题目网址： http://codeforces.com/problemset/problem/414/B
Description
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 Input
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Hint
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].
#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std ;
const int N=2010;
const int mod=1e9+7;
int dp[N][N]; //dp[i][j]表示长度为i，最后一个元素为j的序列数。
int main()
{
int n,k;
while(cin>>n>>k)
{
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
dp[1][i]=1;
for(int t=1;t<k;t++)
{
for(int i=1;i<=n;i++)
{
for(int j=1;i*j<=n;j++)
{
dp[t+1][i*j]+=dp[t][i];//从i=1循环到n，i*j即凡是i的倍数的dp值均加上上一行（序列长度少一）的dp[t][i]值。
dp[t+1][i*j]%=mod;
}
}
}
int set=0;
for(int i=1;i<=n;i++)
{
set+=dp[k][i];
set%=mod;
}
cout<<set<<endl;
}
return 0 ;
}