Codeforces Round #240 (Div. 1)---B.Mashmokh and ACM(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]表示长度为i,第i个数为j的方案数

/*************************************************************************    > File Name: cf-240-div1-B.cpp    > Author: ALex    > Mail: [e-mail]     > Created Time: 2015年06月02日 星期二 12时35分58秒 ************************************************************************/#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <bitset>#include <set>#include <vector>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;static const int mod = 1000000007;LL dp[2010][2010];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 i = 1; i < k; ++i) {            for (int j = 1; j <= n; ++j) {                if (!dp[i][j]) {                    continue;                }                for (int l = j; l <= n; l +=j) {                    dp[i + 1][l] += dp[i][j];                    dp[i + 1][l] %= mod;                }            }        }        LL ans = 0;        for (int i = 1; i <= n; ++i) {            ans += dp[k][i];            ans %= mod;        }        cout << ans << endl;    }    return 0;}