Problem J:Contest Print Server
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65535/32768K (Java/Other)
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 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
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
Sample Output
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：
给出 T （<= 10），代表有 T 组 case。后给出 n（1 ~ 100），s（1 ~ 10007），x（1 ~ 10007），y（1 ~ 10007），mod （1 ~ 10007）代表有 n 个要求，每个要求都有一个 team 要求打印纸张数量，如果要求的这个张数 p <= s，则直接打印 s，若不满足，则先打印出剩下的 s 后，以 s = （ns * x + y）% mod 来更新纸张数量，ns 为最开始时候的原始纸张数量，更新完后再按要求打印出 p。
思路：
模拟。坑在输入，而且s == 0张的时候也会输出的。
AC：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int main () {
int t;
scanf("%d", &t);
while (t--) {
int n, s, x, y, mod, ns;
scanf("%d%d%d%d%d\n", &n, &s, &x, &y, &mod);
ns = s;
while (n--) {
char name[25];
int p;
scanf("%s request %d pages", name, &p);
while (p > s) {
printf("%d pages for %s\n", s, name);
s = (ns * x + y) % mod;
ns = s;
}
printf("%d pages for %s\n", p, name);
s -= p;
}
printf("\n");
}
return 0;
}