hdu1019(求最小公倍数要先除后乘)

最新推荐文章于 2018-04-13 22:34:00 发布

a1647566717

最新推荐文章于 2018-04-13 22:34:00 发布

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分类专栏：数学文章标签： integer output numbers input each ini

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本文介绍了如何通过编程实现计算一组正整数的最小公倍数（LCM），包括输入处理、LCM计算方法和输出结果展示。

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Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12051 Accepted Submission(s): 4410

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

#include<stdio.h>
int gcd(int a,int b)
{
if(b==0)
  return a;
return gcd(b,a%b);
}
int lcm(int a,int b)
{
return a/gcd(a,b)*b;//先除后乘
}
int a[100000];
int main()
{
int t,n,temp,i;
scanf("%d",&t);
while(t--)
{
  scanf("%d",&n);
  for(i=0;i<n;i++)
   scanf("%d",&a[i]);
  temp=a[0];
  for(i=1;i<n;i++)
  {
   temp=lcm(temp,a[i]);
  }
  printf("%d\n",temp);

}
return 0;
}