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：

   1: #include <iostream>

   2: #include <algorithm>

   3: #include <cstring>

   4: using namespace std;

   5: typedef long long ll;

   6:  

   7: ll gcd(ll a,ll b){

   8:     if(a<b) swap(a,b);

   9:     return b==0?a:gcd(b,a%b);

  10: }

  11:  

  12: ll lcm(ll a,ll b){

  13:     return a/gcd(a,b)*b;

  14: }

  15:  

  16: int run()

  17: {

  18:     ll a,b,cas=1,L,T;

  19:     cin>>T;

  20:     while(T--)

  21:     {

  22:         cin>>a>>b>>L;

  23:         ll m=lcm(a,b);

  24:         if(m>L||L%m!=0)

  25:         {

  26:             cout<<"Case "<<cas++<<": "<<"impossible"<<endl;

  27:             continue;

  28:         }

  29:         ll c=L/m,g;

  30:         if(c!=1)

  31:           while((g=gcd(m,c))!=1){

  32:               c*=g,m/=g;

  33:           }

  34:         cout<<"Case "<<cas++<<": "<<c<<endl;

  35:     }

  36:     return 0;

  37: }

  38:  

  39: int main()

  40: {

  41:     ios::sync_with_stdio(0);

  42:     return run();

  43: }