CodeForces 414 B.Mashmokh and ACM（dp）

Description

问满足$1\le b_i\le n$且$b_i|b_{i+1},1\le i<k$的序列$b$的个数

Input

两个整数$n,k(1\le n,k\le 2000)$

Output

输出满足条件的序列个数，结果模$10^9+7$

Sample Input

3 2

Sample Output

5

Solution

令$dp[i][j]$为满足$b_1|b_2|...|b_i=j$的序列$b$的方案数，则有转移$dp[i+1][l]+=dp[i][j],j|l$，答案即为$\sum\limits_{i=1}^ndp[k][i]$

Code

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<ctime>
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
const int INF=0x3f3f3f3f,maxn=2005;
#define mod 1000000007
int n,k,dp[maxn][maxn]; 
int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)dp[1][i]=1;
        for(int i=1;i<k;i++)
            for(int j=1;j<=n;j++)
                for(int l=j;l<=n;l+=j)
                {
                    dp[i+1][l]+=dp[i][j];
                    if(dp[i+1][l]>=mod)dp[i+1][l]-=mod;
                }
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            ans+=dp[k][i];
            if(ans>=mod)ans-=mod;
        }
        printf("%d\n",ans);
    }
    return 0;
}