GCD&LCM逆问题-优快云博客

GCD & LCM Inverse

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5559		Accepted: 993

Description

Given two positive integers a and b, we can easily calculate the greatest common divisor (GCD) and the least common multiple (LCM) of a and b. But what about the inverse? That is: given GCD and LCM, finding a and b.

Input

The input contains multiple test cases, each of which contains two positive integers, the GCD and the LCM. You can assume that these two numbers are both less than 2^63.

Output

For each test case, output a and b in ascending order. If there are multiple solutions, output the pair with smallest a + b.

Sample Input

3 60

Sample Output

12 15

Source

POJ Achilles

大数质因数分解

然后DFS搜索出每一个因子

#include<stdio.h> #include<math.h> #include<string.h> #include<stdlib.h> #include<iostream> #include<time.h> using namespace std; typedef unsigned __int64 ll; const ll MAX=100; const ll inf=(ll)1<<62; ll f0[100],ff,n,tmp,ret,ret1; ll myrandom() { ll a; a=rand(); a*=rand(); a*=rand(); a*=rand(); return a; } ll mulmod(ll a,ll b,ll c) { ll ret=0; while(b) { if(b&1) { ret+=a; if(ret>c) ret=ret-c; } a=a<<1; if(a>c) a=a-c; b=b>>1; } return ret; } ll powmod(ll a,ll b,ll c) { ll ret=1; while(b) { if(b&1) ret=mulmod(ret,a,c); a=mulmod(a,a,c); b=b>>1; } return ret; } int miller(ll base,ll n) { ll m=n-1,k=0; int i; while(m%2==0) { m=m>>1; k++; } if(powmod(base,m,n)==1) return 1; for(i=0;i<k;i++) { if(powmod(base,m<<i,n)==n-1) return 1; } return 0; } int miller_rabin(ll n) { int i; for( i=2;i<100;++i) if(n%i==0) if(n==i) return 1; else return 0; for(i=0;i<MAX;++i) { ll base=myrandom()%(n-1)+1; if(!miller(base,n)) return 0; } return 1; } ll gcd(ll a,ll b) { if(b==0) return a; else return gcd(b,a%b); } ll f(ll a,ll b) { return (mulmod(a,a,b)+1)%b; } ll pollard_rho(ll n) { int i; if(n<=2) return 0; for( i=2;i<100;++i) if(n%i==0) return i; ll p,x,xx; for( i=1;i<MAX;i ++) { x=myrandom()%n; xx=f(x,n); p=gcd((xx+n-x)%n,n); while(p==1) { x=f(x,n); xx=f(f(xx,n),n); p=gcd((xx+n-x)%n,n)%n; } if(p) return p; } return 0; } ll prime(ll a) { if(miller_rabin(a)) return 0; ll t=pollard_rho(a); ll p=prime(t); if(p) return p; return t; } ll ans,ansa,ansb; ll min(ll a,ll b) { if(a>b) return b; else return a; } void dfs(ll n,ll now,ll id) { if(now*now>n)//剪哈枝 return; if(id==ff) { ll a,b; a=now; b=n/now; if(a+b<ans) { ans=a+b; ansa=a; ansb=b; } return; } dfs(n,now*f0[id],id+1); dfs(n,now,id+1); } void swap(ll &a,ll &b) { ll temp; temp=a; a=b; b=temp; } int main() { ll a,b,key,tmp; srand(time(NULL)); while(scanf("%I64u%I64u",&a,&b)!=EOF) { if(a>b) swap(a,b); if(a==b||miller_rabin(b)) { printf("%I64u %I64u\n",a,b); continue; } key=b/a; ff=0,n=key; while(key>1) { if(miller_rabin(key)) break; tmp=prime(key); f0[ff]=tmp; key=key/tmp; while(key%tmp==0) { key=key/tmp; f0[ff]=f0[ff]*tmp; } ff++; } if(key>1) { f0[ff++]=key; } ans=inf; dfs(n,1,0); printf("%I64u %I64u\n",ansa*a,ansb*a); } return 0; }