HDU1130 卡特兰数

How Many Trees?
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3397    Accepted Submission(s):
1964


[align=left]Problem Description[/align]
A binary search tree is a binary tree with root k such
that any node v reachable from its left has label (v) <label (k) and any node
w reachable from its right has label (w) > label (k). It is a search
structure which can find a node with label x in O(n log n) average time, where n
is the size of the tree (number of vertices).

Given a number n, can you
tell how many different binary search trees may be constructed with a set of
numbers of size n such that each element of the set will be associated to the
label of exactly one node in a binary search tree?
 
 

[align=left]Input[/align]
The input will contain a number 1 <= i <= 100 per
line representing the number of elements of the set.
 
 

[align=left]Output[/align]
You have to print a line in the output for each entry
with the answer to the previous question.
 
 

[align=left]Sample Input[/align]

1
2
3

 
 

[align=left]Sample Output[/align]

1
2
5

 
 

[align=left]Source[/align]
UVA

 
 

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http://blog.csdn.net/sunshine_yg/article/details/47685737
卡特兰数

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
using namespace std;
const int base = 10000;
const int N = 100 + 2;
int katelan

;
int main()
{
katelan[1][1] = 1;
katelan[2][1] = 2;
katelan[3][1] = 5;
for (int i = 4; i <= 100; i++)
{
for (int j =1; j<100; j++)//大数乘法
{
katelan[i][j] += katelan[i - 1][j] * (4 * i - 2);
katelan[i][j + 1] += katelan[i][j] / base;
katelan[i][j] %= base;
}
int temp;
for (int j = 100; j > 0; j--)//大数除法
{
temp=katelan[i][j] % (i + 1);
katelan[i][j-1]+=temp* base;
katelan[i][j] /= (i + 1);
}
}
int n;
while (cin >> n)
{
int i = 100;
while (katelan
[i] == 0)i--;
cout << katelan
[i--];
while (i > 0)
printf("%04d", katelan
[i--]);
cout << endl;
}
return 0;
}