Codeforces Round #240 (Div. 1) B---mashmokh and ACM (water 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 is given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer in all. So he wasn ' t able to solve them. That's why he asked the 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, 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 is: [1,?1],? [2,?2],? [3,?3],? [1,?2],? [1,?3].

Dp[i][j] length is I, number of the number of I is J scheme

/************************************************************************* > File Name:CF240D.cpp > Autho R:alex > Mail: [email protected] > Created time:2015 March 17 Tuesday 11:59 48 seconds ********************************* ***************************************/#include <map>#include <set>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace STD;Const DoublePI =ACOs(-1.0);Const intINF =0x3f3f3f3f;Const DoubleEPS =1e-15;typedef Long LongLL;typedefPair <int,int> PLL;Const intMoD =1000000007; LL dp[ .][ .];intMain () {intN, K; while(~scanf("%d%d", &n, &k)) {memset(DP,0,sizeof(DP)); for(inti =1; I <= N; ++i) {dp[1][i] =1; } for(inti =2; I <= K; ++i) { for(intj =1; J <= N; ++J) { for(intK =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(inti =1; I <= N;            ++i) {ans + = dp[k][i];        Ans%= MoD; }printf("%lld\n", ans); }return 0;}

Codeforces Round #240 (Div. 1) B---mashmokh and ACM (water DP)