POJ 2429 GCD & LCM Inverse 因式分解

Description

Input

Output

Sample Input

3 60

Sample Output

12 15

Source

//408K	500MS	
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define Times 11
#define inf ((long long)1<<61)
#define C 201
using namespace std;
long long jl[501],numfactor[501],mini,mina,minb;//jl里面存的是大数的所有质因子，mini为最小的质因数
int ct,num[65];
long long key,gc;
int len;
long long random(long long n)//生成随机数
{
    return (long long)((double)rand()/RAND_MAX*n+0.5);
}
long long gcd(long long a,long long b)//最大公约数
{
    return b==0?a:gcd(b,a%b);
}
long long multi(long long a,long long b,long long m)//a*b%m
{
    long long ret=0;
    while(b>0)
    {
        if(b&1)ret=(ret+a)%m;
        b>>=1;
        a=(a<<1)%m;
    }
    return ret;
}
long long quick_mod(long long a,long long b,long long m)//a^b%m
{
    long long ans=1;
    a%=m;
    while(b)
    {
        if(b&1)
        {
            ans=multi(ans,a,m);
            b--;
        }
        b/=2;
        a=multi(a,a,m);
    }
    return ans;
}
bool Witness(long long a,long long n)
{
    long long m=n-1;
    int j=0;
    while(!(m&1))
    {
        j++;
        m>>=1;
    }
    long long x=quick_mod(a,m,n);
    if(x==1||x==n-1)return false;
    while(j--)
    {
        x=x*x%n;
        if(x==n-1)return false;
    }
    return true;
}
bool miller_rabin(long long n)//素数测试
{
    if(n<2)return false;
    if(n==2)return true;
    if(!(n&1))return false;
    for(int i=1; i<=Times; i++)
    {
        long long a=random(n-2)+1;
        if(Witness(a,n))return false;
    }
    return true;
}
long long pollard_rho(long long n,int c)//整数n分解,c一般为201
{
    long long x,y,d,i=1,k=2;
    x=random(n-1)+1;
    y=x;
    while(1)
    {
        i++;
        x=(multi(x,x,n)+c)%n;
        d=gcd(y-x,n);
        if(1<d&&d<n)return d;
        if(y==x)return n;
        if(i==k)
        {
            y=x;
            k<<=1;
        }
    }
}
void find(long long n,int k)
{
    if(n==1)return;
    if(miller_rabin(n))
    {
        jl[++ct]=n;
        return;
    }
    long long p=n;
    while(p>=n)p=pollard_rho(p,k--);
    find(p,k);
    find(n/p,k);
}
void dfs(int cur,long long value)
{
    long long s=1,a,b;
    if(cur==len+1)
    {
        a=value;
        b=key/value;
        if(gcd(a,b)==1)
        {
            a*=gc;
            b*=gc;
            if(a+b<mini)
            {
                mini=a+b;
                mina=a<b?a:b;
                minb=a<b?b:a;
            }
        }
        return;
    }
    for(int i=0;i<=num[cur];i++)
    {
        if(value*s>=mini) return;
        dfs(cur+1,value*s);
        s*=numfactor[cur];
    }
}
void solve(long long n)
{
    ct=0;
    find(n,C);
    sort(jl+1,jl+ct+1);
    memset(num,0,sizeof(num));
    len=0;
    num[0]=1;
    numfactor[0]=jl[1];
    for(int i=2;i<=ct;i++)
    {
        if(numfactor[len]!=jl[i])numfactor[++len]=jl[i];
        num[len]++;
    }
    dfs(0,1);
    printf("%lld %lld\n",mina,minb);
}
int main()
{
    long long  lc;
    while(scanf("%lld%lld",&gc,&lc)!=EOF)
    {
        if(gc==lc){printf("%lld %lld\n",gc,lc);continue;}
        key=lc/gc;
        mini=inf;
        solve(key);
    }
    return 0;
}