hdoj 1023 Train Problem II 【卡特兰数】

Train Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 6928 Accepted Submission(s): 3750



[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.

代码：

/*
原理：
令h(0)=1,h(1)=1，catalan数满足递推式[1] ：
h(n)= h(0)*h(n-1)+h(1)*h(n-2) + ... + h(n-1)h(0) (n>=2)
例如：h(2)=h(0)*h(1)+h(1)*h(0)=1*1+1*1=2
h(3)=h(0)*h(2)+h(1)*h(1)+h(2)*h(0)=1*2+1*1+2*1=5
另类递推式[2]  ：
h(n)=h(n-1)*(4*n-2)/(n+1);
递推关系的解为：
h(n)=C(2n,n)/(n+1) (n=0,1,2,...)
递推关系的另类解为：
h(n)=c(2n,n)-c(2n,n-1)(n=0,1,2,...)
*/

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[110][110]={0};//二维数组a[输入的n][n的卡特兰数的每一位的值]；
int b[110];//记录位数；
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 n,i;
catalan();
while(scanf("%d",&n)!=EOF)
{
for(i=b
-1;i>=0;i--)
{
printf("%d",a
[i]);
}
printf("\n");
}
return 0;
}