HDU 1023 Train Problem II （卡特兰数问题）

Train Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2830 Accepted Submission(s): 1562


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, you should output how many ways that all the trains can get out of the railway.

[align=left]Sample Input[/align]

1 2 3 10

[align=left]Sample Output[/align]

1 2 5 16796

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

[align=left]Author[/align]
Ignatius.L

简单的卡特兰数问题。
卡特兰数介绍：
/article/4679949.html

这个已经可以做为模版了~~~

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

#include<stdio.h>

//*******************************
//打表卡特兰数
//第 n个 卡特兰数存在a
中，a
[0]表示长度；
//注意数是倒着存的，个位是 a
[1] 输出时注意倒过来。
//*********************************
int a[105][100];
void ktl()
{
int i,j,yu,len;
a[2][0]=1;
a[2][1]=2;
a[1][0]=1;
a[1][1]=1;
len=1;
for(i=3;i<101;i++)
{
yu=0;
for(j=1;j<=len;j++)
{
int t=(a[i-1][j])*(4*i-2)+yu;
yu=t/10;
a[i][j]=t%10;
}
while(yu)
{
a[i][++len]=yu%10;
yu/=10;
}
for(j=len;j>=1;j--)
{
int t=a[i][j]+yu*10;
a[i][j]=t/(i+1);
yu = t%(i+1);
}
while(!a[i][len])
{
len--;
}
a[i][0]=len;
}

}
int main()
{
ktl();
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=a
[0];i>0;i--)
{
printf("%d",a
[i]);
}
puts("");
}
return 0;
}