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 = 210;
  int fac[N], inv[N], ans, n, mod, vis[N][N][N], f[N][N][N];
  inline int ksm(int x, int y) {
    int ret = 1;
    for (; y; y >>= 1, x = 1ll * x * x % mod)
      if (y & 1) ret = 1ll * ret * x % mod;
    return ret;
  }
  inline void init(int n) {
    fac[0] = 1;
    for (int i = 1; i <= n; ++i) fac[i] = 1ll * fac[i - 1] * i % mod;
    inv[n] = ksm(fac[n], mod - 2);
    for (int i = n - 1; i >= 0; --i) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
  }
  inline void inc(int &x, int y) { x = (x + y >= mod) ? x + y - mod : x + y; }
  inline int solve(int a, int x, int s, int k) {
    if (x == n) return 1;
    if (!k) return inv[n - x];
    int &g = f[x][s][k];
    if (vis[x][s][k] == a) return g;
    vis[x][s][k] = a;
    g = 0;
    for (int i = 0; s + i * k <= a && x + i <= n; ++i) {
      int mx = n - (x + i);
      if (k > 1 && mx + 1 - a > k - 1) continue;
      g = (g + 1ll * solve(a, x + i, s + i * k, k - 1) * inv[i]) % mod;
    }
    return g;
  }
  int main() {
    scanf("%d%d", &n, &mod);
    init(n);
    int sq = ceil(sqrt(n) + 0.5);
    for (int i = max(1, n - 2 * sq - 1); i <= n; ++i)
      inc(ans, solve(i, 0, 0, n - i + 1));
    printf("%d\n", 1ll * ans * fac[n] % mod);
    return 0;
  }
  