sdutoj 2624 Contest Print Server

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2624

 

Contest Print Server

 

Time Limit: 1000ms   Memory limit: 65536K  有疑问？点这里^_^

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

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

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

提示

 

来源

 2013年山东省第四届ACM大学生程序设计竞赛

示例程序

 

分析：

按照这个形式输入第A个队伍需要打印B张纸。

然后定义s=(s*x+y)%mod。

当打印的纸张数>=s时，便会重新打印这个队伍的纸张。按照要求输出。

 

 

AC代码：

 1 #include<stdio.h>
 2 #include<string>
 3 #include<iostream>
 4 using namespace std;
 5 struct sa
 6 {
 7     int num;
 8     string name;
 9 }data[1007],cnt;
10 int main()
11 {
12     int t;
13     scanf("%d",&t);
14     while(t--)
15     {
16         int n,s,x,y,mod,i,j,flag;
17         string name,tmp1,tmp2;
18         scanf("%d%d%d%d%d",&n,&s,&x,&y,&mod);
19         getchar();
20         for(i=1;i<=n;i++)
21         {
22             cin>>name>>tmp1>>flag>>tmp2;
23             data[i].name=name;
24             data[i].num=flag;
25         }
26         int count=0,ans=0;
27         for(i=1;i<=n;i++)    
28         {
29             ans=count+data[i].num;
30             if(ans<=s)
31             {
32                 count+=data[i].num;
33                 cnt=data[i];
34             }
35             else 
36             {
37                 cnt.name=data[i].name;
38                 cnt.num=s-count;
39                 count=0;
40                 s=(s*x+y)%mod;
41                 if(s==0)s=(s*x+y)%mod;
42                 i--;
43             }
44             cout<<cnt.num<<" pages for "<<cnt.name<<endl;
45         }
46         cout<<endl;
47     }
48     return 0;
49 }

View Code