Contest Print Server组队第四场J

问题 J:Contest Print Server

时间限制: 1 Sec 内存限制: 128 MB

题目描述

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.

输入

In the first line there is an integer T(T<=10),which indicates the number of test cases.

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.

输出

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.

样例输入 Copy

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

样例输出 Copy

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

题意是：有一台打印机，打印的时候如果累计打印到 s 张还没有将当前任务打印完，那么就会重新打印着一份文件，在此之前 s 会重新生成一个，生成的方式是s = (s * x + y) % mod;

代码没什么东西，但是有一个坑点，虽然开始输入的 s 是一个>=1的数，但是如果重新生成的s为0的话，就会再次重新生成一个 s

code:

int T = read;
    while(T--){
        n=read,s=read,x=read,y=read,mod=read;int sy=s;
        for(int i=1;i<=n;i++){
            char s1[50],s2[50],s3[50];
            int tot;
            scanf("%s%s%d%s",s1,s2,&tot,s3);
            if(tot<=sy){
                printf("%d pages for %s\n",tot,s1);
                sy -= tot;
            }else{
                while(1){
                    if(tot <= sy){
                        printf("%d pages for %s\n",tot,s1); sy-=tot;
                        break;
                    }else{
                        printf("%d pages for %s\n",sy,s1); s=fun();
                        if(s==0) s=(s*x+y)%mod;
                        sy=s;
                    }
                }
            }
        }
        puts("");
    }