HDU1019-多个数的最小公倍数，最大公约数

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23934    Accepted Submission(s): 8980

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

 

Source

East Central North America 2003, Practice

 

Recommend

JGShining

基础的经典题哦！

先两个求出最小公倍数，在把结果和下一个元素求最小公倍数，然后循环下去就好了！

求多数的最大公约数 同理！

注：最小公倍数==两数乘积 / 最大公约数.

#include
using namespace std;
int gcd(int a,int b)
{
	return b==0?a:gcd(b,a%b);
}
int lcm(int a,int b)
{
	return a/gcd(a,b)*b;
}
int main()
{
	int T,i;
	cin>>T;
	while(T--)
	{
		int a[1000];
		int m;
		cin>>m;
		for(i=1;i<=m;i++)
		{
			cin>>a[i];
			if(i!=1)
			{
				a[i]=lcm(a[i],a[i-1]); 
			}
		}
		cout<