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;
  inline long long readint() {
    long long x = 0;
    bool f = 0;
    char c = getchar();
    while (!isdigit(c) && c != '-') c = getchar();
    if (c == '-') {
      f = 1;
      c = getchar();
    }
    while (isdigit(c)) {
      x = x * 10 + c - '0';
      c = getchar();
    }
    return f ? -x : x;
  }
  const int maxn = 200 + 5;
  int n;
  long long m;
  long long ksm(long long a, long long b) {
    long long ans = 1;
    while (b) {
      if (b % 2 == 1) ans = ans * a % m;
      a = a * a % m;
      b /= 2;
    }
    return ans;
  }
  long long fac[maxn], ifac[maxn], f[40][maxn][maxn];
  bool vis[40][maxn][maxn];
  long long dfs(int x, int y, int z, int a) {
    if (x == -1) return y == n ? fac[n] : 0;
    if (vis[x][y][z]) return f[x][y][z];
    vis[x][y][z] = 1;
    f[x][y][z] = 0;
    for (int i = x == 0   ? 1
                 : x == 1 ? 0
                          : max(n + 2 - a - x - y, 0);
         y + i <= n && z + x * i <= a; i++)
      f[x][y][z] =
          (f[x][y][z] + dfs(x - 1, y + i, z + x * i, a) * ifac[i] % m) % m;
    return f[x][y][z];
  }
  int main() {
    n = readint();
    m = readint();
    fac[0] = 1;
    for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % m;
    ifac[n] = ksm(fac[n], m - 2);
    for (int i = n - 1; i >= 0; i--) ifac[i] = ifac[i + 1] * (i + 1) % m;
    long long ans = 0;
    for (int i = max(1, n - 35); i <= n + 1; i++) {
      memset(vis, 0, sizeof(vis));
      ans = (ans + dfs(n + 1 - i, 0, 0, i)) % m;
    }
    printf("%lld\n", ans);
    return 0;
  }
  