description: A sequence of integers b_1, b_2, …, b_m is called good if max(b_1, b_2, …, b_m) ⋅ min(b_1, b_2, …, b_m) ≥ b_1 + b_2 + … + b_m.
  
  A sequence of integers a_1, a_2, …, a_n is called perfect if every non-empty subsequence of a is good.
  
  YouKn0wWho has two integers n and M, M is prime. Help him find the number, modulo M, of perfect sequences a_1, a_2, …, a_n such that 1 ≤ a_i ≤ n + 1 for each integer i from 1 to n.
  
  A sequence d is a subsequence of a sequence c if d can be obtained from c by deletion of several (possibly, zero or all) elements.
  
  Input
  
  The first and only line of the input contains two space-separated integers n and M (1 ≤ n ≤ 200; 10^8 ≤ M ≤ 10^9). It is guaranteed that M is prime.
  
  Output
  
  Print a single integer — the number of perfect sequences modulo M.
  
  Examples
  
  Input
  
  
  2 998244353
  
  
  Output
  
  
  4
  
  
  Input
  
  
  4 100000007
  
  
  Output
  
  
  32
  
  
  Input
  
  
  69 999999937
  
  
  Output
  
  
  456886663
  
  Note
  
  In the first test case, the perfect sequences are [2, 2], [2, 3], [3, 2] and [3, 3].
  
  In the second test case, some of the perfect sequences are [3, 4, 3, 5], [4, 5, 4, 4], [4, 5, 5, 5] etc. One example of a sequence which is not perfect is [2, 3, 3, 4], because, for example, the subsequence [2, 3, 4] is not an good as 2 ⋅ 4 < 2 + 3 + 4.
solution: #include <bits/stdc++.h>
  int n, MOD, fact[205], ifact[205], dp[205][205][205], ans;
  int power(int A, int B) {
    int res = 1;
    while (B) {
      if (B & 1) res = 1ll * res * A % MOD;
      A = 1ll * A * A % MOD;
      B >>= 1;
    }
    return res;
  }
  void ps(int &A, int B) { A = (A + B) % MOD; }
  signed main() {
    std::ios::sync_with_stdio(false);
    std::cin >> n >> MOD;
    fact[0] = 1;
    for (long long i = 1; i <= n; ++i) fact[i] = fact[i - 1] * i % MOD;
    ifact[n] = power(fact[n], MOD - 2);
    for (long long i = n; i; --i) ifact[i - 1] = ifact[i] * i % MOD;
    for (int a1 = n + 1; a1 >= n - 2 * std::sqrt(n); --a1) {
      for (int i = 1; i <= std::min(n, a1); ++i)
        dp[i][0][std::max(i + 2, a1 + 1)] = ifact[i];
      for (int i = 1; i <= n; ++i)
        for (int j = 0; j <= a1; ++j)
          for (int k = a1 + 1; k <= n + 1; ++k)
            for (int l = 0;
                 dp[i][j][k] && k >= i + l + 1 && j + l * (k - a1) <= a1; ++l)
              ps(dp[i + l][j + l * (k - a1)][std::max(i + l + 2, k + 1)],
                 1ll * dp[i][j][k] * ifact[l] % MOD);
      for (int i = 0; i <= a1; ++i)
        ps(ans, 1ll * dp[n][i][n + 2] * fact[n] % MOD);
      memset(dp, 0, sizeof(dp));
    }
    std::cout << ans << std::endl;
    return 0;
  }
  