POJ2234 Matches Game

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powerx_yc

最新推荐文章于 2018-07-21 11:34:24 发布

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原文链接：http://www.cnblogs.com/zjushuiping/archive/2012/07/22/2603714.html

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本文探讨了经典的尼姆博奕游戏，介绍其核心策略并提供了解决方案。通过输入多个堆叠的筹码数量，输出玩家是否能赢得游戏。

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Matches Game

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6537		Accepted: 3739

Description

Here is a simple game. In this game, there are several piles of matches and two players. The two player play in turn. In each turn, one can choose a pile and take away arbitrary number of matches from the pile (Of course the number of matches, which is taken away, cannot be zero and cannot be larger than the number of matches in the chosen pile). If after a player’s turn, there is no match left, the player is the winner. Suppose that the two players are all very clear. Your job is to tell whether the player who plays first can win the game or not.

Input

The input consists of several lines, and in each line there is a test case. At the beginning of a line, there is an integer M (1 <= M <=20), which is the number of piles. Then comes M positive integers, which are not larger than 10000000. These M integers represent the number of matches in each pile.

Output

For each test case, output "Yes" in a single line, if the player who play first will win, otherwise output "No".

Sample Input

2 45 45
3 3 6 9

Sample Output

No
Yes

Source

POJ Monthly,readchild

思路：本题是经典的尼姆博奕（Nimm Game），只要按照尼姆博奕（Nimm Game）的性质，找到必败态和必胜态，进行就OK了。

 1 #include <cstdlib>
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <cmath>
 6 #include <map>
 7 #include <string>
 8 
 9 #define MAXINT 99999999
10 
11 using namespace std;
12 
13 
14 
15 int main(int argc, char *argv[])
16 {int n;
17  int i;
18 
19 while(scanf("%d",&n)!=EOF)
20 {
21                           int sum=0;
22                           int temp;
23                           while(n--)
24                           {
25                                     scanf("%d",&temp);
26                                     sum^=temp;
27                           }
28                           
29                           if(sum==0)
30                           printf("No\n");
31                           else
32                           printf("Yes\n");
33 }
34 
35 
36                                     
37                     
38     
39     //system("PAUSE");
40     return EXIT_SUCCESS;
41 }