首页 > 代码库 > HDU 1019 Least Common Multiple (最小公倍数)
HDU 1019 Least Common Multiple (最小公倍数)
Least Common Multiple
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30285 Accepted Submission(s): 11455
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
Recommend
JGShining
求所给的数字的最小公倍数
1 #include<cstdio> 2 #include<cstring> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 int gcd(int a,int b) 7 { 8 return b?gcd(b,a%b):a; 9 }10 int lcm(int a,int b)11 {12 return a/gcd(a,b)*b;13 }14 int main()15 {16 //freopen("in.txt","r",stdin);17 int kase,m;18 scanf("%d",&kase);19 while(kase--)20 {21 int ans,num;22 scanf("%d",&m);23 for(int i=0;i<m;i++)24 {25 int A;26 scanf("%d",&A);27 if(i==0){ans=A;continue;}28 num=A;29 ans=lcm(ans,num);30 }31 printf("%d\n",ans);32 }33 return 0;34 }
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。