HDU 1133 Buy The Ticket

Problem DescriptionThe "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. Inputthe 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. Outputfor each test case, first print the test number (counting from 1) and then output the number of different W Ays in another line. Sample INPUT3 0 Sample outputtest #1:6Test #2:18Test #3:180

formula: (C (M+n, N)-C (M+n, m+1)) * m! * n! simplification (m+n)! * (m-n+1)/(m+1) by formula: C (M, n) = m! /(n!* (M-N)! ) Derivation (C (M+n, N)-C (M+n, m+1)) * m! * n!: The total arrangement minus the illegal arrangement, and because each person is different multiplied by its own full rank
#include<stdio.h>#include<string.h># define N1010intA[n], M, N;voidFact () {intI, J, s, r =0;  for(i =2; I <= (m+n); i++)    {         for(j =0; J < N; J + +) {s= A[j] * i +R; R= S/Ten; A[J]= s%Ten; }    }} //Find out (m+n)!voidSum () {intI, J, C = m-n+1, d = m+1, B[n], k =0, S, r =0;  for(i =0; i < N; i++) A[i]*=C;  for(i =0; I < n1; i++)    {        if(A[i] >9) {A[i+1] + = A[i]/Ten; A[i]%=Ten; }    } //Find out (m+n)! * (m-n+1)i = n1;  while(A[i] = =0) i--;  for(j = i; J >=0; j--) B[k++] =A[j];  for(i =0; I < K; i++) {s= B[i] +Ten*R; R= s%D; B[i]= S/D; } //Find out (m+n)! * (m-n+1)/(m+1)i =0;  while(B[i] = =0) i++;  for(j = i; J < K; J + +) printf ("%d", B[j]); printf ("\ n");}intMain () {intK =0;  while(SCANF ("%d%d", &m, &n), m+N) {k++; printf ("Test #%d:\n", K); if(M <N) printf ("0\n");//If you take 100 more people than 50, then there's no such platoon .        Else{memset (A,0,sizeof(a)); a[0] =1;            Fact ();        Sum (); }    }    return 0;}

HDU 1133 Buy The Ticket