HDU 3864 D_num (pollard_rho大数素数分解)

D_num

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2046    Accepted Submission(s): 573

Problem Description Oregon Maple was waiting for Bob When Bob go back home. Oregon Maple asks Bob a problem that as a Positive number N, if there are only four Positive number M makes Gcd(N, M) == M then we called N is a D_num. now, Oregon Maple has some Positive numbers, and if a Positive number N is a D_num , he want to know the four numbers M. But Bob have something to do, so can you help Oregon Maple?

Gcd is Greatest common divisor.  

 

Input Some cases (case < 100);

Each line have a numeral N(1<=N<10^18)  

 

Output For each N, if N is a D_NUM, then output the four M (if M > 1) which makes Gcd(N, M) = M. output must be Small to large, else output “is not a D_num”.  

 

Sample Input 6 10 9  

 

Sample Output 2 3 6 2 5 10 is not a D_num  

 

Source
2011 Multi-University Training Contest 3 – Host by BIT  

 

Recommend lcy     就是判断一个long long的数的约数是不是有4个。 用pollard_rho,练习了下模板;

//============================================================================
// Name        : HDU3864.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <string.h>
#include <time.h>
using namespace std;
const int S=2;
long long mult_mod(long long a,long long b,long long c)
{
    a%=c;
    b%=c;
    long long ret=0;
    while(b)
    {
        if(b&1){ret+=a;ret%=c;}
        a<<=1;
        if(a>=c)a%=c;
        b>>=1;
    }
    return ret;
}
long long pow_mod(long long x,long long n,long long mod)
{
    if(n==1)return x%mod;
    x%=mod;
    long long tmp=x;
    long long ret=1;
    while(n)
    {
        if(n&1)ret=mult_mod(ret,tmp,mod);
        tmp=mult_mod(tmp,tmp,mod);
        n>>=1;
    }
    return ret;
}
long long check(long long a,long long n,long long x,long long t)
{
    long long ret=pow_mod(a,x,n);
    long long 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;
}
bool Miller_Rabin(long long n)
{
    if(n<2)return false;
    if(n==2)return true;
    if((n&1)==0)return false;
    long long x=n-1;
    long long t=0;
    while((x&1)==0){x>>=1;t++;}
    for(int i=0;i<S;i++)
    {
        long long a=rand()%(n-1)+1;
        if(check(a,n,x,t))
            return false;
    }
    return true;
}

long long factor[100];
int tol;
long long gcd(long long a,long long b)
{
    if(a==0)return 1;
    if(a<0)return gcd(-a,b);
    while(b)
    {
        long long t=a%b;
        a=b;
        b=t;
    }
    return a;
}

long long Pollard_rho(long long x,long long c)
{
    long long i=1,k=2;
    long long x0=rand()%x;
    long long y=x0;
    while(1)
    {
        i++;
        x0=(mult_mod(x0,x0,x)+c)%x;
        long long d=gcd(y-x0,x);
        if(d!=1&&d!=x)return d;
        if(y==x0)return x;
        if(i==k)
        {
            y=x0;
            k+=k;
        }
    }
}

void findfac(long long n)
{
    if(Miller_Rabin(n))
    {
        factor[tol++]=n;
        return;
    }
    long long p=n;
    while(p>=n)p=Pollard_rho(p,rand()%(n-1)+1);
    findfac(p);
    findfac(n/p);
}

int main()
{
    srand(time(NULL));
    long long n;
    while(scanf("%I64d",&n)==1)
    {
        if(n==1)
        {
            printf("is not a D_num\n");
            continue;
        }
        tol=0;
        findfac(n);
        if(tol!=2 && tol!=3)
        {
            printf("is not a D_num\n");
            continue;
        }
        sort(factor,factor+tol);
        if(tol==2)
        {
            if(factor[0]!=factor[1])
            {
                printf("%I64d %I64d %I64d\n",factor[0],factor[1],factor[0]*factor[1]);
                continue;
            }
            else
            {
                printf("is not a D_num\n");
                continue;
            }
        }
        if(tol==3)
        {
            if(factor[0]==factor[1]&&factor[1]==factor[2])
            {
                printf("%I64d %I64d %I64d\n",factor[0],factor[0]*factor[1],factor[0]*factor[1]*factor[2]);
                continue;
            }
            else
            {
                printf("is not a D_num\n");
                continue;
            }
        }
    }
    return 0;
}

 

    原文作者:算法小白
    原文地址: https://www.cnblogs.com/kuangbin/archive/2012/11/11/2765525.html
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