UPC2226: Contest Print Server

2226: Contest Print Server

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 515  Solved: 119
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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 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.

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

23 7 5 6 177Team1 request 1 pagesTeam2 request 5 pagesTeam3 request 1 pages3 4 5 6 177Team1 request 1 pagesTeam2 request 5 pagesTeam3 request 1 pages

Sample Output

1 pages for Team15 pages for Team21 pages for Team31 pages for Team13 pages for Team25 pages for Team21 pages for Team3

HINT

Source

2013年山东省第四届ACM大学生程序设计竞赛

#include<cstdio>using namespace std;int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,s,x,y,mod;        scanf("%d%d%d%d%d",&n,&s,&x,&y,&mod);        int rest=s;        while(n--)        {            char name[21];            int p;            scanf("%s request %d pages",name,&p);            while(1){                if(p > rest){                    printf("%d pages for %s\n",rest,name);                    s = rest = (s * x + y) % mod;                }                else {                    printf("%d pages for %s\n",p,name);                    rest -= p;                    break;                }            }        }        printf("\n");    }    return 0;}

注意有坑的地方就是当rest为0时，也需要输出。

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