UVA 10303

Problem D

How Many Trees?

Input: standard input

Output: standard output

Memory Limit: 32 MB

A binary search tree is a binary tree with root k such that any node v in the left subtree of k has label (v) <label (k) and any node w in the right subtree of k has label (w) > label (k).

When using binary search trees, one can easily look for a node with a given label x: After we compare x to the label of the root, either we found the node we seek or we know which subtree it is in. For most binary search trees the average time to find one of its n nodes in this way is O(log n).

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?

Input and Output

The input will contain a number 1 <= i <= 1000 per line representing the number of elements of the set. You have to print a line in the output for each entry with the answer to the previous question.

Sample Input

1

2

3

 

Sample Output

1

2

5

题意：给定n个结点，求有几种2叉搜索树。

思路：分别取第n个点做根节点。如此图

代码：

#include <stdio.h>
#include <string.h>
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)
const int N = 1005;
const int MAXBIGN = 1005;


struct bign {
    int s[MAXBIGN];
    int len;
    bign() {
		len = 1;
		memset(s, 0, sizeof(s));
    }
	
    bign operator = (const char *number) {
		len = strlen(number);
		for (int i = 0; i < len; i++)
			s[len - i - 1] = number[i] - '0';
		return *this;
    }
    bign operator = (const int num) {
		char number[N];
		sprintf(number, "%d", num);
		*this = number;
		return *this;
    }
	
    bign (int number) {*this = number;}
    bign (const char* number) {*this = number;}
	
    bign operator + (const bign &c){  
		bign sum;
		int t = 0;
		sum.len = max(this->len, c.len);
		for (int i = 0; i < sum.len; i++) {
			if (i < this->len) t += this->s[i];
			if (i < c.len) t += c.s[i];
			sum.s[i] = t % 10;
			t /= 10;
		}
		
		while (t) {
			sum.s[sum.len++] = t % 10;
			t /= 10;
		}
		
		return sum;  
    }
	
	bign operator * (const bign &c){  
		bign sum; bign zero;
		if (*this == zero || c == zero)
			return zero;
		int i, j;
		sum.len = this->len + c.len;
		for (i = 0; i < this->len; i++) {
			for (j = 0; j < c.len; j ++) {
				sum.s[i + j] += this->s[i] * c.s[j];
			}
		}
		for (i = 0; i < sum.len; i ++) {
			sum.s[i + 1] += sum.s[i] / 10;
			sum.s[i] %= 10;
		}
		sum.len ++;
		while (!sum.s[sum.len - 1]) {
			sum.len --;
		}
		return sum;  
    }	
	bign operator * (const int &num) {
		bign c = num;
		return *this * c;
	}
	bign operator / (const int &num) {
		bign ans; int k = 0;
		ans.len = len;
		for (int i = ans.len - 1; i >= 0; i --) {
			ans.s[i] = (k * 10 + s[i]) / num;
			k = (k * 10 + s[i]) % num;
		}
		while (!ans.s[ans.len - 1]) {
			ans.len --;
		}
		return ans;
	}
    bign operator - (const bign &c) {
		bign ans;
		ans.len = max(this->len, c.len);
		int i;
		
		for (i = 0; i < c.len; i++) {
			if (this->s[i] < c.s[i]) {
				this->s[i] += 10;
				this->s[i + 1]--;
			}
			ans.s[i] = this->s[i] - c.s[i];
		}
		
		for (; i < this->len; i++) {
			if (this->s[i] < 0) {
				this->s[i] += 10;
				this->s[i + 1]--;
			}
			ans.s[i] = this->s[i];
		}
		while (ans.s[ans.len - 1] == 0) {
			ans.len--;
		}
		if (ans.len == 0) ans.len = 1;
		return ans;
    }
	
    void put() {
		if (len == 1 && s[0] == 0) {
			printf("0");
		} else {
			for (int i = len - 1; i >= 0; i--)
				printf("%d", s[i]);
		}
    }
	
    bool operator < (const bign& b) const {
		if (len != b.len)
			return len < b.len;
		
		for (int i = len - 1; i >= 0; i--)
			if (s[i] != b.s[i])
				return s[i] < b.s[i];
			return false;
    }
    bool operator > (const bign& b) const { return b < *this; }
    bool operator <= (const bign& b) const { return !(b < *this); }
    bool operator >= (const bign& b) const { return !(*this < b); }
    bool operator != (const bign& b) const { return b < *this || *this < b;}
    bool operator == (const bign& b) const { return !(b != *this); }
};

bign f[1005];
int n; 

void init() {
	f[1] = 1;
	for (int i = 2; i <= 1000; i ++) {
		f[i] = f[i - 1] * (4 * i - 2) / (i + 1);
	}
}

int main() {
	init();
	while (~scanf("%d", &n) && n) {
		f[n].put();
		printf("
");
	}
	return 0;
}