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>
  using LL = long long;
  const int N = 210;
  using namespace std;
  int n, mod;
  int dp[N][N][40], f[N], inv[N];
  int vis[N][N][40];
  int a1;
  int binpow(int a, int b) {
    int res = 1;
    for (; b; b >>= 1, a = (1LL * a * a) % mod)
      if (b & 1) {
        res = (1LL * res * a) % mod;
      }
    return res;
  }
  int solve(int cur, int sum, int k) {
    if (cur == n) return f[n];
    if (k == 0) return 1LL * f[n] * inv[n - cur] % mod;
    if (vis[cur][sum][k] == a1) return dp[cur][sum][k];
    vis[cur][sum][k] = a1;
    int &res = dp[cur][sum][k];
    res = 0;
    int lim = (a1 - sum) / k;
    for (int cnt = lim; cnt >= 0; cnt--) {
      if (k > 1 && cur + cnt < n - a1 + 2 - k) continue;
      res =
          (res + 1LL * solve(cur + cnt, sum + cnt * k, k - 1) * inv[cnt] % mod) %
          mod;
    }
    return res;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> mod;
    f[0] = inv[0] = 1;
    for (int i = 1; i < N; i++) {
      f[i] = 1LL * f[i - 1] * i % mod;
      inv[i] = binpow(f[i], mod - 2);
    }
    int ans = 0;
    int lim = (2 * sqrt(n) + 1);
    for (a1 = max(1, n - lim); a1 <= n; a1++) {
      ans += solve(0, 0, n + 1 - a1);
      ans %= mod;
    }
    cout << ans;
  }
  