【LightOJ】1215

题目链接：这里写链接内容

1215 - Finding LCM
PDF (English) Statistics Forum
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

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我们肯定是要先求k = LCM（a，b） ，然后判断 l % k ， 如果不能整除，肯定是不满足的。如果可以整除，那么我们得到了 k 中不属于 l 的素因子。但是为了使 k / GCD（ans，k）* ans == l ，我们需要找出 k 和 ans 的公共素因子，并且让 ans 乘以其指数的差，这样才能保证 k / GCD（ans，k）不会把 k 中“有用的”质因子除掉。

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代码如下：

#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<cmath>#include<iostream>using namespace std;typedef long long LL;#define CLR(a,b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3fLL GCD(LL a,LL b){    return b == 0 ? a : GCD(b,a%b);}LL LCM(LL a,LL b){    return a / GCD(a,b) * b;}int main(){    int T;    LL a,b,l;    int Case = 1;    cin >> T;    while (T--)    {        cin >> a >> b >> l;        printf ("Case %d: ",Case++);        LL k = LCM(a,b);        if (l % k != 0)            puts("impossible");        else        {            LL ans = l / k;            while (1)            {                LL g = GCD(k,ans);                if (g == 1)                    break;                ans *= g;                k /= g;            }            cout << ans << endl;        }    }    return 0;}