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=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //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[];   //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=;

 LL gcd(LL a,LL b)
 {
     if(a==)return ;
     if(a<) 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=;
     while(b)
     {
         if(b&){ret+=a;ret%=c;}
         a<<=;
         if(a>=c)a%=c;
         b>>=;
     }
     return ret;
 }

 //计算  x^n %c
 LL pow_mod(LL x,LL n,LL mod)//x^n%c
 {
     if(n==)return x%mod;
     x%=mod;
     LL tmp=x;
     LL ret=;
     while(n)
     {
         if(n&) ret=mult_mod(ret,tmp,mod);
         tmp=mult_mod(tmp,tmp,mod);
         n>>=;
     }
     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=;i<=t;i++)
     {
         ret=mult_mod(ret,ret,n);
         if(ret==&&last!=&&last!=n-) return true;//合数
         last=ret;
     }
     if(ret!=) 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<)return false;
     if(n==)return true;
     if((n&)==) return false;//偶数
     LL x=n-;
     LL t=;
     while((x&)==){x>>=;t++;}
     for(int i=;i<S;i++)
     {
         LL a=rand()%(n-)+;//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=,k=;
     LL x0=rand()%x;
     LL y=x0;
     while()
     {
         i++;
         x0=(mult_mod(x0,x0,x)+c)%x;
         LL d=gcd(y-x0,x);
         if(d!=&&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-)+);
     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==){
             printf("is not a D_num\n");
             continue;
         }
         tol=;
         findfac(n);
         if(tol!= && tol!=){
             printf("is not a D_num\n");
             continue;
         }
         sort(factor,factor+tol);
         if(tol== && factor[]!=factor[]){
             printf("%I64d %I64d %I64d\n",factor[],factor[],n);
         }
         else if(tol== && factor[]==factor[] && factor[]==factor[]){
             printf("%I64d %I64d %I64d\n",factor[],factor[]*factor[],n);
         }
         else printf("is not a D_num\n");
     }
     return ;
 }