Uva 11889 - Benefit（ 数论 ）

Uva 11889 - Benefit（ 数论 ）

题意: calculate the lowest integer B such that LCM(A, B) = C 

分析: LCM(A,B) = C = A*B/GCD(A,B)C*GCD(A,B) = A*BC/A = B/GCD(A,B)如果C%A != 0 无解否则, 令t = C/AB = t * GCD(A,B) 即B 一定是 t 的整数倍从t开始枚举B

#include <cstdio>
typedef long long LL;
typedef unsigned long long ULL;

ULL gcd(ULL a,ULL b)
{
    return b == 0 ? a : gcd(b, a%b);
}

void Orz()
{
    LL A, C, T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld %lld",&A,&C);
        int flag = 1;
        if(C % A == 0)
        {
            LL t = C/A;
            for(int B = t;B <= C; B += t)
            {
                if(B/gcd(B,A) == t)
                {
                    flag = 0;
                    printf("%lld\n",B);
                    break;
                }
            }
        }
        if(flag)
            puts("NO SOLUTION");
    }
}

int main()
{
    Orz();
    return 0;
}

代码君