HDU-3864 D_num Miller_Rabin和Pollard_rho

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3864

　　题意：给定一个数n，求n的因子只有四个的情况。

　　Miller_Rabin和Pollard_rho模板题，复杂度O(n^(1/4))，注意m^3=n的情况。

//STATUS:C++_AC_62MS_232KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef long long LL;
typedef unsigned long long ULL;
//const
const int N=2000010;
const int INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

LL factor[100];   //质因数分解结果（刚返回时是无序的）
int tol;   //质因数的个数。数组小标从0开始
const int S=10;

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;
}

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 mod)//x^n%c
{
    if(n==1)return x%mod;
    x%=mod;
    LL tmp=x;
    LL ret=1;
    while(n)
    {
        if(n&1) ret=mult_mod(ret,tmp,mod);
        tmp=mult_mod(tmp,tmp,mod);
        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;//合数
        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;//偶数
    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;//rand()需要stdlib.h头文件
        if(check(a,n,x,t))
            return false;//合数
    }
    return true;
}

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[tol++]=n;
        return;
    }
    LL p=n;
    while(p>=n)p=Pollard_rho(p,rand()%(n-1)+1);
    findfac(p);
    findfac(n/p);
}

LL n;

int main(){
//    freopen("in.txt","r",stdin);
    srand(time(NULL));
    int i,j;
    LL a,b;
    while(~scanf("%I64d",&n))
    {
        if(n==1){
            printf("is not a D_num
");
            continue;
        }
        tol=0;
        findfac(n);
        if(tol!=2 && tol!=3){
            printf("is not a D_num
");
            continue;
        }
        sort(factor,factor+tol);
        if(tol==2 && factor[0]!=factor[1]){
            printf("%I64d %I64d %I64d
",factor[0],factor[1],n);
        }
        else if(tol==3 && factor[0]==factor[1] && factor[1]==factor[2]){
            printf("%I64d %I64d %I64d
",factor[0],factor[0]*factor[0],n);
        }
        else printf("is not a D_num
");
    }
    return 0;
}