HDU-1108最小公倍数

 1 #include<bits/stdc++.h>
 2 #define inf 0x3f3f3f3f
 3 typedef long long ll;
 4 const int M = int(1e5)*2 + 5;
 5 using namespace std;
 6 
 7 int gcd(int a, int b)
 8 {
 9     return b ?  gcd(b, a%b) : a;
10 }
11 int main()
12 {
13     int n, m;
14     while(~scanf("%lld%lld",&n,&m))
15         cout << n*m/ gcd(n, m)<<endl;
16     return 0;
17 }