hdu 1133 Buy the Ticket

Buy the Ticket

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4185    Accepted Submission(s): 1759

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 03 13 30 0

 

Sample Output

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

 

卡特兰数的应用、组合数学

m个人拿50，n个人拿100 ，可以用0代表50,1代表100；

1、首先当n>m时,显然sum=0;

2、m>=n时，序列总数为：m+n个位置中选出m个放上0，C(m+n, n) *m!*n!;
则合法数目=序列总数-不合法数目；
我们来考虑不合法序列的数目,
因为50元的共m个，当1的个数等于n=m+1是，无论怎样排列总是不合法！！即C(m+n,m+1)*m!*n!;
此序列即是不合法队列。

合法队列=C(m+n, n) *m!*n!-C(m+n,m+1)*m!*n!   化简得 (m+n)!*(m-n+1)/(m+1);

#include"stdio.h"#include"string.h"#define N 100const int M=10000;int main(){int i,j,m,n,cnt=1;int a[N];while(scanf("%d%d",&m,&n),m||n){printf("Test #%d:\n",cnt++);if(m<n){printf("0\n");continue;}memset(a,0,sizeof(a));a[0]=1;for(i=1;i<=n+m;i++){if(n!=0&&i==m+1)continue;for(j=0;j<N;j++){a[j]*=i;}for(j=1;j<N;j++){a[j]+=a[j-1]/M;a[j-1]%=M;}}if(n!=0){for(i=0;i<N;i++)a[i]*=(m-n+1);}for(i=1;i<N;i++){a[i]+=a[i-1]/M;a[i-1]%=M;}int flag=0;for(i=N-1;i>=0;i--){if(flag)printf("%04d",a[i]);else if(a[i]){flag=1;printf("%d",a[i]);}}printf("\n");}return 0;}