Buy the Ticket
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3598 Accepted Submission(s): 1514
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.
3 0 3 1 3 3 0 0
Test #1: 6 Test #2: 18 Test #3: 180AC代码+解释：#include<iostream> #include<cstdio> #include<cstring>//又一个经典的卡特兰数的应用：50元的数量永远不小于100元的数量，m和n最大均为100，第200位卡特兰数是373位 #include<string> #include<algorithm> const int MAX=501; int s[MAX]; using namespace std; int main() { int n,m,q,p=1,c,sum,k,temp,t,i,j; while(cin>>m>>n&&(m||n)) { cout<<"Test #"<<p++<<':'<<endl; if(m<n)//如果m<n那么永远没有方法所以输出0 { cout<<0<<endl; continue; } memset(s,0,sizeof(s)); s[1]=1; sum=0; for(i=2;i<=(m+n);i++)//处理好（n+m）的阶层，如果不懂原理的可以去做HDOJ上的1042题 { for(j=1;j<MAX;j++) { c=s[j]*i+sum; s[j]=c%10; sum=c/10; } } q=m+1-n; sum=0; for(j=1;j<MAX;j++) { c=s[j]*q+sum; s[j]=c%10; sum=c/10; } for(j=MAX-1;j>=1;j--) { if(s[j]!=0) break; } for(i=j;i>=1;i--) for(i=j,sum=0,t=0;i>=1;i--) { temp=sum*10+s[i]; if(t==0)//处理好前导零 { if(temp/(m+1)>0) { t=1; cout<<temp/(m+1); } } else cout<<temp/(m+1); sum=temp%(m+1); } cout<<endl; } return 0; }