Mashmokh and ACM CodeForces - 414B 【DP】

Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.

A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally for all i (1 ≤ i ≤ l - 1).

Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007(109 + 7).

Input

The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).

Output

Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).

    Examples  
      input    
3 2
      output    
5
      input    
6 4
      output    
39
      input    
2 1
      output    
2

Note

In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].

题意：在【1，n】范围内任取k个数，使得a[i+1]%a[i]==0

思路：计数问题加取模，想想就可能是DP了。

dp[i][j]为长度为l,以j为结尾的数量。

思考：我当前的状态可以由那些状态转移而来，很明显，由长度为i-1,结尾为j约数转移而来。

那么有dp[i][j]=sigmadp[i-1][x] 其中x为j的约数，那么预处理出所有约数，处理出dp[1][1~n]=1，递推即可

#include <bits/stdc++.h>

typedef long long ll;
using namespace std;

int MOD=1e9+7;
vector <int> vec[2005];
ll dp[2005][2005];


//dp[i][j]=sigmadp[i-1][j约数]
int main(void){
    ll n,k;
    cin >> n>>k;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=i;j++)
            if(i%j==0)  vec[i].push_back(j);
    for(int i=1;i<=n;i++)   dp[1][i]=1;
    for(int i=2;i<=k;i++){
        for(int j=1;j<=n;j++){
            for(int k=0;k<vec[j].size();k++)
                dp[i][j]=(dp[i][j]%MOD+dp[i-1][vec[j][k]]%MOD)%MOD;
        }
    }
    ll ans=0;
    for(int i=1;i<=n;i++)   ans+=(dp[k][i]%MOD);
    cout << ans%MOD << endl;
}