ZOJ 2604 Little Brackets DP

DP:

边界条件：dp[0][j] = 1
递推公式：dp[i][j] = sum{dp[i-k][j] * dp[k-1][j-1] | 0<k≤i}

i对括号深度不超过j的，能够唯一表示为(X)Y形式，当中X和Y能够为空，设X有k-1对括号，则相应的方案数为dp[i-k][j] * dp[k-1][j-1]

Little Brackets

Time Limit: 2 Seconds Memory Limit: 65536 KB

Consider all regular bracket sequences with one type of brackets. Let us call the depth of the sequence the maximal difference between the number of opening and the number of closing
brackets in a sequence prefix. For example, the depth of the sequence "()()(())" is 2, and the depth of "((()(())()))" is 4.

Find out the number of regular bracket sequences with n opening brackets that have the depth equal to k. For example, for n = 3 and k = 2 there are three such sequences: "()(())", "(()())",
"(())()".

Input

Input file contains several test cases. Each test case is described with n and k (1 <= k <= n <= 50).

Last testcase is followed by two zeroes. They should not be processed.

Output

For each testcase output the number of regular bracket sequences with n opening brackets that have the depth equal to k.

Separate output for different testcases by a blank line. Adhere to the format of the sample output.

Sample Input

3 2
37 23
0 0

Sample Output

Case 1: 3

Case 2: 203685956218528

Author: Andrew Stankevich

Source: Andrew Stankevich's Contest #7

import java.util.*;
import java.math.*;

public class Main
{
static BigInteger dp[][] = new BigInteger[55][55];

static void INIT()
{
for(int i=0;i<55;i++)
for(int j=0;j<55;j++) dp[i][j]=BigInteger.ZERO;

for(int i=0;i<55;i++) dp[0][i]=BigInteger.ONE;

for(int i=1;i<=50;i++)
{
for(int j=1;j<=50;j++)
{
for(int k=1;k<=i;k++)
{
dp[i][j]=dp[i][j].add(dp[i-k][j].multiply(dp[k-1][j-1]));
}
}
}
}

public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
INIT();
int cas=1;
boolean pr = false;
while(in.hasNext())
{
int n=in.nextInt(),k=in.nextInt();
if(n==0&&k==0) break;
if(pr) System.out.println("");
System.out.println("Case "+(cas++)+": "+dp
[k].subtract(dp
[k-1]));
pr=true;
}
}
}