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;
  const int N = 205;
  int n, mod, fc[N], iv[N], ans, a1, dp[N][N][N], v[N][N][N];
  int qp(int a, int b) {
    int ans = 1;
    while (b) {
      if (b & 1) ans = 1ll * ans * a % mod;
      a = 1ll * a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  int dfs(int i, int j, int k) {
    int &f = dp[i][j][k];
    if (v[i][j][k] == a1) return f;
    if (i == n) return f = 1;
    if (!k) return f = iv[n - i];
    v[i][j][k] = a1;
    f = 0;
    for (int c = 0; c <= (a1 - j) / k; c++) {
      if (k > 1 && i + c < n + 1 - a1 - k + 1) continue;
      f = (f + 1ll * dfs(i + c, j + c * k, k - 1) * iv[c]) % mod;
    }
    return f;
  }
  int main() {
    scanf("%d%d", &n, &mod);
    fc[0] = iv[0] = 1;
    for (int i = 1; i <= n; i++)
      fc[i] = 1ll * fc[i - 1] * i % mod, iv[i] = qp(fc[i], mod - 2);
    for (a1 = max(1, (int)(n - 2 * sqrt(n) - 1)); a1 <= n; a1++)
      ans = (ans + dfs(0, 0, n + 1 - a1)) % mod;
    ans = 1ll * ans * fc[n] % mod;
    printf("%d\n", ans);
    return 0;
  }
  