CDZSC_2015寒假新人（2）——数学 A

A - A

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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

 

 1 #include<iostream>

 2 #include<cstdio>

 3 using namespace std;

 4 int LCM(int a,int b)

 5 {

 6     int m=a,n=b;

 7     int c;

 8     while(b!=0)//辗转相除法

 9     { 

10         c=a%b; 

11         a=b; 

12         b=c;

13     }  

14     return m/a*n;//之前写的m*n/a  结果是错的  改成这样之后终于AC了

15 }

16 int main()

17 {

18     int a,n,m,num;

19     scanf("%d",&n);

20     while(n--)

21     {

22         num=1;

23         scanf("%d",&m);

24         while(m--)

25         {

26             scanf("%d",&a);

27             num=LCM(a,num);

28         }

29         printf("%d\n",num);

30     }

31 }

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