Codeforces 414B Mashmokh and ACM(DP） - c++编程基础

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

input

6 4

output

input

2 1

output

Note

In the first sample the good sequences are: [1,?1],?[2,?2],?[3,?3],?[1,?2],?[1,?3].

题目就是有关计数问题的DP; 设dp[i][j]表示以i为结尾长度为j的方案数：dp[i][j]=sum(dp[k][j-1])(i%k==0) 特别的dp[i][1]=1;

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              using namespace std; #define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i ) #define REP( i , n ) for ( int i = 0 ; i < n ; ++ i ) #define CLEAR( a , x ) memset ( a , x , sizeof a ) typedef long long LL; typedef pair
             
              pil; const int INF = 0x3f3f3f3f; const int MOD=1e9+7; const int maxn=2000+100; LL dp[maxn][maxn]; int n,k; int main() { while(cin>>n>>k) { CLEAR(dp,0); REPF(i,1,n) dp[i][1]=1; for(int j=1;j<=n;j++) for(int kk=j;kk<=n;kk+=j) for(int i=2;i<=k;i++) dp[kk][i]=(dp[kk][i]+dp[j][i-1])%MOD; LL ans=0; REPF(i,1,n) ans=(ans+dp[i][k])%MOD;