Minimum Public multiple of HDU 1108

When the onset of laziness cancer occurs, you need to exercise questions.

GCD + LCM calculate GCD and then LCM.

 
Int gcd (int A, int B) {int r = 0; while (B! = 0) {r = A % B; A = B; B = r;} return ;}
 

Example 10 14

10% 14 = 10; A = 14, r = 10, B = 10;

14% 10 = 4, A = 10, r = 4, B = 4;

10% 4 = 2, A = 4, r = 2, B = 2;

4% 2 = 0, A = 2, r = 0, B = 0;

Therefore, gcd = 2;

# Include <cstdio> # include <cstring> # include <string> # include <queue> # include <algorithm> # include <map> # include <stack> # include <iostream> # include <list> # include <set> # include <cmath> # define INF 0x7fffffff # define EPS 1e-6 # define ll long # define clri for (INT I = 0; I <n; I ++) # define clrj for (Int J = 0; j <n; j ++) # define clrk for (int K = 0; k <N; k ++) # define debug puts ("= Fuck ="); # define acfun STD: IOS: sync_with_stdio (Fa LSE) # define Nmax 1001 # define mMax 1001*1001 using namespace STD; int gcd (int A, int B) {int r = 0; while (B! = 0) {r = A % B; A = B; B = r;} return a;} int lcm (int A, int B, int g) {return a * B/g;} int main () {int A, B; while (~ Scanf ("% d", & A, & B) {printf ("% d \ n", lcm (a, B, gcd (, b);} return 0 ;}
 

Minimum Public multiple of HDU 1108