CodeForces 414B Mashmokh and ACM （DP）

B. Mashmokh and ACM

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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).

Sample test(s)

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].

大致题意：

给出一个n和k，让你从 1 -- n 这n个数中，选出 k个（可重复）组成一个序列，使这个序列满足任意一个数都能够整除该序列中它前面的那个数。求这样的序列的个数。

分析：

由于每一个数都与其前面的一个数直接相关，每个数必须是其前面一个数的倍数。那么我们可以定义一个跟末尾数字有关的状态，我们用dp[ i ] [ j ] 表示长度为 i 并且最后一位是 j 的序列 ，那么状态转移方程：dp [ i ] [ j ] = sum（dp [ i-1 ] [ x ] ） x 是 j 前面满足（ j % x == 0）的数。

因为是多组测试，在开始的时候直接打表可以节省时间。

比较雷的是，刚开始写的循环是

for(int i=1;i<2010;i++)

{for(int j=1;j<2010;j++){for(int q=1;q<=j;q++){if(j%q==0){dp[i][j] += dp[i-1][q];dp[i][j] %= MOD;}}}}

显然会T的，不能这么暴力那么无知的做。。

换成这样会好一点。至少就不会T了。。对于每一个 i ，都有从1开始的 j 去控制的 q ，来增加dp [ i ] [ ] ，这样一圈循环下来，可以完全更新~

for(int i=1;i<2010;i++){for(int j=1;j<2010;j++){for(int q=j;q<2010;q=q+j){dp[i][q] += dp[i-1][j];dp[i][q] %= MOD; }}}