poj 2429 GCD & LCM Inverse

http://poj.org/problem?id=2429

这题能1A感觉很好！Pollard_Rho 很神奇呀，能把这么大的数分解成素因子,然后dfs 找到其 ,lcm/gcd 的所有的因子就行了，取最优的答案

#include<iostream>
#include<cstdio>
#include<ctime>
#include<cstring>
#include<cstdlib>
#include<algorithm>

#define C 240
#define TIME 10
#define LL unsigned long long
using namespace std;
int cnt;
LL ans;
LL prime[1010],a,b;

LL gcd(LL a, LL b)
{
    if(a==0) return 1;
    if(a<0) return gcd(-a,b);
    return b==0?a:gcd(b,a%b);
}
LL MultMod(LL a,LL b,LL n)
{
    a%=n;
    b%=n;
    LL ret=0;
    while(b)
    {
        if(b&1)
        {
            ret+=a;
            if(ret>=n) ret-=n;
        }
        a=a<<1;
        if(a>=n) a-=n;
        b=b>>1;
    }
    return ret;
}
LL PowMod(LL a,LL n,LL m)
{
    LL ret=1;
    a=a%m;
    while(n>=1)
    {
        if(n&1)
            ret=MultMod(ret,a,m);
        a=MultMod(a,a,m);
        n=n>>1;
    }
    return ret;
}
bool Witness(LL a,LL n)
{
    LL t=0,u=n-1;
    while(!(u&1))
    {
        t++;
        u/=2;
    }
    LL x0=PowMod(a,u,n);
    for(int i=1; i<=t; i++)
    {
        LL x1=MultMod(x0,x0,n);
        if(x1==1&&x0!=1&&x0!=(n-1))
            return true;
        x0=x1;
    }
    if(x0!=1)
        return true;
    return false;
}
bool Miller_Rabin(LL n,int t)
{
    if(n==2) return true;
    if((n&1)==0)  return false;
    srand(time(NULL));
    for(int i=0; i<t; i++)
    {
        LL a=rand()%(n-1)+1;
        if(Witness(a,n))
            return false;
    }
    return true;
}
LL Pollard_Rho(LL n,LL c)
{
    LL i=1,x=rand()%n,y=x,k=2;
    while(1)
    {
        i++;
        x=(MultMod(x,x,n)+c)%n;
        LL d=gcd(y-x,n);
        if(d!=1&&d!=n)
            return d;
        if(x==y)
            return n;
        if(i==k)
        {
            y=x;
            k*=2;
        }
    }
}
void get_small(LL n,LL c)
{
    if(n==1) return;
    if(Miller_Rabin(n,TIME))
    {
        prime[cnt++]=n;
        return;
    }
    LL p=n;
    while(p>=n) p=Pollard_Rho(p,c--);
    get_small(p,c);
    get_small(n/p,c);
}
LL factor[20],ans1,ans2;
int fac_n;

void dfs(int id,LL mul)
{
    //cout<<id<<" "<<mul<<endl;
    if(id==fac_n)
    {
        if(ans1==-1||ans1+ans2 > a*mul+b/mul)  ans1=a*mul,ans2=b/mul;
        if(ans1>ans2) swap(ans1,ans2);
        return;
    }
    dfs(id+1,mul*factor[id]);
    dfs(id+1,mul);
}
int main()
{
    while(scanf("%llu %llu",&a,&b)==2)
    {
        cnt=0;
        get_small(b/a,C);
        sort(prime,prime+cnt);
        /*for(int i=0;i<cnt;i++)
          cout<<prime[i]<<" ";
        cout<<endl;*/
        fac_n=0;
        for(int i=0;i<cnt;i++)
        {
             factor[fac_n]=prime[i];
             while(i<cnt-1&&prime[i]==prime[i+1]) factor[fac_n]*=prime[i++];
             fac_n++;
        }
       /* for(int i=0;i<fac_n;i++)
          cout<<factor[i]<<" ";
        cout<<endl;*/
        ans1=-1,ans2=-1;
        dfs(0,1);
        printf("%llu %llu\n",ans1,ans2);
    }
    return 0;
}
//3 60