CF414B Mashmokh and ACM

思路：

dp。

实现：

1.O(n^5/2)

#include <iostream>
#include <cstdio>
using namespace std;

const int MOD = 1e9 + 7;

int n, k, dp[2005][2005];

int solve()
{
    for (int i = 1; i <= n; i++) dp[0][i] = 1;
    for (int i = 1; i < k; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            for (int p = 1; p * p <= j; p++)
            {
                if (j % p == 0) 
                {
                    dp[i][j] = (dp[i][j] + dp[i - 1][p]) % MOD;
                    if (p * p != j) dp[i][j] = (dp[i][j] + dp[i - 1][j / p]) % MOD;
                }
            }
        }
    }
    int sum = 0;
    for (int i = 1; i <= n; i++) sum = (sum + dp[k - 1][i]) % MOD;
    return sum;
}

int main()
{
    cin >> n >> k;
    cout << solve() << endl;
    return 0;
}

2.O(n²log(n))

#include <iostream>
#include <cstdio>
using namespace std;

const int MOD = 1e9 + 7;

int n, k, dp[2005][2005];

int solve()
{
    for (int i = 1; i <= n; i++) dp[0][i] = 1;
    for (int i = 1; i < k; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            for (int p = j; p <= n; p += j)
            {
                dp[i][p] = (dp[i][p] + dp[i - 1][j]) % MOD;
            }
        }
    }
    int sum = 0;
    for (int i = 1; i <= n; i++) sum = (sum + dp[k - 1][i]) % MOD;
    return sum;
}

int main()
{
    cin >> n >> k;
    cout << solve() << endl;
    return 0;
}