hdoj1023 Train Problem II

Train Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7333    Accepted Submission(s): 3946

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


   
   

    
    

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

   
   
   
    
  
  

 

解题思路：由测试用例可看出是这是卡特兰数

卡特兰数又称卡塔兰数，英文名Catalan number，是组合数学中一个常出现在各种计数问题中出现的数列。以比利时的数学家欧仁·查理·卡塔兰 (1814–1894)的名字来命名，其前几项为 : 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, ...


令h(0)=1,h(1)=1，catalan数满足递推式
h(n)= h(0)*h(n-1)+h(1)*h(n-2) + ... + h(n-1)h(0) (n>=2)
另类递推式

h(n)=h(n-1)*(4*n-2)/(n+1);

我是用的这个递推式做的

首先将h(n-1)和（4*n-2）看做大数乘法，所得值保留到h(n-1)中，然后将h(n-1)和(n+1)看做大数除法。最后即为得数。

#include <stdio.h>
#include <string.h>
int a[10000],count;
void bigMul(int y)//大数除法
{    int i,car = 0;
  for(i=0;i<=count;i++)
    {    a[i] = a[i] * y + car;
         car = a[i] / 10;
         a[i] %= 10;
    }
    if(car > 0)
    {        
        a[++count] = car;   
    }                   
}
void bigDiv(int y)//大数乘法
{
	int i,temp,ret = 0;
	for(i=count;i>=0;i--)
	{
		temp = a[i] + ret;
		a[i] = temp / y;
		ret = temp % y * 10;
	}
}
int main()
{    
    int n,i,j;
    while(scanf("%d",&n)!=EOF){
    if(n == 0 || n == 1)
    {
  	   printf("1\n");
  	   continue;
  	}
    memset(a,0,sizeof(a));
    a[0] = 1;
    count = 0;
    for(i=2;i<=n;i++) {
        bigMul((4*i-2));
        bigDiv((i+1));
    }
    if(a[count] != 0)//处理第一个为零的情况
    {
        printf("%d",a[count]);
    }
    for(i=count-1;i>=0;i--)
    {
        if(!a[i])//数的前面为0不输出
            continue;
        while(i>=0)
            printf("%d",a[i--]);
    }
    printf("\n");
}
return 0;}