codeforces 604 E. Lieges of Legendre (sg函数)

codeforces 604 E. Lieges of Legendre (sg函数)

标签： algorithm博弈

2015-12-06 21:29  378人阅读  评论(0)  收藏  举报

  分类：

博弈（5） 

版权声明：by whai

E. Lieges of Legendre

time limit per test

 2 seconds

memory limit per test

 256 megabytes

input

 standard input

output

 standard output

Kevin and Nicky Sun have invented a new game called Lieges of Legendre. In this game, two players take turns modifying the game state with Kevin moving first. Initially, the game is set up so that there are n piles of cows, with the i-th pile containing ai cows. During each player's turn, that player calls upon the power of Sunlight, and uses it to either:

1. Remove a single cow from a chosen non-empty pile.
2. Choose a pile of cows with even size 2·x (x > 0), and replace it with k piles of x cows each.

The player who removes the last cow wins. Given n, k, and a sequence a1, a2, ..., an, help Kevin and Nicky find the winner, given that both sides play in optimal way.

Input

The first line of the input contains two space-separated integers n and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ 109).

The second line contains n integers, a1, a2, ... an (1 ≤ ai ≤ 109) describing the initial state of the game.

Output

Output the name of the winning player, either "Kevin" or "Nicky" (without quotes).

Sample test(s)

input

2 1
3 4

output

Kevin

input

1 2
3

output

Nicky

Note

In the second sample, Nicky can win in the following way: Kevin moves first and is forced to remove a cow, so the pile contains two cows after his move. Next, Nicky replaces this pile of size 2 with two piles of size 1. So the game state is now two piles of size 1. Kevin then removes one of the remaining cows and Nicky wins by removing the other.

思路：
挺常规的一道sg函数的题目，打个表就清楚多了

[cpp]  view plain  copy

1. /*====================================================== 
2. # Author: whai 
3. # Last modified: 2015-12-04 14:22 
4. # Filename: e.cpp 
5. ======================================================*/  
6. #include <iostream>  
7. #include <cstdio>  
8. #include <vector>  
9. #include <algorithm>  
10. #include <cstring>  
11. #include <string>  
12. #include <cmath>  
13. #include <set>  
14. #include <map>  
15. #include <queue>  
16. #include <stack>  
17.   
18. using namespace std;  
19.   
20. #define LL __int64  
21. #define PB push_back  
22. #define P pair<int, int>  
23. #define X first  
24. #define Y second  
25.   
26. const int N = 1e5 + 5;  
27.   
28. int a[N];  
29.   
30. int get_sg(int x, int k) {  
31.     bool vis[3] = {0};  
32.     if(x == 0) return 0;  
33.     if(x == 1) return 1;  
34.     if(x >= 5 && x % 2 == 1) return 0;  
35.     vis[get_sg(x - 1, k)] = 1;  
36.     if(x % 2 == 0) {  
37.         if(k % 2 == 0) vis[0] = 1;  
38.         else vis[get_sg(x / 2, k)] = 1;  
39.     }  
40.     int ret = 0;  
41.     while(vis[ret]) ++ ret;  
42.     return ret;  
43. }  
44.   
45. void gao(int n, int k) {  
46.     int ans = 0;  
47.   
48.     for(int i = 0; i < n; ++i) {  
49.         ans ^= get_sg(a[i], k);  
50.     }  
51.       
52.     if(ans == 0) {  
53.         puts("Nicky");  
54.     } else {  
55.         puts("Kevin");  
56.     }  
57. }  
58.   
59. /*int sg[105]; 
60. bool vis[105]; 
61.  
62. void bf(int n, int k) { 
63.     sg[0] = 0; 
64.     for(int i = 1; i <= 100; ++i) { 
65.         memset(vis, 0, sizeof(vis)); 
66.         vis[sg[i - 1]] = 1; 
67.         if(i % 2 == 0) { 
68.             if(k % 2 == 0) vis[0] = 1; 
69.             else vis[sg[i / 2]] = 1; 
70.         } 
71.         int tmp = 0; 
72.         while(vis[tmp]) ++tmp; 
73.         sg[i] = tmp; 
74.     } 
75.     for(int i = 0; i <= 100; ++i) { 
76.         cout<<sg[i]<<' '; 
77.     } 
78.     cout<<endl; 
79.     for(int i = 0; i <= 100; ++i) { 
80.         cout<<get_sg(i, k)<<' '; 
81.     } 
82.     cout<<endl; 
83. }*/  
84.   
85. int main() {  
86.     int n, k;  
87.   
88.     while(scanf("%d%d", &n, &k) != EOF) {  
89.         //bf(n, k);  
90.         for(int i = 0; i < n; ++i) {  
91.             scanf("%d", &a[i]);  
92.         }  
93.         gao(n, k);  
94.     }  
95.   
96.     return 0;  
97. }