CodeForces 415D Mashmokh and ACM

$dp$。

记$dp[i][j]$表示已经放了$i$个数字，并且第$i$个数字放了$j$的方案数。那么$dp[i][j] = sumlimits_{k|j}^{}  {dp[i - 1][k]}$。答案为$sumlimits_{k=1}^{m}  {dp[n][k]}$

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-6;
void File()
{
    freopen("D:\in.txt","r",stdin);
    freopen("D:\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c = getchar();
    x = 0;
    while(!isdigit(c)) c = getchar();
    while(isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
}

const int maxn=2020;
LL dp[maxn][maxn],mod=1e9+7;
int n,m;

int main()
{
    scanf("%d%d",&m,&n);
    for(int i=1;i<=m;i++) dp[1][i]=1;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            for(int k=j;k<=m;k=k+j)
            {
                dp[i+1][k]=(dp[i+1][k]+dp[i][j])%mod;
            }
        }
    }
    LL ans=0;
    for(int i=1;i<=m;i++) ans=(ans+dp[n][i])%mod;
    cout<<ans<<endl;
    return 0;
}