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>
  struct R {
    int mod;
    double inv;
    R(int m) : mod(m), inv(1. / m) {}
    uint64_t operator()(uint64_t x) {
      x -= uint64_t(x * inv + .5) * mod;
      return x < mod ? x : x + mod;
    }
  } A(2);
  int mod;
  inline int inc(int x, int y) { return x += y - mod, x += x >> 31 & mod; }
  inline int dec(int x, int y) { return x -= y, x += x >> 31 & mod; }
  inline int mul(int x, int y) { return A(uint64_t(x) * y); }
  int qpow(int a, int b) {
    int c = 1;
    for (; b; b >>= 1, a = mul(a, a))
      if (b & 1) c = mul(a, c);
    return c;
  }
  const int maxn = 205;
  int n, fac[maxn], fic[maxn], low[maxn], dp[maxn][maxn][maxn];
  inline void upd(int &x, int y) { x = inc(x, y); }
  void init() {
    fac[0] = 1;
    for (int i = int(1); i <= int(n); i++) fac[i] = mul(fac[i - 1], i);
    fic[n] = qpow(fac[n], mod - 2);
    for (int i = int(n); i >= int(1); i--) fic[i - 1] = mul(fic[i], i);
  }
  int solve(int m) {
    memset(low, 0, sizeof(low));
    int s = 0;
    for (int i = int(1); i <= int(n - m); i++) {
      s += (low[m + i] = i + 1);
    }
    if (s > m) return 0;
    memset(dp, 0, sizeof(dp));
    for (int i = int(1); i <= int(m); i++) {
      dp[0][i][0] = fic[i];
    }
    for (int i = int(0); i <= int(n - m); i++)
      for (int j = int(0); j <= int(n); j++)
        for (int k = int(0); k <= int(m); k++)
          if (k + n - j <= m) {
            for (int l = int(0); l <= int(n - j); l++)
              if (i + 1 >= low[j + l])
                upd(dp[i + 1][j + l][k + n - j], mul(dp[i][j][k], fic[l]));
          }
    int x = 0;
    for (int i = int(0); i <= int(m); i++) {
      upd(x, dp[n - m + 1][n][i]);
    }
    return mul(x, fac[n]);
  }
  int main() {
    scanf("%d %d", &n, &mod), A = R(mod);
    init();
    int s = 1;
    for (int m = int(n / 2 + 1); m <= int(n); m++) {
      s = inc(s, solve(m));
    }
    printf("%d\n", s);
    return 0;
  }
  