HDU 3864 D_num(pollard_rho大数素数分解)
18271 ワード
D_num
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2046 Accepted Submission(s): 573
Problem Description
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.
Input
Some cases (case < 100);
Each line have a numeral N(1<=N<10^18)
Output
For each N, if N is a D_NUM, then output the four M (if M > 1) which makes Gcd(N, M) = M. output must be Small to large, else output “is not a D_num”.
Sample Input
6 10 9
Sample Output
2 3 6 2 5 10 is not a D_num
Source
2011 Multi-University Training Contest 3 - Host by BIT
Recommend
lcy
一つのlong longの数の約数が4つあるかどうかを判断することです.
pollard_でrho、テンプレートを練習しました.
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2046 Accepted Submission(s): 573
Problem Description
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.
Input
Some cases (case < 100);
Each line have a numeral N(1<=N<10^18)
Output
For each N, if N is a D_NUM, then output the four M (if M > 1) which makes Gcd(N, M) = M. output must be Small to large, else output “is not a D_num”.
Sample Input
6 10 9
Sample Output
2 3 6 2 5 10 is not a D_num
Source
2011 Multi-University Training Contest 3 - Host by BIT
Recommend
lcy
一つのlong longの数の約数が4つあるかどうかを判断することです.
pollard_でrho、テンプレートを練習しました.
//============================================================================
// Name : HDU3864.cpp
// Author :
// Version :
// Copyright : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <string.h>
#include <time.h>
using namespace std;
const int S=2;
long long mult_mod(long long a,long long b,long long c)
{
a%=c;
b%=c;
long long ret=0;
while(b)
{
if(b&1){ret+=a;ret%=c;}
a<<=1;
if(a>=c)a%=c;
b>>=1;
}
return ret;
}
long long pow_mod(long long x,long long n,long long mod)
{
if(n==1)return x%mod;
x%=mod;
long long tmp=x;
long long ret=1;
while(n)
{
if(n&1)ret=mult_mod(ret,tmp,mod);
tmp=mult_mod(tmp,tmp,mod);
n>>=1;
}
return ret;
}
long long check(long long a,long long n,long long x,long long t)
{
long long ret=pow_mod(a,x,n);
long long 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;
last=ret;
}
if(ret!=1)return true;
return false;
}
bool Miller_Rabin(long long n)
{
if(n<2)return false;
if(n==2)return true;
if((n&1)==0)return false;
long long x=n-1;
long long t=0;
while((x&1)==0){x>>=1;t++;}
for(int i=0;i<S;i++)
{
long long a=rand()%(n-1)+1;
if(check(a,n,x,t))
return false;
}
return true;
}
long long factor[100];
int tol;
long long gcd(long long a,long long b)
{
if(a==0)return 1;
if(a<0)return gcd(-a,b);
while(b)
{
long long t=a%b;
a=b;
b=t;
}
return a;
}
long long Pollard_rho(long long x,long long c)
{
long long i=1,k=2;
long long x0=rand()%x;
long long y=x0;
while(1)
{
i++;
x0=(mult_mod(x0,x0,x)+c)%x;
long long d=gcd(y-x0,x);
if(d!=1&&d!=x)return d;
if(y==x0)return x;
if(i==k)
{
y=x0;
k+=k;
}
}
}
void findfac(long long n)
{
if(Miller_Rabin(n))
{
factor[tol++]=n;
return;
}
long long p=n;
while(p>=n)p=Pollard_rho(p,rand()%(n-1)+1);
findfac(p);
findfac(n/p);
}
int main()
{
srand(time(NULL));
long long n;
while(scanf("%I64d",&n)==1)
{
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)
{
if(factor[0]!=factor[1])
{
printf("%I64d %I64d %I64d
",factor[0],factor[1],factor[0]*factor[1]);
continue;
}
else
{
printf("is not a D_num
");
continue;
}
}
if(tol==3)
{
if(factor[0]==factor[1]&&factor[1]==factor[2])
{
printf("%I64d %I64d %I64d
",factor[0],factor[0]*factor[1],factor[0]*factor[1]*factor[2]);
continue;
}
else
{
printf("is not a D_num
");
continue;
}
}
}
return 0;
}