Codeforces Round #240 (Div. 1)B

B. Mashmokh and ACM

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 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].

链接http://codeforces.com/problemset/problem/414/B

一道简单的dp题，一开始想复杂了，搞了好久，后来才发现其实很简单的，对枚举比较熟悉的话一下子就能想出来的

主要的状态转移过程：当前位置的数到下一位置只有可能是当前数的任意倍数且不大于n

状态转移方程：（i是当前位置，j是当前位置的数字，l是不大于n的j的倍数）dp[i+1][l]=dp[i+1][l]+dp[i][j]

ac代码

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int M=1000000007;int n,k;long long dp[2222][2222];int main(){long long ans=0;int i,j,l;memset(dp,0,sizeof(dp));scanf("%d%d",&n,&k);for(i=1;i<=n;++i){dp[0][i]=1;                                                     //不要忘记初始化赋值}for(i=0;i<k;++i){for(j=1;j<=n;++j){for(l=j;l<=n;l+=j)<span style="white-space:pre"></span>//三重枚举{dp[i+1][l]+=dp[i][j];                            //是累加不是等于，同一个值有多个约数dp[i+1][l]%=M;<span style="white-space:pre"></span>//不要忘记mod，在这里wa了一次}}}for(i=1;i<=n;++i){ans+=dp[k-1][i];ans%=M;}printf("%I64d\n",ans);return 0;}