Codeforces Round #240 (Div. 1)B---Mashmokh and ACM(水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).
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].

dp[i][j] 长度为i,第i个数为j的方案数

/*************************************************************************
    > File Name: CF240D.cpp
    > Author: ALex
    > Mail: [email protected] 
    > Created Time: 2015年03月17日 星期二 11时59分48秒
 ************************************************************************/

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int mod = 1000000007;
LL dp[2010][2010];

int main ()
{
    int n, K;
    while(~scanf("%d%d", &n, &K))
    {
        memset (dp, 0, sizeof(dp));
        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 = 1; k * k <= j; ++k)
                {
                    if (j % k == 0)
                    {
                        dp[i][j] += dp[i - 1][k] + (k * k == j ? 0 : dp[i - 1][j / k]);
                        dp[i][j] %= mod;
                    }
                }
            }
        }
        LL ans = 0;
        for (int i = 1; i <= n; ++i)
        {
            ans += dp[K][i];
            ans %= mod;
        }
        printf("%lld\n", ans);
    }
    return 0;
}