hdu3187 HP Problem

Problem Description

In mathematics, the greatest common divisor (gcd), of two non-zero integers, is the largest positive integer that divides both two numbers without a remainder. For example, gcd( 10, 15 ) = 5, gcd( 5, 4 ) = 1. If gcd( k, n ) == 1 , then we say k is co-prime to n ( also , n is co-prime to k ), the totient function H(n) of a positive integer n is defined to be the number of positive integers not greater than n that are co-prime to n. In particular H(1) = 1 since 1 is co-prime to itself (1 being the only natural number with this property). For example, H (9) = 6 since the six numbers 1, 2, 4, 5, 7 and 8 are co-prime to 9. Also, we define the number of different prime of n is P (n). For example, P (4) = 1 (4 = 2*2), P (10) = 2(10 = 2*5), P (60) = 3(2*2*3*5). Now, your task is, give you a positive integer n not greater than 2^31-1, please calculate the number of k (0 < k < 2^31) satisfied that H (k) = n and P (k) <= 3(So we called HP Problem).

 

Input

Each line will contain only one integer n. Process to end of file.

 

Output

For each case, output the answer in one line.

 

Sample Input

6

 

Sample Output

4

思路：给定n为在[1,2^31-1]范围内有多少个数满足 H(k)=n&& P(k)<=3;

H（k)为k的欧拉函数；P(K)为k中所含质因子的种类个数；

欧拉公式：如果k=p1^q1 *P2^q2*....*pm^qm;

则H(k)=k*(p1-1)*(p2-1)...(pm-1)/(p1*p2*p3*p4...pm)

如果已知H（k）以及k中质因子（p1, p2, p3, p4,..pm)则可以唯一确定一个k=H(k)*(p1*p2*...*pm)/((p1-1)*(p2-1)...*(pm-1))

于是对于n，如果找到了质因子集合{p1,p2,...,pm}( m<=3)则等于找到了一个k；

同时这样的质因子集合满足

n%(pi-1)==0&&n 不含集合外的其他质因子；同时求得的K应小于2^31;

由于m<=3可以枚举具有上述性质的质因子集合来得到答案；

n为1时，有2个，H(1)=1这种特殊情况要考虑。

(真心看不懂啊……）

#include<stdio.h>
#include<queue>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<memory>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<malloc.h>
#include<iostream>
using namespace std;
typedef __int64 ll;
#define N 100000
#define M_F 1000
const ll base=2147483648;
int tag[N];
ll prime[N/10],cnt;
ll p[M_F],q[M_F],num;
ll a[M_F],m;
ll ans,n,pr[10],t;
void get_prime()
{
    int i,j;
    memset(tag,0,sizeof(tag));
    for(i=2;i*i<N;i++)
    if(!tag[i])
    {
        for(j=2;j*i<N;j++)
        tag[j*i]=1;
    }
    for(i=2,cnt=0;i<N;i++)
    if(tag[i]==0)prime[cnt++]=i;
}

void factor()
{
    int i;
    ll tmp=n;
    for(num=0,i=0;i<cnt&&prime[i]*prime[i]<=n;i++)
    if(n%prime[i]==0)
    {
        p[num]=prime[i];
        q[num]=0;
        while(n%prime[i]==0)
        {
            q[num]++;
           n/=prime[i];
        }
        num++;
    }
    if(n>1)
    {
        p[num]=n;
        q[num++]=1;
    }
    n=tmp;
}

int Is_Ok(ll cur)
{
    int i;
    cur++;
    for(i=0;i<cnt&&prime[i]*prime[i]<=cur;i++)
    if(cur%prime[i]==0)return 0;
    return 1;
}

void dfs(ll cur,int idx)
{
    ll tmp=1;
    if(idx>=num)
    {
        if(Is_Ok(cur))a[m++]=cur;
        return ;
    }
    for(int i=0;i<=q[idx];i++)
    {
        dfs(cur*tmp,idx+1);
       tmp=tmp*p[idx];
   }
}

int judge()
{
    ll ret,tmp;
   ret=n;
   int i;
    for(i=0;i<t;i++)
    {
        if(ret%(pr[i]-1))return 0;
        else ret/=(pr[i]-1);
    }
    for(i=0;i<t;i++)
    {
    ret=ret*pr[i];
    if(ret>=base||ret<0)return 0;
    }
    //tmp=ret;
    ret=n;
    for(i=0;i<t;i++)
    {
        ret/=(pr[i]-1);
    }
    for(i=0;i<t;i++)
    while(ret%pr[i]==0)ret/=pr[i];
    //if(ret==1)cout<<tmp<<endl;
    return (ret==1);
}

void dfs1(int idx,int k)
{
    if(k>3)return ;
    else
   {
        if(k)
        {
            t=k;
            ans+=judge();
        }
        int i;
        for(i=idx;i<m;i++)
         {
             pr[k]=a[i]+1;
             dfs1(i+1,k+1);
         }
    }
}

int main()
{
   get_prime();
   while(scanf("%I64d",&n)!=EOF)
   {

    factor();
    m=0;
    dfs(1,0);
     ans=0;
    dfs1(0,0);
    if(n==1)ans++;
    printf("%I64d\n",ans);
    }
return 0;
}