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 }

View Code