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 + ;

 int n, k, dp[][];

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

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

2.O(n²log(n))

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

 const int MOD = 1e9 + ;

 int n, k, dp[][];

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

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