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;
  int n, mod, ans;
  int dp[205][205][205], fac[205], ifac[205];
  int qpow(int a, int b) {
    int ans = 1;
    while (b) {
      if (b & 1) ans = 1LL * ans * a % mod;
      a = 1LL * a * a % mod;
      b >>= 1;
    }
    return ans;
  }
  void init(int n) {
    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 >= 0; --i) {
      ifac[i] = 1LL * ifac[i + 1] * (i + 1) % mod;
    }
  }
  int a1;
  int dfs(int i, int j, int k) {
    if (!k) return !j ? ifac[i] : 0;
    if (n - i < n + 1 - a1 - k) return 0;
    if (~dp[i][j][k]) return dp[i][j][k];
    int &tmp = dp[i][j][k];
    tmp = 0;
    for (int cnt = 0; k * cnt <= j && cnt <= i; ++cnt) {
      tmp = (tmp + 1LL * ifac[cnt] * dfs(i - cnt, j - k * cnt, k - 1)) % mod;
    }
    return tmp;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> mod;
    init(n);
    for (a1 = max(1, n - 25); a1 <= n + 1; ++a1) {
      memset(dp, -1, sizeof(dp));
      for (int sum = 1; sum <= a1; ++sum) {
        ans = (ans + dfs(n, sum, n + 1 - a1)) % mod;
      }
    }
    cout << (1LL * ans * fac[n] + 1) % mod << '\n';
    return 0;
  }
  