Problem Description
After AC all the hardest problems in the world , the ACboy 8006 now has nothing to do . One day he goes to an old library to find a part-time job .It is also a big library which has N books and M users.The user's id is from 1 to M , and the book id is from 1 to N . According to the library rules , every user are only allowed to borrow 9 books .But what surprised him is that there is no computer in the library , and everything is just recorded in paper ! How terrible , I must be crazy after working some weeks , he thinks .So he wants to change the situation .
In the other hand , after 8006's fans know it , they all collect money and buy many computers for the library .
Besides the hardware , the library needs a management program . Though it is just a piece of cake for 8006 , the library turns to you , a excellent programer,for help .
What they need is just a simple library management program . It is just a console program , and have only three commands : Borrow the book , Return the book , Query the user .
1.The Borrow command has two parameters : The user id and the book id
The format is : "B ui bi" (1<=ui<=M , 1<=bi<=N)
The program must first check the book bi wether it's in the library . If it is not , just print "The book is not in the library now" in a line .
If it is , then check the user ui .
If the user has borrowed 9 books already, print "You are not allowed to borrow any more" .
Else let the user ui borrow the book , and print "Borrow success".
2.The Return command only has one parameter : The book id
The format is : "R bi" (1<=bi<=N)
The program must first check the book bi whether it's in the library . If it is , just print "The book is already in the library" . Otherwise , you can return the book , and print "Return success".
3.The Query command has one parameter : The user id
The format is : "Q ui" (1<=ui<=M)
If the number of books which the user ui has borrowed is 0 ,just print "Empty" , otherwise print the books' id which he borrows in increasing order in a line.Seperate two books with a blank.
Input
The input file contains a series of test cases . Please process to the end of file . The first line contains two integers M and N ( 1<= M <= 1000 , 1<=N<=100000),the second line contains a integer C means the number of commands(1<=C<=10000). Then it comes C lines . Each line is a command which is described above.You can assum all the books are in the library at the beginning of each cases.
Output
For each command , print the message which described above .
Please output a blank line after each test.
If you still have some questions , see the Sample
Sample Input
5 10 9 R 1 B 1 5 B 1 2 Q 1 Q 2 R 5 Q 1 R 2 Q 1 5 10 9 R 1 B 1 5 B 1 2 Q 1 Q 2 R 5 Q 1 R 2 Q 1
Sample Output
The book is already in the library Borrow success Borrow success 2 5 Empty Return success 2 Return success Empty The book is already in the library Borrow success Borrow success 2 5 Empty Return success 2 Return success Empty
//Huge input, the function scanf() may work better than cin
Hint
Hint
纯粹的模拟题
#include<iostream> #include<algorithm> #include<cstring> using namespace std; int book[101000]; struct node { int x[10]; int y; } p[1010]; int main() { int n,m,b,u; char a; while(cin>>n>>m) { memset(book,0,sizeof(book)); for(int i=0; i<=n; i++) p[i].y=0; int t; cin>>t; while(t--) { cin>>a; if(a=='B') { cin>>u>>b; if(book[b]) puts("The book is not in the library now"); else if(p[u].y>=9) puts("You are not allowed to borrow any more"); else { p[u].x[p[u].y++]=b; book[b]=u; puts("Borrow success"); } } else if(a=='R') { cin>>b; if(book[b]==0) puts("The book is already in the library"); else { int poit; u=book[b]; book[b]=0; for(int i=0; i<p[u].y; i++) { if(p[u].x[i]==b) { poit=i; break; } } for(int i=poit; i<p[u].y-1; i++) p[u].x[i]=p[u].x[i+1]; p[u].y--; puts("Return success"); } } else if(a=='Q') { cin>>u; if(p[u].y==0) puts("Empty"); else { int b[10]; for(int i=0; i<p[u].y; i++) b[i]=p[u].x[i]; sort(b,b+p[u].y); for(int i=0; i<p[u].y; i++) { if(i==0) cout<<b[i]; else cout<<" "<<b[i]; } cout<<endl; } } } cout<<endl; } return 0; }