LOJ Finding LCM(math)

1215 - Finding LCM

Time Limit: 2 second(s)
Memory Limit: 32 MB

LCM is an abbreviation used for Least Common Multiple in Mathematics. We say LCM (a, b, c) = L if and only if L is the least integer which is divisible by a, b and c.

You will be given a, b and L. You have to find c such that LCM (a, b, c) = L. If there are several solutions, print the one where c is as small as possible. If there is no solution, report so.

Input

Input starts with an integer T (≤ 325), denoting the number of test cases.

Each case starts with a line containing three integers a b L (1 ≤ a, b ≤ 10⁶, 1 ≤ L ≤ 10¹²).

Output

For each case, print the case number and the minimum possible value of c. If no solution is found, print 'impossible'.

Sample Input

3

3 5 30

209475 6992 77086800

2 6 10

Output for Sample Input

Case 1: 2

Case 2: 1

Case 3: impossible

题意：lcm(a,b,c)=L;现已知a,b,L的值，求是否存在c满足lcm(a,b,c)=L。

::首先求出a,b的最小公倍数m,则c必包含因子t=L/m;

令g=gcd(c,m);

假设c=t，c*m/g=L,当且仅当gcd(c,m)=1等式成立;

如果gcd(c,m)>1；

那么令(c*g)*(m/g)/gcd(c*g,m/g)=L;当且仅当gcd(c*g,m/g)=1;

如果gcd(c*g,m/g)>1重复上述操作；

例：a=2,b=3,L=12;

则m=6,L=12,t=2;

令c=t;判断gcd(6,2)==2,令c=c*2(==4),m=m/2(==3)

gcd(c,m)==1,故c=4;

代码：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;

ll gcd(ll a,ll b){
    if(a<b) swap(a,b);
    return b==0?a:gcd(b,a%b);
}

ll lcm(ll a,ll b){
    return a/gcd(a,b)*b;
}

int run()
{
    ll a,b,cas=1,L,T;
    cin>>T;
    while(T--)
    {
        cin>>a>>b>>L;
        ll m=lcm(a,b);
        if(m>L||L%m!=0)
        {
            cout<<"Case "<<cas++<<": "<<"impossible"<<endl;
            continue;
        }
        ll c=L/m,g;
        if(c!=1)
          while((g=gcd(m,c))!=1){
              c*=g,m/=g;
          }
        cout<<"Case "<<cas++<<": "<<c<<endl;
    }
    return 0;
}

int main()
{
    ios::sync_with_stdio(0);
    return run();
}