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 namespace std;
  enum { N = 209 };
  int jc[N], ijc[N], p, a1, f[N][N][27], b[N][N][27], n;
  int qp(int a, int b) {
    int r = 1;
    for (; b; b >>= 1, a = a * 1ll * a % p)
      if (b & 1) r = r * 1ll * a % p;
    return r;
  }
  int dfs(int i, int s, int k) {
    if (!k) return jc[n] * 1ll * ijc[n - i] % p;
    if (b[i][s][k] == a1) return f[i][s][k];
    int &w = f[i][s][k], c = (a1 - s) / k;
    for (w = 0, b[i][s][k] = a1; c >= 0 && (k < 2 || i + c >= n + 2 - a1 - k);
         --c)
      w = (w + dfs(i + c, s + c * k, k - 1) * 1ll * ijc[c]) % p;
    return w;
  }
  int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int i, w = 0;
    for (cin >> n >> p, jc[0] = i = 1; i < N; ++i)
      jc[i] = jc[i - 1] * 1ll * i % p;
    for (ijc[i = N - 1] = qp(jc[N - 1], p - 2); i; --i)
      ijc[i - 1] = ijc[i] * 1ll * i % p;
    for (a1 = max(1, n - (int)sqrt(n / 2) * 2 - 2); a1 <= n; ++a1)
      w = (w + dfs(0, 0, n + 1 - a1)) % p;
    cout << w;
    return 0;
  }
  