Hdoj/hdu 1133 Buy The Ticket (number theory ~ Cattleya ~ large number ~)

Problem Description
The "Harry Potter and the Goblet of Fire" would be on show in the next few days. As a crazy fan of Harry Potter, you'll go to the cinema and has the first sight, won ' t you?

Suppose the cinema only have one ticket-office and the price for per-ticket are dollars. The queue for buying the tickets are consisted of M + N persons (m persons each have the 50-dollar bill and N persons E Ach only has the 100-dollar bill).

Now, the problem for are to calculate the number of different ways of the "the" and 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 is stopped on the occasion that the Ticket-office have no 50-dollar bill but the first person of th E queue only had the 100-dollar bill.

Input
The input file contains several test cases. Each test case was made up of 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) and then output the number of different ways in Another line.

Sample Input
3 0
3 1
3 3
0 0

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

Test instructions

is m+n individual to buy tickets, the price of 50 yuan, M personal only with 50 yuan a note, n individuals with only 100 yuan a note, the conductor began to have no money in hand.
Ask, how to let the conductor in the case can be all the change, arrange these people buy tickets order ~

In other words: to 100 yuan people before the ticket, the conductor must have a 50 yuan in hand.

The derivation process is as follows:
M personal take 50,n 1001,
If n > M, then the number of sorting methods is 0, it is easy to think clearly
2, now we assume that 50 of the people with ' 0 ', take 100 of the people with ' 1 '. If there is such a sequence 0101101001001111.
When the number of 1 in the K position is more than 0, it is an illegal sequence. Suppose that a sequence of m=4,n=3 is: 0110100.
This means that any illegal sequence (m 0,n 1) can be obtained by another sequence (n-1 0 and m+1 1).
In addition, we know that a sequence is either legal or illegal.
Therefore, the number of legal sequences = total number of sequences-the total amount of an illegal sequence.
The total number of sequences can be calculated like this, m+n position, select n position to fill 1, so is C (m+n,n).
The number of illegal sequences is: m+n a position, select m+1 position to fill 1, so is C (m+n,m+1). And then everyone is not the same, so you need to arrange the whole m! * n!.

So the final formula is: (C (M+n,n)-C (m+n,m+1)) * m! * n! Simplification is: (m+n)!* (m-n+1)/(m+1)

If you do not understand: You can refer to my previous analysis of Cattleya number:
http://blog.csdn.net/qq_26525215/article/details/51453493

ImportJava.math.BigInteger;ImportJava.util.Scanner;/** * @author Chen Haoxiang * * 2016-5-19 * * Public  class Main{     Public Static void Main(string[] args) {Scanner SC =NewScanner (system.in);intt=0; while(Sc.hasnext ()) {intM =sc.nextint ();intn =sc.nextint ();if(n==0&&m==0){return; } System.out.println ("Test #"+ (++T) +":");if(n>m) {System.out.println (0);Continue; } BigInteger A =NewBigInteger ("1"); for(intI=2; i<=m+n;i++) {a=a.multiply (NewBigInteger (""+i)); } a=a.multiply (NewBigInteger (""+ (m-n+1))); A=a.divide (NewBigInteger (""+ (m+1)));        System.out.println (a); }    }}

Hdoj/hdu 1133 Buy The Ticket (number theory ~ Cattleya ~ large number ~)