Contest Print Server

Description

    In ACM/ICPC on-site contests ,3 students share 1 computer,so you can print your source code any time. Here you need to write a contest print server to handle all the requests.

Input

In each case,the first line contains 5 integers n,s,x,y,mod (1<=n<=100, 1<=s,x,y,mod<=10007), and n lines of requests follow. The request is like "Team_Name request p pages" (p is integer, 0<p<=10007, the length of "Team_Name" is no longer than 20), means the team "Team_Name" need p pages to print, but for some un-know reason the printer will break down when the printed pages counter reached s(s is generated by the function s=(s*x+y)%mod ) and then the counter will become 0. In the same time the last request will be reprint from the very begin if it isn't complete yet(The data guaranteed that every request will be completed in some time).
    You can get more from the sample.

Output

    Every time a request is completed or the printer is break down,you should output one line like "p pages for Team_Name",p is the number of pages you give the team "Team_Name".

    Please note that you should print an empty line after each case.

Sample Input

2
3 7 5 6 177
Team1 request 1 pages
Team2 request 5 pages
Team3 request 1 pages
3 4 5 6 177
Team1 request 1 pages
Team2 request 5 pages
Team3 request 1 pages

Sample Output

1 pages for Team1
5 pages for Team2
1 pages for Team3

1 pages for Team1
3 pages for Team2
5 pages for Team2
1 pages for Team3

题目大义：有一台打印机，每隔s张都会清空计数器并重新打印。

简单代码：

#include<iostream>
using namespace std;
int main()
{
    int c;
    cin>>c;
    while(c--)
    {
        int n,s,x,y,mod;
        cin>>n>>s>>x>>y>>mod;
        string name,re,pa;
        int fl;
        int cot=0;
        while(n--)
        {
            cin>>name>>re>>fl>>pa;
            //cout<<fl<<endl;

            if(cot+fl<=s)
            {
                cout<<fl<<" pages for "<<name<<endl;
                cot+=fl;
            }
            else
            {
                    cout<<s-cot<<" pages for "<<name<<endl;
                    int n = cot;
                    s=(s*x+y)%mod;
                    cot =0;
                    while(fl>s){
                            cout<<s<<" pages for "<<name<<endl;
                            cot=0;
                            s=(s*x+y)%mod;
                        }
                    cout<<fl<<" pages for "<<name<<endl;
                    cot+=fl;
            }
        }
        cout<<endl;
    }
}