暑假期间，小龙报名了一个模拟野外生存作战训练班来锻炼体魄，训练的第一个晚上，教官就给他们出了个难题。由于地上露营湿气重，必须选择在高处的树屋露营。小龙分配的树屋建立在一颗高度为N+1尺（N为正整数）的大树上，正当他发愁怎么爬上去的时候，发现旁边堆满了一些空心四方钢材（如图1.1），经过观察和测量，这些钢材截面的宽和高大小不一，但都是1尺的整数倍，教官命令队员们每人选取N个空心钢材来搭建一个总高度为N尺的阶梯来进入树屋，该阶梯每一步台阶的高度为1尺，宽度也为1尺。如果这些钢材有各种尺寸，且每种尺寸数量充足，那么小龙可以有多少种搭建方法？（注：为了避免夜里踏空，钢材空心的一面绝对不可以向上。）

 1 //It is made by Awson on 2017.10.29
 2 #include <set>
 3 #include <map>
 4 #include <cmath>
 5 #include <ctime>
 6 #include <stack>
 7 #include <queue>
 8 #include <vector>
 9 #include <string>
10 #include <cstdio>
11 #include <cstdlib>
12 #include <cstring>
13 #include <iostream>
14 #include <algorithm>
15 #define LL long long
16 #define Min(a, b) ((a) < (b) ? (a) : (b))
17 #define Max(a, b) ((a) > (b) ? (a) : (b))
18 #define Abs(x) ((x) < 0 ? (-(x)) : (x))
19 using namespace std;
20 const int N = 10005;
21
22 int n;
23 struct BIG_NUM {
24     int a[N+5], len;
25     BIG_NUM() {
26     }
27     BIG_NUM(int* _a, int _len) {
28         len = _len;
29         for (int i = 1; i <= _len; i++) a[i] = _a[i];
30     }
31     BIG_NUM &operator *= (const int &b) {
32         for (int i = 1; i <= len; i++) a[i] *= b;
33         for (int i = 1; i <= len; i++) a[i+1] += a[i]/10, a[i] %= 10;
34         int loc = len+1;
35         while (a[loc]) {
36             a[loc+1] += a[loc]/10; a[loc] %= 10; loc++;
37         }
38         len = loc-1;
39     }
40     BIG_NUM &operator /= (const int &b) {
41         int r = 0, i;
42         BIG_NUM ans;
43         for (i = len; i >= 1; i--) {
44             r = r*10+a[i];
45             if (r >= b) break;
46         }
47         ans.len = i;
48         for (i-- ; i >= 1; i--) {
49             ans.a[i+1] = r/b; r = r%b*10+a[i];
50         }
51         ans.a[1] = r/b;
52         len = ans.len; for (int i = 1; i <= len; i++) a[i] = ans.a[i];
53     }
54     void print() {
55         for (int i = len; i >= 1; i--) printf("%d", a[i]);
56         printf("\n");
57     }
58 }A;
59
60 void work() {
61     scanf("%d", &n);
62     A.len = A.a[1] = 1;
63     for (int i = n+2; i <= (n<<1); i++) A *= i;
64     for (int i = 2; i <= n; i++) A /= i;
65     A.print();
66 }
67 int main() {
68     work();
69     return 0;
70 }