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

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25035    Accepted Submission(s): 9429

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 int gb(long m,long n)
{
if(m>&&n>)
{
long a,b,r;
if(n>m)
{
r=m;
m=n;
n=r;
}
a=m;
b=n;
while(r!=)
{
r=a%b;
a=b;
b=r;
}
return m/a*n;
}
else
return ;
}
int main()
{
int i,n,a,b,t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
scanf("%d",&a);
for(i=;i<n;i++)
{
scanf("%d",&b);
a=gb(a,b);
}
printf("%d\n",a);
}
}
#include"stdio.h"
int mcm( long x, long y)
{
long t;
while(y)
{
t=x%y;
x=y;
y=t;
}
return x;
}
int main()
{
long n,m;
long x,y,t1;
while(scanf("%lld",&n)!=EOF)
{
while(n--)
{
scanf("%ld",&m);
scanf("%ld",&x); while(--m)
{
scanf("%ld",&y);
t1=mcm(x,y);
x=x*y/t1;
}
printf("%ld\n",x);
}
}
return ;
}

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