ZOJ Problem Set - 1797
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: East Central North America 2003, Practice
long long i;
if(a==0||b==0) //如果a和b之中有一个是0,则最大公约数就是0
return 0;
if(a<b) //进行交换将a和b中大的给a小的给b
{
i=a;
a=b;
b=i;
}
while(a%b) //辗转相除的核心,求最大公约数
{
i=a%b;
a=b;
b=i;
}
return b; //b就是最大公约数
}
int main()
{
int t,n,i;
long long s,b;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
scanf("%lld",&s);
for(i=1;i<n;i++)
{
scanf("%lld",&b); //输入一个b
long long temp=gcd(s,b);
if(temp==0) //如果最大公约数是0,那么最小公倍数就是0
s=0;
else
s=s*b/temp; //最大公约数是temp,那么最小公倍数就是a*b/temp(求a和b的最小公倍数)
}
printf("%lld\n",s);
}
return 0;
}