codeforces--414B--Mashmokh and ACM

Mashmokh and ACM

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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, ..., b_l(1 ≤ b₁ ≤ b₂ ≤ ... ≤ b_l ≤ 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 Input

Input

3 2

Output

5

Input

6 4

Output

39

Input

2 1

Output

2

Hint

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

 

给出n和k，要求k个数中的每个数都在1到n内，并且a(i)/a(i-1)整除。共有多少中。

暴搜就可以

 

 

#include <cstdio>#include <cstring>#include <algorithm>using namespace std ;#define MOD 1000000007#define LL __int64LL dp[2][2100] ;int main(){    LL n , m , k , temp , l , i , j ;    while( scanf("%I64d %I64d", &n, &m) != EOF )    {        memset(dp,0,sizeof(dp)) ;        temp = 0 ;        for(j = 1 ; j <= n ; j++)            dp[temp][j] = 1 ;        for(i = 2 ; i <= m ; i++)        {            l = 1 - temp ;            for(j = 1 ; j <= n ; j++)            {                k = j ;                while( k <= n )                {                    dp[l][k] += dp[temp][j] ;                    dp[l][k] %= MOD ;                    k += j ;                }            }            memset(dp[temp],0,sizeof(dp[temp])) ;            temp = l ;        }        LL ans = 0 ;        for(j = 1 ; j <= n ; j++)            ans = ( ans + dp[temp][j] ) % MOD ;        printf("%I64d\n", ans) ;    }    return 0;}