【HDU】1133 - Buy the Ticket（BigDecimal & 组合数学 & 递推）
【HDU】1133 - Buy the Ticket（BigDecimal & 组合数学 & 递推）
点击打开题目
Buy the Ticket
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6176 Accepted Submission(s): 2601
Problem Description
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.
Input
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.
Output
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.
Sample Input
3 0
3 1
3 3
0 0Sample Output
Test #1:
6
Test #2:
18
Test #3:
180Author
HUANG, Ninghai
上一篇博客的推理很明白了，这道题再乘一个 n！和m！就行了。记得特判 n 和 m 大小关系。
点击链接看推理过程：点击打开链接
代码如下：
import java.math.BigDecimal;
import java.util.Scanner;
public class Main
{
public static void main(String[] args)
{
int Case = 1;
Scanner sc = new Scanner(System.in);
BigDecimal Cat[] = new BigDecimal [111];
BigDecimal one = new BigDecimal(1);
BigDecimal two = new BigDecimal(2);
BigDecimal four = new BigDecimal(4);
Cat[1] = Cat[0] = new BigDecimal(1);
for (int i = 2 ; i <= 100 ; i++) //打印卡特兰数
{
BigDecimal t = new BigDecimal(i);
Cat[i] = Cat[i-1].multiply(four.multiply(t).subtract(two)).divide(t.add(one));
}
BigDecimal dp[][] = new BigDecimal [111][111];
for (int i = 1 ; i <= 100 ; i++)
{
dp[i][0] = new BigDecimal(1);
}
for (int i = 1 ; i <= 100 ; i++) //n==m时，结果为其对应的卡特兰数
{
dp[i][i] = Cat[i];
}
for (int i = 2 ; i <= 100 ; i++)
{
for (int j = 1 ; j < i ; j++)
{
dp[i][j] = dp[i-1][j].add(dp[i][j-1]);
}
}
while (true) //输入
{
int m = sc.nextInt();
int n = sc.nextInt();
if (m==0 && n==0)
{
break;
}
System.out.println("Test #"+Case+":");
Case++;
if (n > m)
System.out.println(0);
else
System.out.println(dp[m][n].multiply(cal(m)).multiply(cal(n)));
}
}
public static BigDecimal cal(int x) //计算阶乘
{
BigDecimal ans = new BigDecimal(1);
for (int i = 2 ; i <= x ; i++)
{
BigDecimal t = new BigDecimal(i);
ans = ans.multiply(t);
}
return ans;
}
}腾讯云开发者
