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
求连续多个数的最小公倍数。
#include <iostream> #include <algorithm> using namespace std; int main() { long long t, n, x; cin>>t; while(t--) { cin>>n; n--; cin>>x; long long f = x; while(n--) { cin>>x; f = f*x/(__gcd(f, x)); } cout<<f<<endl; } return 0; }