Mashmokh and ACM

朴灿烈づ我的快乐病毒、 2022-08-18 00:49 29阅读 0赞

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

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。