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本文链接：https://blog.youkuaiyun.com/zjfuyuyang/article/details/52766211

本文探讨了一个经典的组合数学问题TrainProblemII，通过使用卡特兰数解决火车进出站的不同排列方式。文章提供了详细的C++实现代码，并解释了如何处理大数运算。

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Train Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8534 Accepted Submission(s): 4563

Problem Description

As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasing order, how many orders that all the trains can get out of the railway.

Input

The input contains several test cases. Each test cases consists of a number N(1<=N<=100). The input is terminated by the end of file.

Output

For each test case, you should output how many ways that all the trains can get out of the railway.

Sample Input

Sample Output

  
  
   
   1
2
5
16796


   
   
    
    
     
     Hint
    
    
The result will be very large, so you may not process it by 32-bit integers.

Author

Ignatius.L

本题主要为卡特兰数的应用，若想具体了解可以翻翻一些大牛的博客：

http://www.cnblogs.com/kuangbin/archive/2012/03/21/2410519.html

http://blog.youkuaiyun.com/niushuai666/article/details/6936859

此处我就贴出自己的C++代码，此代码也参考了他人的，自己稍做了改变，其主要涉及的还是大数的乘除法：

#include<iostream> #include<iomanip> #define MAX 101 #define BASE 10000 //以四位数为基存储数据到数组 using namespace std; void multiply(int a[], int len, int b) //卡特兰数递推中的乘法（4*i-2） { for (int i = len - 1, carry = 0; i >= 0; i--) //从最后一位开始相乘 { carry += b*a[i]; a[i] = carry%BASE; carry /= BASE; } } void divide(int a[], int len, int b) //卡特兰数递推中的除法（i+1） { for (int i = 0, div = 0; i < len; i++) { div = div*BASE + a[i]; // 除法是从高位开始，同一维数组中的相连两数相差一个基（base） a[i] = div / b; // 故当往后移动一个数时，需要乘以一个基 div %= b; } } int main() { int i, j, h[101][MAX]; memset(h[1], 0, MAX*sizeof(int)); for (i = 2, h[1][MAX - 1] = 1; i <= 100; i++) { memcpy(h[i], h[i - 1], MAX*sizeof(int)); multiply(h[i], MAX, 4 * i - 2); divide(h[i], MAX, i + 1); } while (cin >> i&&i >= 1 && i <= 100) //首先列举100以内的卡特兰数 { for (j = 0; j < MAX; j++) //此处是寻找对应维中的非0元素 { if (h[i][j] != 0) break; } cout << h[i][j++]; //找到后开始打印第一个数，即最高位（该数可能4位数） for (; j < MAX; j++)//之后的数除以（i+1）后可能没有4位，所以也就需要往左补0 { if (h[i][j] < BASE) cout << setw(4) << setfill('0') << h[i][j]; else cout << h[i][j]; } cout << endl; } return 0; } </iomanip></iostream>