题目传送门
1 /*
2 题意：每支队伍需求打印机打印n张纸，当打印纸数累计到s时，打印机崩溃，打印出当前打印的纸数，s更新为(s*x+y)%mod
3 累计数清空为0，重新累计
4 模拟简单题：关键看懂题意
5 注意：打印机一张一张纸打印，当某支队伍打印完正好累计到s时，要输出0，坑点！
6 */
7 #include <cstdio>
8 #include <cmath>
9 #include <algorithm>
10 #include <iostream>
11 #include <cstring>
12 using namespace std;
13
14 int cnt[110];
15 char ss[110][22];
16
17 int main(void) //J Contest Print Server
18 {
19 //freopen ("J.txt", "r", stdin);
20
21 int t;
22 while (scanf ("%d", &t) == 1)
23 {
24 while (t--)
25 {
26 int n, s, x, y, mod;
27 scanf ("%d%d%d%d%d", &n, &s, &x, &y, &mod);
28
29 for (int i=1; i<=n; ++i)
30 scanf ("%s request %d pages", &ss[i], &cnt[i]);
31
32 int num = 0, j;
33 int ok = 0;
34 for (int i=1; i<=n; ++i)
35 {
36 if (ok == 1) printf ("%d pages for %s
", 0, ss[i]);
37 ok = 0;
38 for (j=1; j<=cnt[i]; ++j)
39 {
40 num++;
41 if (num == s)
42 {
43 printf ("%d pages for %s
", j, ss[i]);
44 s = (s * x + y) % mod;
45 if (s == 0) s = (s * x + y) % mod;
46 num = 0; break;
47 }
48 }
49 if (j < cnt[i]) i--;
50 else if (j == cnt[i]) ok = 1;
51 else printf ("%d pages for %s
", cnt[i], ss[i]);
52 }
53 if (t) puts ("");
54 }
55
56 }
57
58
59 return 0;
60 }
61
62 /*
63 1 pages for Team1
64 5 pages for Team2
65 1 pages for Team3
66
67 1 pages for Team1
68 3 pages for Team2
69 5 pages for Team2
70 1 pages for Team3
71 */