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>
  const int N = 205;
  int MOD;
  inline int mval(int x) { return x >= MOD ? x - MOD : x; }
  inline void inc(int &x, int a) { x = mval(x + a); }
  int n, fac[N], inv[N], f[N][N][N], ans;
  inline int barret(unsigned long long c) {
    static __int128 x = -1ull / MOD;
    int r = c - (unsigned long long)((x * c) >> 64) * MOD;
    return mval(r);
  }
  int main() {
    scanf("%d%d", &n, &MOD);
    fac[0] = inv[0] = inv[1] = 1;
    for (int i = 2; i <= n; ++i)
      inv[i] = MOD - 1ll * (MOD / i) * inv[MOD % i] % MOD;
    for (int i = 1; i <= n; ++i)
      fac[i] = 1ll * i * fac[i - 1] % MOD,
      inv[i] = 1ll * inv[i - 1] * inv[i] % MOD;
    f[n + 2][0][0] = 1;
    for (int i = n + 1; 1ll * (n + 1 - i) * (n - i) / 2 <= n; --i) {
      for (int j = 0; j < n; ++j)
        for (int s = 0; s <= n; ++s)
          if (f[i + 1][j][s]) {
            int val = f[i + 1][j][s];
            for (int t = 0; t + j <= n && s + t + j <= n; ++t)
              if (t + j >= n + 2 - i) {
                inc(f[i][j + t][s + t + j], barret(1ll * val * inv[t]));
              }
            if (s <= i) {
              inc(ans, barret(1ll * inv[n - j] * val));
            }
            if (j + i - 1 == n && i > 1 && s + j <= i - 1) {
              inc(ans, barret(1ll * inv[n - j] * val));
            }
          }
    }
    printf("%d\n", (1ll * ans * fac[n]) % MOD);
    return 0;
  }
  