Topcoder/SRM565

SRM565 DIV II:

Problem Statement

    

A divisible sequence with starting value N and length H is a sequence of positive integers with the following properties:

The sequence has H elements, labeled A[0] through A[H-1].

A[0] equals N.

For each i between 0 and H-2, inclusive, A[i+1] divides A[i].

You are given the ints N and H. Let C be the count of all divisible sequences with starting value N and length H. Return the value (C modulo 1,000,000,009).

Definition

    

Class:

DivisibleSequence

Method:

count

Parameters:

int, int

Returns:

int

Method signature:

int count(int N, int H)

(be sure your method is public)

    

Constraints

-

N will be between 1 and 1,000,000,000, inclusive.

-

H will be between 1 and 1,000,000,000, inclusive.

Examples

0)

    

5

3

Returns: 3

The three possible sequences are: 5, 5, 5 5, 5, 1 5, 1, 1

1)

    

6

3

Returns: 9

The nine possible sequences are: 6, 6, 6 6, 6, 3 6, 6, 2 6, 6, 1 6, 3, 3 6, 3, 1 6, 2, 2 6, 2, 1 6, 1, 1

2)

    

10

2

Returns: 4

A[1] can be equal to each of the divisors of the number 10. That is, A[1] can be 1, 2, 5, or 10.

3)

    

1

10000

Returns: 1

The only possible sequence consists of all ones.

4)

    

30

4

Returns: 64

5)

    

25883

100000

Returns: 991000009

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我的代码：

#include <deque>

#include <vector>

#include <algorithm>

#include <functional>

#include <iterator>

#include <map>

#include <memory>

#include <numeric>

#include <queue>

#include <set>

#include <utility>

#include <stack>

#include <iostream>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#define PI 3.1415926535898

#define LL long long

 

using namespace std;

class DivisibleSequence

{

 int c;

 int cnt;

 void count_process(const vector<int>& factor,int cur,int H)

 {

  if( H == 1)

  {

   for(int i =cur;i>=0;i--)

   {

    if( factor[cur]%factor[i] == 0)

     c++;

   }

  }

  else

  {

   for(int i = cur;i>=0;i--)

   {

    if ( factor[cur]%factor[i] == 0)

     count_process(factor,i,H-1);

   }

  }

 }

 public:

  int count(int N, int H)

  {

   cnt = 1;

   c = 0;

   if (( N == 1) || ( H == 1))

   {

    return 1;

   }

   vector <int> factor(N/2+1,0);

   factor[0] = 1;

   for(int i = 2;i<=N/2;i++)

   {

    if( N%i == 0)

    {

     factor[cnt] = i;

     cnt++;

    }

   }

   H--;

   factor[cnt] = N;

   cnt++;

   if ( H == 1)

    c = cnt;

   else

    count_process(factor,cnt-1,H);

   return (c%1000000009);

  }

};

开始得了999，后来发现第六组数据时效性太差。。。果断把自己给Challenge了。（结果真可以Challenge,汗。。)