HDU 1019 Least Common Multiple （最小公倍数_水题）

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
   
   
   
   

 

int gcd(int a,int b)    // 最大公约数模板
{
 if(a<b)    return gcd(b,a);
 int r;
 while(b) {r=a%b;a=b;b=r;}
 return a;
}


int lcm(int a,int b) {return a/gcd(a,b)*b;}  //最小公倍数模板

此题为模板题。。。

#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;

int gcd(int a,int b) 
{
 if(a<b)    return gcd(b,a);
 int r;
 while(b) {r=a%b;a=b;b=r;}
 return a;
}

int lcm(int a,int b) {return a/gcd(a,b)*b;}

int main()
{
	int t,n,i,j,num,sum;
	scanf("%d",&t);
	while(t--) {
		scanf("%d%d",&n,&sum);
		for(i=2;i<=n;i++){
			scanf("%d",&num);
			if(sum%num) {
				sum=lcm(sum,num);
			}
		}
		printf("%d\n",sum);
	}
	return 0;
}