poj 3421 X-factor Chains

给定一个数X.

1=X0, X1, X2.....Xm = X 是X的因数

求一串因数,要求Xi | Xi+1,即上一个因数能整除下一个因数,

问这条串就的最长长度,和有多少条这样长度的串.

X = p1^a1 * p2^a2 ... pn^an

 Xi = p1^b2 * p2^b2 ...pk^bk... pn^bn,

Xi+1 = p1^b2 * p2^b2 ...pk^(bk)... pn^bn,

 要使length最长,只要从1开始，每次只乘以X的一个质因数即可,即length = (a1+a2+...an)

而方法数就是X的质因数的重排列数,way = (a1+a2+...an)!/(a1!a2!...an!)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include<cstring>
#include<vector>
#include<string>
#define LL long long
using namespace std;
int prime[90000],cnt=0;
bool hash[550024]={0};
void Prime( )
{
    int t1 = 1 << 10,t2=1<<20;
    for( int i = 3 ; i <= t1; i += 2 )
    {
         if( !hash[i>>1] )
         {
             int x = i << 1;
             for( int j = i * i ; j <= t2 ; j += x )
                  hash[j>>1] = true;        
         }    
    }    
    prime[cnt++] = 2;
    t1 = 1 << 19;
    for( int i = 1 ; i <= t1; i ++ )
    {
         if( !hash[i] )
         {
             prime[cnt++] = i<<1|1;
         }    
    }
}
LL A( int n )
{
   LL ans = 1;
   while( n )
   {
        ans *= n;
        n--;        
   }
   return ans;    
}
void Solve( int n )
{
    int num[24]={0},count=0,len=0;
    for( int i = 0; i < cnt; i ++ )
    {
         if( prime[i] > n ) break;
         if( n % prime[i] == 0 )
         {
             while( n % prime[i] == 0 )
             {
                   num[count]++;
                   n /= prime[i];        
             }
             len += num[count];
             count++;
         }    
    }    
    LL ans = A( len );
    for( int i = 0 ; i < count ; i ++ )
        ans /= A( num[i] );
    printf( "%d %I64d\n",len,ans );    
}
int main(  )
{
    int n;
    Prime();
    while( scanf( "%d",&n )==1 )
    {
          Solve( n );    
    }
    //system( "pause" );
    return 0;
}