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AcWing 1317. 树屋阶梯
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[题目传送门](https://www.acwing.com/problem/content/description/1319/)
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### 一、题目分析
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考虑以阶梯左下角那个点为第一个钢材的左下角，那么第一个钢材摆放情况便如下图（以 $n = 5$ 为例）
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<center><img src='https://img-blog.csdnimg.cn/8e09382d5838423fa239a81e00ab40f2.png?x-oss-process=image/watermark,type_ZHJvaWRzYW5zZmFsbGJhY2s,shadow_50,text_Q1NETiBAeWV6enou,size_20,color_FFFFFF,t_70,g_se,x_16'></center>
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对每种情况分别讨论，那么问题都被分成了两个子问题，设$f[n]$表示摆放高度为$n$的台阶的方法数，那么：
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$$\large f[5]=f[4]*f[0]+f[3]*f[1]+f[2]*f[2]+f[1]*f[3]+f[0]*f[4]=\sum_{k=1}^5f(5-k)*f(k-1)$$
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泛化一下，就是：
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$$\large f[n]=f[n-1]*f[0]+f[n-2]*f[1]+f[n-3]*f[2]+...+f[1]*f[3]+f[0]*f[n-1]=\sum_{k=1}^nf(n-k)*f(k-1)$$
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显然，这就是**卡特兰数**了
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但是这题很烦，还要用高精乘:
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```c++
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 2010;
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int primes[N], cnt;
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bool st[N];
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int a[N], b[N];
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void get_primes(int n) {
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for (int i = 2; i <= n; i++) {
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if (!st[i]) primes[cnt++] = i;
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for (int j = 0; primes[j] * i <= n; j++) {
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st[i * primes[j]] = true;
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if (i % primes[j] == 0) break;
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}
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}
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}
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int get(int n, int p) { // n!中p的次数
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int s = 0;
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while (n) n /= p, s += n;
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return s;
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}
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void mul(int a[], int b, int &len) {
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int t = 0;
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for (int i = 1; i <= len; i++) {
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t += a[i] * b;
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a[i] = t % 10;
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t /= 10;
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}
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while (t) {
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a[++len] = t % 10;
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t /= 10;
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}
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//去前导0
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while (len > 1 && !a[len]) len--;
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}
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int C(int a, int b, int c[]) {
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int len = 1;
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c[1] = 1;
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for (int i = 0; i < cnt; i++) {
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int p = primes[i];
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int s = get(a, p) - get(b, p) - get(a - b, p);
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while (s--) mul(c, p, len);
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}
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return len;
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}
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void sub(int a[], int b[], int &len) {
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int t = 0;
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for (int i = 1; i <= len; i++) {
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t = a[i] - b[i] - t;
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a[i] = (t + 10) % 10;
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t < 0 ? t = 1 : t = 0;
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}
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while (len > 1 && !a[len]) len--;
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}
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int main() {
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//加快读入
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ios::sync_with_stdio(false);
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cin.tie(0);
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cout.tie(0);
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get_primes(N - 10);
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int n;
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cin >> n;
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int a1 = C(n + n, n, a);
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int b1 = C(n + n, n - 1, b); // bl下面没有用到，原因是两数相减，我们知道a>b,按着a的长度来就行了
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sub(a, b, a1);
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for (int i = a1; i >= 1; i--) printf("%d", a[i]);
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return 0;
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}
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```
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