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 mod;
  using namespace std;
  int n, K;
  int fac[210], ifac[210];
  int f[210][210][210];
  int ksm(int a, int k) {
    int res = 1;
    for (; k; k >>= 1, a = 1ll * a * a % mod)
      if (k & 1) res = 1ll * res * a % mod;
    return res;
  }
  void qmo(int &x) { x += (x >> 31) & mod; }
  int work(int a1) {
    memset(f, 0, sizeof(f));
    for (int i = 1; i <= n; i++) f[0][i][0] = ifac[i];
    for (int i = 1; i <= n + 1; i++)
      for (int j = n - n / i; j <= n && j - i + 1 <= a1; j++) {
        for (int k = 0; k <= n; k++)
          for (int l = 0; l <= j && l <= k / i; l++) {
            f[i][j][k] =
                (f[i][j][k] + 1ll * f[i - 1][j - l][k - i * l] * ifac[l]) % mod;
          }
      }
    int res = (a1 >= n ? ifac[n] : 0);
    for (int i = 1; i <= a1; i++) qmo(res += f[n + 1 - a1][n][i] - mod);
    return res;
  }
  int main() {
    scanf("%d%d", &n, &mod);
    int res = 0;
    fac[0] = ifac[0] = 1;
    for (int i = 1; i <= n; i++)
      fac[i] = 1ll * i * fac[i - 1] % mod, ifac[i] = ksm(fac[i], mod - 2),
      assert(1ll * fac[i] * ifac[i] % mod == 1);
    for (int a1 = n - 17; a1 <= n + 1; a1++) qmo(res += work(a1) - mod);
    printf("%d\n", 1ll * res * fac[n] % mod);
  }
  