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

 1 #include <cstdio>
 2
 3 typedef long long LL;
 4
 5 LL gcd(LL a, LL b)
 6 { return b == 0 ? a : gcd(b, a%b); }
 7
 8 int main()
 9 {
10     int T;
11     scanf("%lld", &T);
12     LL a, c;
13     while(T--)
14     {
15         scanf("%lld%lld", &a, &c);
16         if(c % a == 0)
17         {
18             LL b = c / a;
19             LL g = gcd(a, b);
20             LL t = 1;
21             while(g != 1)
22             {
23                 a /= g;
24                 t *= g;
25                 g = gcd(a, b);
26             }
27             printf("%lld\n", b*t);
28         }
29         else puts("NO SOLUTION");
30     }
31
32     return 0;
33 }

代码君