首页 > ACM题库 > HDU-杭电 > HDU 3864-D_num-数论-[解题报告]HOJ
2015
04-13

HDU 3864-D_num-数论-[解题报告]HOJ

D_num

问题描述 :

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.

输入:

Some cases (case < 100);
Each line have a numeral N(1<=N<10^18)

输出:

Some cases (case < 100);
Each line have a numeral N(1<=N<10^18)

样例输入:

6
10
9

样例输出:

2 3 6
2 5 10
is not a D_num

题意:题意很简单,就是求n是否只有4个因子,如果是就输出除1外的所有因子。

题解:这题n太大(n<=1e18),所以不能直接求解,需要用到Pollard_rho算法和Miller_Rabin算法。Miller_Rabin算法的作用是判断一个数是否是个素数,算法速度很快,虽然是概率算法,有一定误判概率,不过可以多次运算大幅度减少误判,误判概率与运算次数t有关,为2^(-t);当t够大时,误判的可能性就很小了。Pollard_rho算法作用是求一个数的因子,这个复杂度为O(sqrt(p)),p为这个数的因子。两者结合才能解决这道题。具体算法说明如下,更详细的请百度,google

算法链接:http://blog.csdn.net/a601025382s/article/details/12168043

耗时:78MS

#include <cstdio>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
using namespace std;
typedef __int64 LL;
const LL NUM=10;//运算次数,Miller_Rabin算法为概率运算,误判率为2^(-NUM);
LL t,f[100];
LL mul_mod(LL a,LL b,LL n)//求a*b%n,由于a和b太大,需要用进位乘法
{
    a=a%n;
    b=b%n;
    LL s=0;
    while(b)
    {
        if(b&1)
            s=(s+a)%n;
        a=(a<<1)%n;
        b=b>>1;
    }
    return s;
}
LL pow_mod(LL a,LL b,LL n)//求a^b%n
{
    a=a%n;
    LL s=1;
    while(b)
    {
        if(b&1)
            s=mul_mod(s,a,n);
        a=mul_mod(a,a,n);
        b=b>>1;
    }
    return s;
}
bool check(LL a,LL n,LL r,LL s)
{
    LL ans,p,i;
    ans=pow_mod(a,r,n);
    p=ans;
    for(i=1;i<=s;i++)
    {
        ans=mul_mod(ans,ans,n);
        if(ans==1&&p!=1&&p!=n-1)return true;
        p=ans;
    }
    if(ans!=1)return true;
    return false;
}
bool Miller_Rabin(LL n)//Miller_Rabin算法,判断n是否为素数
{
    if(n<2)return false;
    if(n==2)return true;
    if(!(n&1))return false;
    LL i,r,s,a;
    r=n-1;s=0;
    while(!(r&1)){r=r>>1;s++;}
    for(i=0;i<NUM;i++)
    {
        a=rand()%(n-1)+1;
        if(check(a,n,r,s))
            return false;
    }
    return true;
}
LL gcd(LL a,LL b)
{
    return b==0?a:gcd(b,a%b);
}
LL Pollard_rho(LL n,LL c)//Pollard_rho算法,找出n的因子
{
    LL i=1,j,k=2,x,y,d,p;
    x=rand()%n;
    y=x;
    while(true)
    {
        i++;
        x=(mul_mod(x,x,n)+c)%n;
        if(y==x)return n;
        if(y>x)p=y-x;
        else p=x-y;
        d=gcd(p,n);
        if(d!=1&&d!=n)return d;
        if(i==k)
        {
            y=x;
            k+=k;
        }
    }
}
void find(LL n)//找出n的所有因子
{
    if(Miller_Rabin(n))
    {
        f[t++]=n;//保存所有因子
        return;
    }
    LL p=n;
    while(p>=n)p=Pollard_rho(p,rand()%(n-1)+1);//由于p必定为合数,所以通过多次求解必定能求得答案
    find(p);
    find(n/p);
}
int main()
{
    srand(time(NULL));//随机数设定种子
    LL n;
    while(cin>>n)
    {
        if(n==1){cout<<"is not a D_num"<<endl;continue;}//特判
        t=0;
        find(n);
        if(t!=2&&t!=3){cout<<"is not a D_num"<<endl;continue;}
        sort(f,f+t);
        if(t==2)
        {
            if(f[0]!=f[1])cout<<f[0]<<" "<<f[1]<<" "<<n<<endl;
            else cout<<"is not a D_num"<<endl;
        }
        else//n是一个素数的三次方
        {
            if(f[0]==f[1]&&f[1]==f[2])cout<<f[0]<<" "<<f[0]*f[0]<<" "<<n<<endl;
            else cout<<"is not a D_num"<<endl;
        }
    }
    return 0;
}

版权声明:本文为博主原创文章,未经博主允许不得转载。

参考:http://blog.csdn.net/a601025382s/article/details/12177549


  1. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮