Contest Print Server (2013年山东省第四届ACM大学生程序设计竞赛)：

本文介绍了一道关于ACM竞赛中模拟打印服务器处理请求的题目。解析了题目要求及给出的示例输入输出，通过代码实现了解决方案，强调了在处理打印请求时打印机状态更新的正确逻辑。

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Contest Print Server

Time Limit: 1000MS Memory limit: 65536K

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 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.

示例输入

23 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

示例输出

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

分析：这是13年省赛里的一道水题，在重现赛的时候做的时候，一直过不去，根据题意，是判断打印纸张是否够用，够用的话就把需要的打印张数打印出来，如果不够用，就把剩下的纸张打印出来，然后S(s is generated by the function s=(s*x+y)%mod )根据函数式变化，count清零，再判断纸张是否够用，并把没打印完整的重新打印一遍。

当时我自己认为纸张刚好用完的时候，下个队伍申请打印的时候应该S自动变化，不再打印0张的时候，可事实上我是错的，队友给了纠正，交了一遍就AC了

代码如下：

#include <stdio.h>
int main()
{
	int T;
	int n,s,x,y,mod,temp,count;
	int i,j,p;
	char str[25];
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d %d %d %d %d",&n,&s,&x,&y,&mod);
		count=s;
		for(i=1;i<=n;i++)
		{
			if(temp)
				scanf("%s request %d pages",str,&p);
			temp=1;
			if(count>=p)
			{
				printf("%d pages for %s\n",p,str);
				count-=p;
			}
			else
			{
				printf("%d pages for %s\n",count,str);
				s=(s*x+y)%mod;
				count=s;
				temp=0;
				i--;
			}
		}
        printf("\n");
	}
	return 0;
}