UVA 10157 - Expressions（dp卡特兰数高精度）

Problem A: EXPRESSIONS 

Let X be the set of correctly built parenthesis expressions. The elements of X are strings consisting only of the characters �(� and �)�. The set X is defined as follows:

• an empty string belongs to X
• if A belongs to X, then (A) belongs to X
• if both A and B belong to X, then the concatenation AB belongs to X.

For example, the following strings are correctly built parenthesis expressions (and therefore belong to the set X):

()(())()

(()(()))

The expressions below are not correctly built parenthesis expressions (and are thus not in X):

(()))(()

())(()

Let E be a correctly built parenthesis expression (therefore E is a string belonging to X).

The length of E is the number of single parenthesis (characters) in E.

The depth D(E) of E is defined as follows:

               ì 0 if E is empty 
D(E)= í D(A)+1 if E = (A), and A is in X 
               î max(D(A),D(B)) if E = AB, and A, B are in X

For example, the length of �()(())()� is 8, and its depth is 2.

What is the number of correctly built parenthesis expressions of length n and depth d, for given positive integers n and d?

Task 
Write a program which

• reads two integers n and d
• computes the number of correctly built parenthesis expressions of length n and depth d;

Input data 
Input consists of lines of pairs of two integers - n and d, at most one pair on line, 2 £ n £ 300, 1 £ d£ 150. 
The number of lines in the input file is at most 20, the input may contain empty lines, which you don't need to consider.

Output data 
For every pair of integers in the input write single integer on one line - the number of correctly built parenthesis expressions of length n and depth d.

Example 
Input data                                   Output data 
6 2                 3 
300 150             1 

There are exactly three correctly built parenthesis expressions of length 6 and depth 2: 
(())() 
()(()) 
(()()) 

题意：求长度n，深度最多为d的括号种数。

思路：dp，dp[i][j]表示长度i，深度不超过j的种数，对于每种情况，取最左边的括号和右边每个位置进行匹配，剩下的为内部和外部两部分，种数为dp[k][j - 1]*dp[i - k - 1][j]。然后把所有加起来即可

代码：

#include <stdio.h>
#include <string.h>
#include <math.h>
const int MAXN = 155;

#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)

const int MAXSIZE = 1005;

struct bign {
	int s[MAXSIZE];
	bign ()	{memset(s, 0, sizeof(s));}
	bign (int number) {*this = number;}
	bign (const char* number) {*this = number;}
    
	void put();
	bign mul(int d);
	void del();
    
	bign operator =  (char *num);
	bign operator =  (int num);

	bool operator <  (const bign& b) const;
	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 operator + (const bign& c);
	bign operator * (const bign& c);
	bign operator - (const bign& c);
	int  operator / (const bign& c);
	bign operator / (int k);
	bign operator % (const bign &c);
	int  operator % (int k);
	void operator ++ ();
	bool operator -- ();
};

bign bign::operator = (char *num) {
	s[0] = strlen(num);
	for (int i = 1; i <= s[0]; i++)
		s[i] = num[s[0] - i] - '0';
	return *this;
}

bign bign::operator = (int num) {
	char str[MAXSIZE];
	sprintf(str, "%d", num);
	return *this = str;
}

bool bign::operator < (const bign& b) const {
	if (s[0] != b.s[0])
		return s[0] < b.s[0];
	for (int i = s[0]; i; i--)
		if (s[i] != b.s[i])
			return s[i] < b.s[i];
	return false;
}

bign bign::operator + (const bign& c) {
	int sum = 0;
	bign ans;
	ans.s[0] = max(s[0], c.s[0]);

	for (int i = 1; i <= ans.s[0]; i++) {
		if (i <= s[0]) sum += s[i];
		if (i <= c.s[0]) sum += c.s[i];
		ans.s[i] = sum % 10;
		sum /= 10;
	}
	while (sum) {
		ans.s[++ans.s[0]] = sum % 10;
		sum /= 10;
	}
	return ans;
}

bign bign::operator * (const bign& c) {
	bign ans;
	ans.s[0] = 0; 

	for (int i = 1; i <= c.s[0]; i++){  
		int g = 0;  

		for (int j = 1; j <= s[0]; j++){  
			int x = s[j] * c.s[i] + g + ans.s[i + j - 1];  
			ans.s[i + j - 1] = x % 10;  
			g = x / 10;  
		}  
		int t = i + s[0] - 1;

		while (g){  
			++t;
			g += ans.s[t];
			ans.s[t] = g % 10;
			g = g / 10;  
		}  

		ans.s[0] = max(ans.s[0], t);
	}  
	ans.del();
	return ans;
}

bign bign::operator - (const bign& c) {
	bign ans = *this; int i;
	for (i = 1; i <= c.s[0]; i++) {
		if (ans.s[i] < c.s[i]) {
			ans.s[i] += 10;
			ans.s[i + 1]--;;
		}
		ans.s[i] -= c.s[i];
	}

	for (i = 1; i <= ans.s[0]; i++) {
		if (ans.s[i] < 0) {
			ans.s[i] += 10;
			ans.s[i + 1]--;
		}
	}

	ans.del();
	return ans;
}

int bign::operator / (const bign& c) {
	int ans = 0;
	bign d = *this;
	while (d >= c) {
		d = d - c;
		ans++;
	}
	return ans;
}

bign bign::operator / (int k) {
	bign ans; 
	ans.s[0] = s[0];
	int num = 0;  
	for (int i = s[0]; i; i--) {  
		num = num * 10 + s[i];  
		ans.s[i] = num / k;  
		num = num % k;  
	}  
	ans.del();
	return ans;
}

int bign:: operator % (int k){  
	int sum = 0;  
	for (int i = s[0]; i; i--){  
		sum = sum * 10 + s[i];  
		sum = sum % k;  
	}  
	return sum;  
} 

bign bign::operator % (const bign &c) {
	bign now = *this;
	while (now >= c) {
		now = now - c;
		now.del();
	}
	return now;
}

void bign::operator ++ () {
	s[1]++;
	for (int i = 1; s[i] == 10; i++) {
		s[i] = 0;
		s[i + 1]++;
		s[0] = max(s[0], i + 1);
	}
}

bool bign::operator -- () {
	del();
	if (s[0] == 1 && s[1] == 0) return false;

	int i;
	for (i = 1; s[i] == 0; i++)
		s[i] = 9;
	s[i]--;
	del();
	return true;
}

void bign::put() {
	if (s[0] == 0)
		printf("0");
	else
		for (int i = s[0]; i; i--)
			printf("%d", s[i]);
}

bign bign::mul(int d) {
	s[0] += d; int i;
	for (i = s[0]; i > d; i--)
		s[i] = s[i - d];
	for (i = d; i; i--)
		s[i] = 0;
	return *this;
}

void bign::del() {
	while (s[s[0]] == 0) {
		s[0]--;
		if (s[0] == 0) break;
	}
}

int n, d;
bign f[MAXN][MAXN];

void init() {
	int i, j, k;
	for (i = 0; i <= 150; i ++)
		f[0][i] = 1;
	for (i = 1; i <= 150; i ++) {
		for (j = 1; j <= 150; j ++) {
			for (k = 0; k < i; k ++) {
				f[i][j] = f[k][j - 1] * f[i - k - 1][j] + f[i][j];
			}
		}
	}
}

int main() {
	init();
	while (~scanf("%d%d", &n, &d)) {
		(f[n / 2][d] - f[n / 2][d - 1]).put();
		printf("\n");
	}
	return 0;
}