Buy the Ticket
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4981 Accepted Submission(s): 2085
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.
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 0
Sample Output
Test #1:
6
Test #2:
18
Test #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++;
}
}
//从网上找到了公式 即：结果等于 (m+n)!*(m-n+1)/(m+1)
//那么这题就是高精度问题了
