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, fac[N], ifac[N];
  inline void inc(int& x, int y) { x = x + y >= mod ? x + y - mod : x + y; }
  inline int qpow(int x, int y) {
    int res = 1;
    for (; y; y >>= 1, x = 1ll * x * x % mod)
      (y & 1) && (res = 1ll * res * x % mod);
    return res;
  }
  inline void pre() {
    fac[0] = 1;
    for (int i = 1; i <= n; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
    ifac[n] = qpow(fac[n], mod - 2);
    for (int i = n - 1; ~i; --i) ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;
  }
  int f[N][N][30], ans;
  inline void work(int a1) {
    memset(f, 0, sizeof f);
    int mx = n + 1 - a1;
    f[0][0][0] = 1;
    for (int i = 0; i < n; ++i)
      for (int j = 0; j <= a1; ++j)
        for (int k = 0; k <= mx; ++k) {
          int& d = f[i][j][k];
          if (!d) continue;
          for (int ct = 1; ct <= n - i; ++ct) {
            if (i + ct > k + a1) break;
            if (i + ct == k + a1 && i) break;
            if (i + ct > a1 && k + a1 < i + ct + 1) break;
            int sum = j + ct * k;
            if (i + ct == n) {
              inc(ans, 1ll * d * ifac[ct] % mod);
              break;
            }
            for (int t = k + 1; t <= mx && sum + (n - ct - i) * t <= a1; ++t)
              inc(f[i + ct][sum][t], 1ll * d * ifac[ct] % mod);
          }
        }
  }
  int main() {
    scanf("%d%d", &n, &mod);
    pre();
    for (int i = max(1, n - 2 * (int)(sqrt(n + 0.01))); i <= n + 1; ++i) work(i);
    ans = 1ll * ans * fac[n] % mod;
    printf("%d\n", ans);
    return 0;
  }
  