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 == 0 ? 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 == 0)
        {
            LL b = c / a;
            LL g = gcd(a, b);
            LL t = 1;
            while(g != 1)
            {
                a /= g;
                t *= g;
                g = gcd(a, b);
            }
            printf("%lld\n", b*t);
        }
        else puts("NO SOLUTION");
    }

    return 0;
}