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\n", 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\n", 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\n", 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 */

 

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