UVa Problem Solution: 10050 - Hartals

At the first look, the hartals can be counted with the principle of inclusion and exclusion. However, this may cause the calculations to expand to O(2^numberof(parties)). In another way, simply simulation will also do the job.

Code:

1. /*************************************************************************
2.  * Copyright (C) 2008 by liukaipeng                                      *
3.  * liukaipeng at gmail dot com                                           *
4.  *************************************************************************/
6. /* @JUDGE_ID 00000 10050 C "Hartals" */
8. #include <stdio.h>
9. #include <stdlib.h>
10. #include <string.h>
11. #include <strings.h>
12.                
13. int lost(int hp[], int nparty, int nday)
14. {
15.   int hartals[3651] = {0};
16.   int nlost = 0;
17.   int i, j;
18.   for (i = 0; i < nparty; ++i) {
19.     for (j = hp[i]; j <= nday; j += hp[i]) {
20.       int weekday = j % 7;
21.       if (weekday != 6 && weekday != 0 && !hartals[j]) {
22.         hartals[j] = 1;
23.         nlost += 1;
24.       }
25.     }
26.   }
27.   return nlost;
28. }
30. int main(int argc, char *argv[])
31. {
32. #ifndef ONLINE_JUDGE
33.   char in[256];
34.   char out[256];
35.   strcpy(in, argv[0]);
36.   strcat(in, ".in");
37.   freopen(in, "r", stdin);
38.   strcpy(out, argv[0]);
39.   strcat(out, ".out");
40.   freopen(out, "w", stdout);
41. #endif
43.   int ncase;
44.   scanf("%d/n", &ncase);
45.   while (ncase-- > 0) {
46.     int nday;
47.     int nparty, i;
48.     int hp[100];
49.     scanf("%d/n", &nday);
50.     scanf("%d/n", &nparty);
51.     for (i = 0; i < nparty; ++i)
52.       scanf("%d/n", &hp[i]);
53.     printf("%d/n", lost(hp, nparty, nday));
54.   }
55.   return 0;
56. }
57.