【HDU-1023】 Train Problem II 【卡特兰数+高精度】

传送门：http://acm.split.hdu.edu.cn/showproblem.php?pid=1023

描述：

Train Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8612    Accepted Submission(s): 4613

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

12310

 

Sample Output

12516796
Hint
The result will be very large, so you may not process it by 32-bit integers.
 

 

Author

Ignatius.L

 

Recommend

We have carefully selected several similar problems for you:  1133 1022 1130 1131 1134 

题意：

给你一个数n，表示有n辆火车，编号从1到n，从远方驶过来，问你有多少种出站的可能。

思路：

模拟栈的问题而已。

这道题其实也是组合数学卡特兰数的一个典型应用而已。

卡特兰数的介绍推荐看ACdreamer的http://blog.csdn.net/acdreamers/article/details/7628667

和kuangbin的http://www.cnblogs.com/kuangbin/archive/2012/03/21/2410516.html  这两篇博客

当然100以内你也可以打表：

可以去戳这个网站，里面有1到100的卡特兰数http://blog.csdn.net/lttree/article/details/29392541

代码一（C++）：

#include <iostream>#include <stdio.h>using namespace std;int a[101][101]={0};int main(){  int n,i,j,len,r,temp,t;  int b[101];  a[1][0] = 1;  len = 1;  b[1] = 1;  for(i=2;i<=100;i++){    t = i-1;    for(j=0;j<len;j++) //乘法      a[i][j] = a[i-1][j]*(4*t+2);    for(r=j=0;j<len;j++) { //处理相乘结果      temp = a[i][j] + r;      a[i][j] = temp % 10;      r = temp / 10;    }    while(r){ //进位处理      a[i][len++] = r % 10;      r /= 10;    }    for(j=len-1,r=0;j>=0;j--){ //除法      temp = r*10 + a[i][j];      a[i][j] = temp/(t+2);      r = temp%(t+2);    }    while(!a[i][len-1]) //高位零处理      len --;      b[i] = len;    }    while(cin>>n){         for(j=b[n]-1;j>=0;j--)        printf("%d",a[n][j]);      puts("");    }  return 0;}

代码二（java处理高精度）：

import java.io.*;  import java.util.*;  import java.math.BigInteger;      public class Main  {      public static void main(String args[])      {                 BigInteger[] a = new BigInteger[101];          a[0] = BigInteger.ZERO;          a[1] = BigInteger.valueOf(1);          for(int i = 2; i <= 100; ++i)              a[i] = a[i - 1].multiply(BigInteger.valueOf(4 * i - 2)).divide(BigInteger.valueOf(i+1));              Scanner in = new Scanner(System.in);              int n;              while(in.hasNext())              {                  n = in.nextInt();                  System.out.println(a[n]);              }      }  }