第四届 山东省ACM Contest Print Server

Contest Print Server

Time Limit: 1000MS Memory Limit: 65536KB
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Problem 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.

Example 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

Example Output

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

Hint

 

Author

 2013年山东省第四届ACM大学生程序设计竞赛

#include <iostream>
#include <algorithm>
#include <stdio.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define maxn 105

int n,s,x,y,mod;
struct str
{
char name[25];
int p;
};

str tt[maxn];

int main()
{
//read;
int cas;
scanf("%d",&cas);
while(cas--)
{
scanf("%d%d%d%d%d",&n,&s,&x,&y,&mod);
char s1[25],s2[25];
for(int i=1; i<=n; ++i)
scanf("%s%s%d%s",tt[i].name,s1,&tt[i].p,s2);
int now=s;
for(int i=1; i<=n; ++i)
{
if(tt[i].p<=now)
{
printf("%d pages for %s\n",tt[i].p,tt[i].name);
now-=tt[i].p;
}
else
{
printf("%d pages for %s\n",now,tt[i].name);
s=(x*s+y)%mod;
now=s;
--i;
}
}
printf("\n");
}
return 0;
}