HDU 1564 Play a game （找规律博弈）
Play a game
Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1030Accepted Submission(s): 838
Problem Description
New Year is Coming!
ailyanlu is very happy today! and he is playing a chessboard game with 8600.
The size of the chessboard is n*n. A stone is placed in a corner square. They play alternatively with 8600 having the first move. Each time, player is allowed to move the stone to an unvisited neighbor square horizontally or vertically. The one who can't make a move will lose the game. If both play perfectly, who will win the game?
ailyanlu is very happy today! and he is playing a chessboard game with 8600.
The size of the chessboard is n*n. A stone is placed in a corner square. They play alternatively with 8600 having the first move. Each time, player is allowed to move the stone to an unvisited neighbor square horizontally or vertically. The one who can't make a move will lose the game. If both play perfectly, who will win the game?
Input
The input is a sequence of positive integers each in a separate line.
The integers are between 1 and 10000, inclusive,(means 1 <= n <= 10000) indicating the size of the chessboard. The end of the input is indicated by a zero.
The integers are between 1 and 10000, inclusive,(means 1 <= n <= 10000) indicating the size of the chessboard. The end of the input is indicated by a zero.
Output
Output the winner ("8600" or "ailyanlu") for each input line except the last zero.
No other characters should be inserted in the output.
No other characters should be inserted in the output.
Sample Input
2
0
Sample Output
8600
Author
ailyanlu
Source
Happy 2007
Recommend
8600
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从一个n*n的角落出发，每次移动到相邻的，而且没有经过的格子上。谁不能操作了谁输。
结论就是n为偶数，先手赢，奇数，后手赢。
可以看下图
S表示起点。
如果n为偶数，那么所有格子可以被2*1的砖块覆盖掉。
这样先手每次都移动到当前1*2的另外一块。先手必赢。
如果n为奇数。出了起始那个店，其余点都可以被覆盖。
所有后手赢
#include <string.h> #include <stdio.h> #include <algorithm> #include <iostream> using namespace std; int main() { int n; while(scanf("%d",&n)==1 &&n) { if(n%2==0)printf("8600\n"); else printf("ailyanlu\n"); } return 0; }优质内容筛选与推荐>>
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