杭电 1019 简单数学题

 就是一道简单数学题，求多个数的最小公倍数。。。题目：

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14953    Accepted Submission(s): 5562

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
    
    
    
    

 

ac代码：

#include <iostream>
#include <cstdio>
using namespace std;
__int64 max(__int64 a,__int64 b){
  return a>b?a:b;
}
__int64 min(__int64 a,__int64 b){
  return a<b?a:b;
}
__int64 gcd(__int64 x,__int64 y){
  __int64 mmax=max(x,y);
  __int64 mmin=min(x,y);
  __int64 r;
  while(mmin!=0){
    r=mmax%mmin;
	mmax=mmin;
	mmin=r;
  }
  return mmax;
}
int main(){
 //freopen("1.txt","r",stdin);
  int numcase;
  scanf("%d",&numcase);
  while(numcase--){
    __int64 n,x,y=1,z=1;
	scanf("%I64d",&n);
	while(n--){
	  scanf("%I64d",&x);
	  int gg=gcd(x,z);
	  y=x*z/gg;
	  z=y;
	}
	printf("%I64d\n",y);
  }
  return 0;
}