Least Common Multiple

最新推荐文章于 2023-12-18 15:05:55 发布

原创最新推荐文章于 2023-12-18 15:05:55 发布 · 1k 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#ACM #优化 #算法 #HDU #数据

ACM 同时被 2 个专栏收录

521 篇文章

订阅专栏

HDU

171 篇文章

订阅专栏

本文提供了一种高效的方法来解决HDU在线评测平台上的1019题，即计算一组整数的最小公倍数。通过使用欧几里得算法求最大公约数并进行相应的数学运算，本文详细解释了如何实现这一过程，包括输入解析、算法应用以及输出格式。适合初学者理解和实践。

Problem Link：http://acm.hdu.edu.cn/showproblem.php?pid=1019

原题：

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30183 Accepted Submission(s): 11417

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

AC C code：

--------------------------------------------------------------------------------------------------------------------------------------------

#include<stdio.h>
int gcd(int a,int b)//欧几里德求最大公约数
{
if(b==0)
return a;
return gcd(b,a%b);
}
int main()
{
int cas,n,x,y,t,i;
while(scanf("%d",&cas)!=EOF)
{
while(cas--)
{
scanf("%d%d",&n,&x);
for(i=2;i<=n;i++)
{
scanf("%d",&y);
if(y>x)
{
t=y;
y=x;
x=t;
}
x=x/gcd(x,y)*y;//最小公倍数=两数之积/最大公约数(注意：这里换一下顺序，先除后乘，以免数据溢出）
}
printf("%d\n",x);
}
}
return 0;
}