1 #include<iostream> 2 #include<cstdio> 3 using namespace std; 4 int n,m; 5 //求解n,m的gcd和lcm 6 7 int gcd(int a,int b) 8 { 9 while(1) 10 { 11 if(a<b) {int t=a;a=b;b=t;} 12 if(b==0) return a; 13 int x=a%b;a=b;b=x; 14 } 15 } 16 17 int lcm(int a,int b) 18 { 19 return a/gcd(a,b)*b; 20 } 21 22 int main() 23 { 24 scanf("%d %d",&n,&m); 25 cout<<gcd(n,m)<<endl; 26 cout<<lcm(n,m); 27 return 0; 28 }
希望自己下次不要再忘记算法啦~~~