• 序列中找子序列的dp


    题目网址: http://codeforces.com/problemset/problem/414/B

    Description

    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 Input

    Input
    3 2
    Output
    5
    Input
    6 4
    Output
    39
    Input
    2 1
    Output
    2

    Hint

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

    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    
    using namespace std ;
    const int N=2010;
    const int mod=1e9+7;
    int dp[N][N];        //dp[i][j]表示长度为i,最后一个元素为j的序列数。
    
    int main()
    {
        int n,k;
        while(cin>>n>>k)
        {
            memset(dp,0,sizeof(dp));
            for(int i=1;i<=n;i++)
            dp[1][i]=1;
            for(int t=1;t<k;t++)
            {
                for(int i=1;i<=n;i++)
                {
                    for(int j=1;i*j<=n;j++)
                    {
                        dp[t+1][i*j]+=dp[t][i];//从i=1循环到n,i*j即凡是i的倍数的dp值均加上上一行(序列长度少一)的dp[t][i]值。
                        dp[t+1][i*j]%=mod;
                    }
                }
            }
            int set=0;
            for(int i=1;i<=n;i++)
            {
                set+=dp[k][i];
                set%=mod;
            }
            cout<<set<<endl;
        }
        return 0 ;
    }
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  • 原文地址:https://www.cnblogs.com/chen9510/p/5017197.html
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