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;
  int n, Mod, x;
  long long fac[210], inv[210];
  int dp[210][210][210];
  int solve(int i, int j, int k) {
    if (!i) return (k ? 0 : (!j ? 0 : inv[j]));
    if (j > x + i - 1) return 0;
    if (~dp[i][j][k]) return dp[i][j][k];
    int res = 0;
    for (int t = 0; t <= j && i * t <= k; t++)
      res = (res + solve(i - 1, j - t, k - i * t) * inv[t]) % Mod;
    dp[i][j][k] = res;
    return res;
  }
  int main() {
    scanf("%d%d", &n, &Mod);
    fac[0] = 1;
    for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % Mod;
    inv[1] = 1;
    for (int i = 2; i <= n; i++) inv[i] = (Mod - Mod / i) * inv[Mod % i] % Mod;
    inv[0] = 1;
    for (int i = 1; i <= n; i++) inv[i] = inv[i - 1] * inv[i] % Mod;
    int ans = 0;
    for (x = max(1, n - 17); x <= n + 1; x++) {
      memset(dp, -1, sizeof(dp));
      for (int k = 0; k <= x; k++) ans = (ans + solve(n + 1 - x, n, k)) % Mod;
    }
    printf("%lld\n", ans * fac[n] % Mod);
    return 0;
  }
  