UVa 11889 (GCD) Benefit

好吧，被大白书上的入门题给卡了。=_=||

已知LCM(A, B) = C，已知A和C，求最小的B

一开始我想当然地以为B = C / A，后来发现这时候的B不一定满足gcd(A, B) = 1

A要不断地除去gcd(A, B)，直到满足gcd(A, B) = 1

B最后就应该乘上A除去的值

 #include <cstdio>

 typedef long long LL;

 LL gcd(LL a, LL b)
 { return b ==  ? a : gcd(b, a%b); }

 int main()
 {
     int T;
     scanf("%lld", &T);
     LL a, c;
     while(T--)
     {
         scanf("%lld%lld", &a, &c);
         if(c % a == )
         {
             LL b = c / a;
             LL g = gcd(a, b);
             LL t = ;
             while(g != )
             {
                 a /= g;
                 t *= g;
                 g = gcd(a, b);
             }
             printf("%lld\n", b*t);
         }
         else puts("NO SOLUTION");
     }

     return ;
 }

代码君