POJ2234 Matches Game
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 }