Codeforces 414B 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 $技术分享$ 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].

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;

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
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<int,int>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;
        cout<<ans<<endl;
    }
    return 0;
}