Least Common Multiple

最新推荐文章于 2020-07-31 14:23:58 发布

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最新推荐文章于 2020-07-31 14:23:58 发布

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本篇博客探讨了如何解决一组正整数的最小公倍数问题，并提供了两种求解方法：一种是通过递归计算最大公约数来求解最小公倍数，另一种则是使用数组方法直接遍历计算。

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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

分析：本题是求每组数据中的最小公倍数是多少；

代码：

#include<stdio.h>
#include<math.h>
int gcd(int a,int b)
{
	 return !b?a:gcd(b,a%b);
	
}
int main()
{
	int n,t,m,i,b;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%d%d",&t,&m);
	  for(i=1;i<t;i++)	
	 {
	  scanf("%d",&b); 
	  	m=m/gcd(m,b)*b;
	  }
	  	printf("%d\n",m);
	}
	return 0;
}

数组方法：

#include<stdio.h>
#include<math.h>
int a[1000000];
int gcd(int a,int b)
{
<span style="white-space:pre">	</span> return !b?a:gcd(b,a%b);
<span style="white-space:pre">	</span>
}
int main()
{
<span style="white-space:pre">	</span>int n,t,m,i;
<span style="white-space:pre">	</span>scanf("%d",&n);
<span style="white-space:pre">	</span>while(n--)
<span style="white-space:pre">	</span>{
<span style="white-space:pre">		</span>scanf("%d",&t);
<span style="white-space:pre">	</span>  for(i=0;i<t;i++)<span style="white-space:pre">	</span>
<span style="white-space:pre">	</span>  scanf("%d",&a[i]);
<span style="white-space:pre">	</span>  m=a[0];
<span style="white-space:pre">	</span>  //printf("%d\n",m);
<span style="white-space:pre">	</span>  for(i=1;i<t;i++)
<span style="white-space:pre">	</span>  <span style="white-space:pre">	</span>m=m/gcd(m,a[i])*a[i];
<span style="white-space:pre">	</span>  <span style="white-space:pre">	</span>printf("%d\n",m);
<span style="white-space:pre">	</span>}
<span style="white-space:pre">	</span>return 0;
}