Least Common Multiple（最小公倍数）

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

解析：给一个整数串，求出这些数的最小公倍数。

思路：两个两个的求。

#include"stdio.h"
int mcm(long long x,long long y)
//find the greatest common divisor of two numbers
{
long long
t;
while(y)
{
t=x%y;
x=y;
y=t;
}
return x;
}
int main()
{
long long
n,m;
long long
x,y,t1;
while(scanf("%lld",&n)!=EOF)
{
while(n--)
{
scanf("%lld",&m);
scanf("%lld",&x);
while(--m)
{
scanf("%lld",&y);
t1=mcm(x,y);
x=x*y/t1;
//least common multiple is two numbers multiply divide their greatest common divisor
}
printf("%lldn",x);
}
}
return 0;
}

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