2015
04-13

D_num

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.

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

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

6
10
9

2 3 6
2 5 10
is not a D_num

#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;
}


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