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
如下为一般方法写的代码
#include<iostream>
using namespace std;
int s[];
__int64 x;
int cmp ( const void *a , const void *b )
{
return *(int *)b - *(int *)a;
}
void f(int s[],int n)
{
int p=;
bool flag;
while(){
flag=true;
x=s[]*p;
for(int i=;i<n;i++){
if(x%s[i]!=){
flag=false;
break;
}
}
if(flag==true)break;
p++;
}
cout<<x<<endl;
}
int main()
{
int n;
cin>>n;
while(n--){
int a;
cin>>a;
for(int i=;i<a;i++)cin>>s[i];
qsort(s,a,sizeof(s[]),cmp);
f(s,a);
}
return ;
}
如下为运用辗转相除法写的代码
#include<iostream>
using namespace std;
int s[];
__int64 x,y;
int g(int a,int b)
{
int i,t;
y=a*b;
if(a<b){
t=a;
a=b;
b=t;
}
while(b){
t=b;
b=a%b;
a=t;
}
return y/a;
}
void f(int s[],int n)
{
x=s[];
for(int i=;i<n;i++){
x=g(s[i],x);
}
cout<<x<<endl;
}
int main()
{
int n;
cin>>n;
while(n--){
int a;
cin>>a;
for(int i=;i<a;i++)cin>>s[i];
f(s,a);
}
//system("pause");
return ;
}