POJ 2429 long long 质因数分解

Posted 2020-09-29 LuZhiyuan

GCD & LCM Inverse

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16206		Accepted: 3008

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

题意：

给出gcd和lcm求和最小的两个数a,b

代码：

//a*b=gcd*lcm,x=a/gcd,y=b/gcd,s=lcm/gcd => x*y=s;
//想要a+b小 => x*gcd+y*gcd小 => x*y小，x和y越接近sqrt(S)时x+y越小(对勾函数)
//所以求出s的素因子然后搞一搞就行了
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<algorithm>
using namespace std;
typedef long long ll;
const int S=20;//随机算法判定次数，S越大判错概率越小
//计算(a*b)%c;a,b,c<2^63
ll mult_mod(ll a,ll b,ll c){
    a%=c;
    b%=c;
    ll ret=0;
    while(b){
        if(b&1){
            ret+=a;
            ret%=c;
        }
        a<<=1;
        if(a>=c) a%=c;
        b>>=1;
    }
    return ret;
}
//计算(x^n)%c
ll pow_mod(ll x,ll n,ll c){
    if(n==1) return x%c;
    x%=c;
    ll tmp=x;
    ll ret=1;
    while(n){
        if(n&1) ret=mult_mod(ret,tmp,c);
        tmp=mult_mod(tmp,tmp,c);
        n>>=1;
    }
    return ret;
}
//以a为基,n-1=x*2^t      a^(n-1)=1(mod n)  验证n是不是合数
//一定是合数返回true,不一定返回false
bool check(ll a,ll n,ll x,ll t){
    ll ret=pow_mod(a,x,n);
    ll last=ret;
    for(int i=1;i<=t;i++){
        ret=mult_mod(ret,ret,n);
        if(ret==1&&last!=1&&last!=n-1) return true;
        last=ret;
    }
    if(ret!=1) return true;
    return false;
}
// Miller_Rabin()算法素数判定
//是素数返回true.(可能是伪素数，但概率极小)
//合数返回false;
bool Miller_Rabin(ll n){
    if(n<2) return false;
    if(n==2) return true;
    if((n&1)==0) return false;
    ll x=n-1;
    ll t=0;
    while((x&1)==0){
        x>>=1;
        t++;
    }
    for(int i=0;i<S;i++){
        ll a=rand()%(n-1)+1;
        if(check(a,n,x,t)) return false;
    }
    return true;
}
ll gcd(ll a,ll b){
    if(a==0) return 1;
    if(a<0) return gcd(-a,b);
    while(b){
        ll t=a%b;
        a=b;
        b=t;
    }
    return a;
}
//Pollard_rho质因数分解算法
ll factor[1000];//质因数分解结果(无序的)
int tot;//质因数的个数，数组下标从0开始
ll Pollard_rho(ll x,ll c){
    ll i=1,k=2;
    ll x0=rand()%x;
    ll y=x0;
    while(1){
        i++;
        x0=(mult_mod(x0,x0,x)+c)%x;
        ll d=gcd(y-x0,x);
        if(d!=1&&d!=x) return d;
        if(y==x0) return x;
        if(i==k){
            y=x0;
            k+=k;
        }
    }
}
//对n进行素因子分解
void findfac(ll n){
    if(Miller_Rabin(n)){//素数
        factor[tot++]=n;
        return;
    }
    ll p=n;
    while(p>=n)
        p=Pollard_rho(p,rand()%(n-1)+1);
    findfac(p);
    findfac(n/p);
}
ll ans,fa[100];
int nu;
void dfs(int st,ll x,ll maxx)
{
    if(st>nu){
        if(x<=maxx&&x>ans)
            ans=x;
        return;
    }
    dfs(st+1,x,maxx);
    dfs(st+1,x*fa[st],maxx);
}
int main()
{
    //srand(time(NULL));
    ll a,b;
    while(scanf("%lld%lld",&a,&b)==2){
        if(a==b){
            printf("%lld %lld\n",a,b);
            continue;
        }
        tot=nu=0;
        ll s=b/a;
        findfac(s);
        sort(factor,factor+tot);
        fa[0]=factor[0];
        for(int i=1;i<tot;i++){//合并相同的素因子
            if(factor[i]==factor[i-1]) fa[nu]*=factor[i];
            else
                fa[++nu]=factor[i];
        }
        ll x=(ll)sqrt(s*1.0);
        ans=1;
        dfs(0,1,x);
        printf("%lld %lld\n",ans*a,s/ans*a);
    }
    return 0;
}