HDU2028_Lowest Common Multiple Plus

Lowest Common Multiple Plus

http://acm.hdu.edu.cn/showproblem.php?pid=2028

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 54570    Accepted Submission(s): 22595

Problem Description

求n个数的最小公倍数。

 

Input

输入包含多个测试实例，每个测试实例的开始是一个正整数n，然后是n个正整数。

 

Output

为每组测试数据输出它们的最小公倍数，每个测试实例的输出占一行。你可以假设最后的输出是一个32位的整数。

 

Sample Input

  

   2 4 6
3 2 5 7
  

 

Sample Output

  

   12
70
  

 

Author

lcy

 

Source

C语言程序设计练习（五）

#include <stdio.h>

int GCD(long long int n,long long int m)
{
	if (m <= n)
		if (n%m == 0)
			return(m);
		else
			return (GCD(m, n%m));
	else
		return(GCD(m, n));
}

int main()
{
	int n, i;
	long long int a[100] = { 0 }, k = 0, s = 1;
	while (scanf("%d", &n) != EOF)
	{
		for (i = 0;i<n;i++)
			scanf("%lld", &a[i]);
		s = 1;
		for (i = 0;i < n;i++)
		{
			k = GCD(s, a[i]);
			if (k == 1)
				s = s*a[i];
			else
				s = s*(a[i] / k);
		}
		printf("%lld\n", s);
	}
}