UVa Problem Solution: 847 - A Multiplication Game

We can easily observe that if the drawn number is in range [2, 9], Stan will win, and if it is in range [10, 18], Ollie will win. Suppose that one of Span's win range is [m, n], it can be extend to the next win range of [2*m+1, 2*9*n)]. Thus, the range width expands by a factor of 18. Then we can reduce the drawn number to the initial win ranges to check who will win the game.

Code:

1. /*************************************************************************
2.  * Copyright (C) 2008 by liukaipeng                                      *
3.  * liukaipeng at gmail dot com                                           *
4.  *************************************************************************/
6. /* @JUDGE_ID 00000 847 C++ "A Multiplication Game" */
8. #include <algorithm>
9. #include <cstdio>
10. #include <cstring>
11. #include <deque>
12. #include <fstream>
13. #include <iostream>
14. #include <list>
15. #include <map>
16. #include <queue>
17. #include <set>
18. #include <stack>
19. #include <string>
20. #include <vector>
22. using namespace std;
23.           
24. int main(int argc, char *argv[])
25. {
26. #ifndef ONLINE_JUDGE
27.   freopen((string(argv[0]) + ".in").c_str(), "r", stdin);
28.   freopen((string(argv[0]) + ".out").c_str(), "w", stdout);
29. #endif
31.   for (int n; scanf("%d", &n) == 1; ) { 
32.     for (; n > 18; n = (n + 17) / 18) {}
33.     puts(n <= 9 ? "Stan wins." : "Ollie wins.");
34.   }
36.   return 0;
37. }