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Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 4981 Accepted Submission (s): 2085

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 03 13 30 0

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

Author

#include <cstdio> #include <cstring>int a[390];int n,m,leap;void bfact (int n) {int i,j;        for (i=2; i<=n; i++) {if (leap&&i==m+1) continue;        int c=0,s;            for (j=0; j<380; J + +) {s=i*a[j]+c;            a[j]=s%10;        C=S/10;    }}}void bx (int n) {int J;    int s,c=0;        for (j=0; j<380; J + +) {s=n*a[j]+c;        a[j]=s%10;    C=S/10;    }}int Main () {//freopen ("case.in", "R", stdin);    int i,j,c=1;        while (scanf ("%d%d", &m,&n)!=-1) {if (!m&&!n) break;        Leap=0;        memset (A,0,sizeof (a));        A[0]=1;        printf ("Test #%d:\n", c);            if (m<n) {printf ("0\n");            C + +;        Continue        } if ((m-n+1)% (m+1) ==0) bx ((m-n+1)/(m+1));            else {bx (m-n+1);        Leap=1;        } bfact (N+m);        for (i=380; i>=0; i--) if (a[i]) break; for (j=i; j>= 0;        j--) printf ("%d", a[j]);        printf ("\ n");    C + +; }}//found the formula on the Internet: The result is equal to (m+n)!* (m-n+1)/(m+1)//Then the problem is high precision

　　

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