题目意思

题目规定的还是比较正统的BST，那么就左右子树差值不超过给定的第二个数字，高度为第一个数字

解题思路

直接动态规划是最简单的解法：
我们想想假设待求得状态为，代表高度为i的所求节点个数
那么就有代表着高度为和得左右子树所求节点个数之和再加上根节点
最终就有左子数为满二叉树，右子树为满足题目高度只差不超过m的二叉树，减掉即可！注意别取到负数的下标了

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const int MOD = 1e9 + 7;
const ll INF = 0x3f3f3f3f;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e5 + 7;
ll dp[100];

int main() {
    int n = read(), m = read();
    dp[1] = 1;
    for (int i = 2; i <= n; ++i)
        dp[i] = dp[i - 1] + dp[max(i - m - 1, 0)] + 1;
    ll ans = 1ll << n - 1;
    ans -= dp[n - m - 1];
    cout << ans - 1 << endl;
    return 0;
}