码迷,mamicode.com
首页 > 其他好文 > 详细

杭电1019Least Common Multiple

时间:2016-04-04 14:38:58      阅读:161      评论:0      收藏:0      [点我收藏+]

标签:

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

题目:

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
 
思路:最小公倍数(lcm),最大公约数(gcd)。
  求gcd(a,b),用辗转相除法。
  代码:
long gcd(long a,long b)
{

    long r=1;
    while (r>0)
    {
        r=a%b;
        a=b;
        b=r;
    }
    return a;}
lcm(a,b)=a*b/gcd(a,b);
代码:
  
long lcm(long a,long b)
{
    return a/gcd(a,b)*b;
}
ac代码:
技术分享
 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cmath>
 5 #include <cstring>
 6 #include <queue>
 7 #include <stack>
 8 #include <map>
 9 #include <vector>
10 
11 #define PI acos((double)-1)
12 #define E exp(double(1))
13 using namespace std;
14 
15 long gcd(long a,long b)
16 {
17 
18     long r=1;
19     while (r>0)
20     {
21         r=a%b;
22         a=b;
23         b=r;
24     }
25     return a;
26 //    if(b)
27 //        while( (a%=b)  && (b %= a));
28 //    return a+b;
29 }
30 
31 long lcm(long a,long b)
32 {
33     return a/gcd(a,b)*b;
34 }
35 int main (void)
36 {
37     int t,n;
38     cin>>t;
39     while(t--)
40     {
41         long temp,m;
42         cin>>n;
43         scanf("%ld",&temp);
44         for(int i = 1;i<n;i++)
45         {
46             scanf("%ld",&m);
47             temp = lcm(temp,m);
48         }
49         cout<<temp<<endl;
50 
51     }
52     return 0;
53 }
View Code

 

 

杭电1019Least Common Multiple

标签:

原文地址:http://www.cnblogs.com/weeping/p/5351874.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!