Least Common Multiple
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
Recommend
JGShining
AC code:
#include "stdio.h"
__int64 s[1000];
__int64 hcf(__int64 a,__int64 b)
{
__int64 r=0;
while(b!=0)
{
r=a%b;
a=b;
b=r;
}
return(a);
}
__int64 lcd(__int64 u,__int64 v,__int64 h)
{
return(u*v/h);
}
int main(int argc, char* argv[])
{
int n,m,i;
while(scanf("%d",&n)==1)
{
while(n--)
{
scanf("%d",&m);
for(i=0;i<=m-1;i++)
scanf("%I64d",&s[i]);
for(i=0;i<=m-2;i++)
{
s[i+1]=lcd(s[i],s[i+1],hcf(s[i],s[i+1]));
}
printf("%I64d\n",s[m-1]);
}
}
return 0;
__int64 s[1000];
__int64 hcf(__int64 a,__int64 b)
{
__int64 r=0;
while(b!=0)
{
r=a%b;
a=b;
b=r;
}
return(a);
}
__int64 lcd(__int64 u,__int64 v,__int64 h)
{
return(u*v/h);
}
int main(int argc, char* argv[])
{
int n,m,i;
while(scanf("%d",&n)==1)
{
while(n--)
{
scanf("%d",&m);
for(i=0;i<=m-1;i++)
scanf("%I64d",&s[i]);
for(i=0;i<=m-2;i++)
{
s[i+1]=lcd(s[i],s[i+1],hcf(s[i],s[i+1]));
}
printf("%I64d\n",s[m-1]);
}
}
return 0;
}