hdoj1019 Least Common Multiple(多个数求最小公倍数)

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10917    Accepted Submission(s): 3935

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
   
   
   
   

  #include<stdio.h> long LCM(long p1,long p2)//求最小公倍数 { long n1=p1,n2=p2; if(n1<n2) { long t=n1; n1=n2; n2=t; } long m=n1%n2; while(m!=0) { n1=n2; n2=m; m=n1%n2; } return p1/n2*p2;//不容易溢出 } int main() { long t,n; scanf("%ld",&t); while(t--) { long a=1,b=1; scanf("%ld",&n); while(n--) { scanf("%ld",&a); b=LCM(a,b); } printf("%ld/n",b); } return 0; }