Mashmokh and ACM

题目描述：
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Status
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, …, 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 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].

代码实现：

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>using namespace std;int dp[2005][2005];int main(){    int n,k;    int ans;    while(scanf("%d%d",&n,&k)!=EOF)    {        memset(dp,0,sizeof(dp));        ans=0;        for(int i=1;i<=n;i++)        {            dp[1][i]=1;        }        for(int i=1; i<=n; i++)//i每变动一次dp[]数组元素就会更新一次        {            for(int t=1; t<k; t++)-            {                for(int j=1; i*j<=n; j++)                {                    dp[t+1][i*j]+=dp[t][i];                    dp[t+1][i*j]=dp[t+1][i*j]%(int)(1e9+7);                }            }        }        for(int i=1; i<=n; i++)        {            ans+=dp[k][i];            ans=ans%(int)(1e9+7);        }        printf("%d\n",ans);    }    return 0;}

写完代码，去睡觉。