首页 > ACM题库 > HDU-杭电 > HDU 1133 Buy the Ticket-动态规划-[解题报告] java
2013
11-27

HDU 1133 Buy the Ticket-动态规划-[解题报告] java

Buy the Ticket

问题描述 :

The "Harry Potter and the Goblet of Fire" will be on show in the next few days. As a crazy fan of Harry Potter, you will go to the cinema and have the first sight, won’t you?

Suppose the cinema only has one ticket-office and the price for per-ticket is 50 dollars. The queue for buying the tickets is consisted of m + n persons (m persons each only has the 50-dollar bill and n persons each only has the 100-dollar bill).

Now the problem for you is to calculate the number of different ways of the queue that the buying process won’t be stopped from the first person till the last person.
Note: initially the ticket-office has no money.

The buying process will be stopped on the occasion that the ticket-office has no 50-dollar bill but the first person of the queue only has the 100-dollar bill.

输入:

The input file contains several test cases. Each test case is made up of two integer numbers: m and n. It is terminated by m = n = 0. Otherwise, m, n <=100.

输出:

For each test case, first print the test number (counting from 1) in one line, then output the number of different ways in another line.

样例输入:

3 0
3 1
3 3
0 0

样例输出:

Test #1:
6
Test #2:
18
Test #3:
180

题意:买票的时候,顾客有50元的钱或者100元的钱,,只能用顾客的钱找环给顾客。。所以50每一次,拿50元的要不少于拿100元的顾客。。

思路:动态规划。

方程:dp[i][j] = X*dp[i-1][j] + Y*dp[i][j-1]  .,X,Y都是表示,选择第i个人时  ,有(n-i+1)种选择方式。dp[i][j] 表示当前已经有i个拿50的顾客,j个拿100元的顾客买到票了。

import java.util.*;
import java.math.*;
public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner cin = new Scanner(System.in);
        BigInteger dp[][] = new BigInteger[109][109] ;
        
        int a,b;
        int T = 1;
        while(true){
            a = cin.nextInt();b=cin.nextInt();
            int ta =a,tb = b;
            if(a==0&&b==0) break;
            for(int i=0;i<109;i++)
                for(int j=0;j<109;j++) dp[i][j] = BigInteger.ZERO;
            System.out.println("Test #"+T+":");T++;
            if(a<b){
                System.out.println("0");
                continue;
            }
            dp[0][0] = BigInteger.ONE;
            for(int j=1;j<=a;j++)
                {dp[0][j] = dp[0][j-1].multiply(BigInteger.valueOf(ta));ta--;}
            for(int i=1;i<=b;i++){
                for(int j=i;j<=a;j++){
                    dp[i][j]=dp[i-1][j].multiply(BigInteger.valueOf(b+1-i));
                    dp[i][j]=dp[i][j].add(dp[i][j-1].multiply(BigInteger.valueOf(a+1-j)));
                }
            } 
            System.out.println(dp[b][a]);
        }
    }

}


  1. 第二块代码if(it != mp.end())应改为if(it != mp.end() && (i+1)!=(it->second +1));因为第二种解法如果数组有重复元素 就不正确

  2. Thanks for taking the time to examine this, I really feel strongly about it and love studying a lot more on this topic. If possible, as you acquire experience