LightOj 1215 Finding LCM
Discription
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
Sample Output
Case 1: 2
Case 2: 1
Case 3: impossible
本来感觉这个题太水了就不想写博客了2333,但是考虑到我马上要做的那个题可能要用到这个题 一个东西(但是回头再看这句话时突然打脸),所以还是写一下。
可以先求出lcm(a,b)=p,然后本质就是求一个 最小的 X 使得 lcm(X,p) = L。
无解很好判,只要p不是L的约数就无解。
考虑到lcm是对指数取max,而我们的目的是让X最小,所以我们可以让X在p和L次数相同的质因子上的次数取0,而在其他的质因子上取L在这上面的次数。
所以我们可以直接对 L/p 质因子分解, 然后 这里的质因子就是所有X要和L次数一样的质因子,只要把p和L/p在这上面的指数加起来就好啦。
#include<bits/stdc++.h> #define ll unsigned long long using namespace std; int T,d[100],num=0; ll a,b,L,ans=1; ll gcd(ll x,ll y){ return y?gcd(y,x%y):x; } inline void dvd(ll x){ for(int i=2;i*(ll)i<=x;i++) if(!(x%i)){ d[++num]=i; ll pre=x; while(!(x%i)) x/=i; ans*=pre/x; if(x==1) break; } if(x!=1) d[++num]=x,ans*=x; } inline void solve(){ for(int i=1;i<=num;i++){ ll pre=a; while(!(a%d[i])) a/=d[i]; ans*=pre/a; } printf("%llu\n",ans); } int main(){ scanf("%d",&T); for(int i=1;i<=T;i++){ scanf("%llu%llu%llu",&a,&b,&L); a=a*b/gcd(a,b); printf("Case %d: ",i); num=0,ans=1; if(L%a) puts("impossible"); else{ dvd(L/a); solve(); } } return 0; }