HDOJ1019Least Common Multiple

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

也就是求最小公倍数

/**题意：
求输入的所有数最小公倍数。

思路：
先用 欧几里德定理
求两个数的最小公倍数，所得的公倍数再与下一个数求最小公倍数。
**/
#include <stdio.h>
#include <stdlib.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 t,n,m,i,a,b; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); a=m; for(i=1;i<n;i++) { scanf("%d",&m); if(a<m) { b=a;a=m;m=b; } a=a/gcd(a,m)*m;//最小公倍数=两数之积/最大公约数 } printf("%d\n",a); } return 0;
}

  

 

  
1
  
2
  
3
  
4
  
5
  
6
  
7
  
8
  
9
  
10
  
11
  
12
  
13
  
14
  
15
  
16
  
17
  
18
  
19
  
20
  
21
  
22
  
23
  
24
  
25
  
26
  
27
  
28
  
29
  
30
  
31
  
32
  
33
  
34
  
35
  
36
  
37
 

文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/49005049