zoukankan      html  css  js  c++  java
  • 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[n][k].subtract(dp[n][k-1]));
                pr=true;
            }
        }
    }
    


  • 相关阅读:
    joomla
    笨兔兔的故事(下)
    wine乱码
    笨兔兔的故事(中)
    gedit中文乱码
    php不常用函数
    android配置开发环境ubuntu
    笨兔兔的故事(上)
    关于linux编码
    linux常用命令小记
  • 原文地址:https://www.cnblogs.com/yxysuanfa/p/6783045.html
Copyright © 2011-2022 走看看