Codeforces Round #240 (Div. 1) B. Mashmokh and ACM DP

B. Mashmokh and ACM

time limit per test

1 second

memory limit per test

256 megabytes

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

Note

In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].

题意：给你一个N，K, 表示从1到n，选取长度为k的序列A满足 Ai整除Ai-1

题解：dp[i][j]表示长度为j是最大为i的序列方案数

//

#include<iostream>

#include<cstdio>

#include<cstring>

#include<string>

#include<algorithm>

#include<queue>

#include<cmath>

#include<map>

#include<bitset>

#include<set>

#include<vector>

using namespace std ;

typedef long long ll;

#define mem(a) memset(a,0,sizeof(a))

#define meminf(a) memset(a,127,sizeof(a));

#define memfy(a) memset(a,-1,sizeof(a))

#define TS printf("111111\n");

#define FOR(i,a,b) for( int i=a;i<=b;i++)

#define FORJ(i,a,b) for(int i=a;i>=b;i--)

#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)

#define mod 1000000007

#define maxn 2005

inline ll read()

{

    ll x=,f=;

    char ch=getchar();

    while(ch<''||ch>'')

    {

        if(ch=='-')f=-;

        ch=getchar();

    }

    while(ch>=''&&ch<='')

    {

        x=x*+ch-'';

        ch=getchar();

    }

    return x*f;

}

//****************************************

ll  dp[maxn][maxn];

int main()

{

    int n=read(),m=read();

    mem(dp);

    FOR(i,,n)dp[i][]=;

    FOR(k,,m-)

    FOR(i,,n)

    { if(dp[i][k])

         for(int j=i;j<=n;j+=i)

         {

             dp[j][k+]=(dp[j][k+]+dp[i][k])%mod;

         }

    }

    ll ans=;

    FOR(i,,n)ans=(ans+dp[i][m])%mod;

    cout<<ans<<endl;

    return ;

}

代码

巴特西

Codeforces Round #240 (Div. 1) B. Mashmokh and ACM DP

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