【九度】题目1439：Least Common Multiple

最新推荐文章于 2019-03-01 11:16:09 发布

原创最新推荐文章于 2019-03-01 11:16:09 发布 · 344 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种求解多个正整数最大公倍数(LCM)的有效算法，并提供了完整的C++实现代码。该算法利用辗转相除法计算两数的最大公约数(GCD)，进而求得最小公倍数。通过迭代此过程，可以找到一组数的最小公倍数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

时间限制：1 秒

内存限制：128 兆

特殊判题：否

提交：4287

解决：1251

题目描述：

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.

输入：

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.

输出：

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

样例输入：

2
3 5 7 15
6 4 10296 936 1287 792 1

样例输出：

105
10296

// 多个数字的最大公倍数.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <string.h>
int gcd(int a,int b){
	return b!=0? gcd(b,a%b):a;
}

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

}
int _tmain(int argc, _TCHAR* argv[])
 { int n;
    
while(scanf("%d",&n)){
 
	while (n--)

	{  int buf[10000] ; 
		int len;
		scanf("%d",&len);

		for (int i = 0; i < len; i++)
		{
           scanf("%d",&buf[i]);
		}
	int ans=buf[0];
	printf("buf[0]=%d\n",ans);

	for (int i=1 ; i<len; i++)
	{
		ans=lcm(ans,buf[i]);
	}
	printf("%d\n",ans);
	
	}
	


}
	return 0;

	}