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 ;
}