hdu 1023 Train Problem II（大数卡特兰数-出栈次序问题）-优快云博客

本文介绍了一个关于火车出站顺序的经典问题，并通过卡特兰数进行解答。问题要求计算不同数量的火车可能的出站顺序组合，涉及高精度的大数计算实现。

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Train Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2111 Accepted Submission(s): 1238

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

  1 2 3 10
 

Sample Output

  1 2 5 16796 
 

这题的意思就是和那个经典的出栈次序问题一模一样，就是卡特兰数，不过它的N的范围很大，肯定要用到高精度，就是大数了，直接套大数模板来求卡特兰数。就是一道经典的卡特兰数问题。

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1023

代码：

#include <iostream>
#include <stdio.h>
#include <cmath>
using namespace std;

int a[105][105];    //大数卡特兰数
int b[105];         //卡特兰数的长度

void catalan()  //求卡特兰数
{
    int i, j, len, carry, temp;
    a[1][0] = b[1] = 1;
    len = 1;
    for(i = 2; i <= 100; i++)
    {
        for(j = 0; j < len; j++)    //乘法
            a[i][j] = a[i-1][j]*(4*(i-1)+2);
        carry = 0;
        for(j = 0; j < len; j++)    //处理相乘结果
        {
            temp = a[i][j] + carry;
            a[i][j] = temp % 10;
            carry = temp / 10;
        }
        while(carry)    //进位处理
        {
            a[i][len++] = carry % 10;
            carry /= 10;
        }
        carry = 0;
        for(j = len-1; j >= 0; j--) //除法
        {
            temp = carry*10 + a[i][j];
            a[i][j] = temp/(i+1);
            carry = temp%(i+1);
        }
        while(!a[i][len-1])     //高位零处理
            len --;
        b[i] = len;
    }
}

int main()
{
    int i, n;
    catalan();
    while(scanf("%d", &n) != EOF)
    {
        for(i = b[n]-1; i>=0; i--)
        {
            printf("%d", a[n][i]);
        }
        printf("\n");
    }

    return 0;
}