cout<<a[i]<<endl;    }    return 0;}