CodeForces - 414B Mashmokh and ACM（dp）

题目链接：https://vjudge.net/problem/CodeForces-414B
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 b 1, b 2, …, b l (1 ≤ b 1 ≤ b 2 ≤ … ≤ 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 (10^9 + 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 (10^9 + 7).
Input

3 2

Output

5

Input

6 4

Output

39

Input

2 1

Output

2

翻译：

给定两个数n和k。
求有多少种长度为k的子序列，满足：

1. $b_1,b_2,...,b_k(1\le b_1\le b_2\le ...\le b_k\le n)$
2. 对于任意的$1\le i\le k-1$，满足$b_{i+1}$能整除$b_i$

最后结果对mod取余

分析：
定义：:
$dp[i][j]$为长度为i，最高位为j的方案数

可以得出：
$dp[i][j]=dp[i的所有因子][j-1]$

完整代码：

#include
#include
#include
using namespace std;
#define mod 1000000007
const int N=2*1e3+10;
int n,k;
int dp[N][N];///dp[i][j]:长度为i，最高位为j的方案数
//dp[i][j]=dp[i-1][j的因子]
int main()
{
     
	memset(dp,0,sizeof(dp));
	scanf("%d%d",&n,&k);
	for(int i=0; i<=n; i++)
		dp[1][i]=1;

	for(int i=1; i<k; i++)
	{
     
		for(int j=1; j<=n; j++)
		{
     
			for(int m=j; m<=n; m+=j)
				dp[i+1][m]=(dp[i+1][m]+dp[i][j])%mod;
		}
	}
	int res=0;
	for(int i=1; i<=n; i++)
		res=(res+dp[k][i])%mod;
	printf("%d\n",res);
	return 0;
}