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!)

View Code

#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;

}