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 M = 205;
  long long read() {
    long long x = 0, f = 1;
    char c;
    while ((c = getchar()) < '0' || c > '9') {
      if (c == '-') f = -1;
    }
    while (c >= '0' && c <= '9') {
      x = (x << 3) + (x << 1) + (c ^ 48);
      c = getchar();
    }
    return x * f;
  }
  long long n, MOD, a1, ans, fac[M], inv[M], f[M][M][M], v[M][M][M];
  long long qkpow(long long a, long long b) {
    long long r = 1;
    while (b > 0) {
      if (b & 1) r = r * a % MOD;
      a = a * a % MOD;
      b >>= 1;
    }
    return r;
  }
  long long dfs(long long i, long long s, long long k) {
    if (i == n) return fac[n];
    if (k == 0) return fac[n] * inv[n - i] % MOD;
    if (v[i][s][k] == a1) return f[i][s][k];
    long long &r = f[i][s][k];
    r = 0;
    v[i][s][k] = a1;
    for (long long j = (a1 - s) / k; j >= 0; j--) {
      if (k > 1 && i + j < n - a1 + 1 - k + 1) continue;
      r = (r + dfs(i + j, s + j * k, k - 1) * inv[j]) % MOD;
    }
    return r;
  }
  signed main() {
    n = read();
    MOD = read();
    fac[0] = 1;
    for (long long i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % MOD;
    for (long long i = 0; i <= n; i++) inv[i] = qkpow(fac[i], MOD - 2);
    for (a1 = max(1ll, n - 30); a1 <= n; a1++)
      ans = (ans + dfs(0, 0, n + 1 - a1)) % MOD;
    printf("%lld\n", ans);
  }
  