kiki's game Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 40000/10000 K (Java/Others)Total Submission(s): 10208    Accepted Submission(s): 6161 Problem Description Recently kiki has nothing to do. While she is bored, an idea appears in his…

http://acm.hdu.edu.cn/showproblem.php?pid=2147 kiki's game Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 40000/10000 K (Java/Others)Total Submission(s): 12133    Accepted Submission(s): 7393 Problem Description Recently kiki has nothing to…

这个是纳什博弈?不知道怎么看的 依据PN图,从左下角開始推 左下角P 最后一行都是PNPNPN 第一列都是 P N P N P 完了填完即可了 #include<cstdio> int main() { int n,m; while(scanf("%d%d",&n,&m)&&n&&m){ if(n&1&&m&1) printf("What a pity!\n"); else…

Recently kiki has nothing to do. While she is bored, an idea appears in his mind, she just playes the checkerboard game.The size of the chesserboard is n*m.First of all, a coin is placed in the top right corner(1,m). Each time one people can move the…

Problem Description: 最近kiki无事可做,于是他想玩棋盘游戏.棋盘的大小是n * m.首先,棋子放置在右上角(1,m). 每次可以将棋子向左方,下方或左下方移动一个位置.当移动到(n,1)时就无法移动,无法移动的人失败. kiki和bibi一起玩.游戏总是从kiki开始. 如果两者都完美发挥,谁会赢得比赛? Input: 输入包含多个测试用例. 每行包含两个整数n,m(0 <n,m <= 2000). 当n = 0和m = 0时输入终. Output: 如果kiki赢,输…

Rabbit and Grass Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2145    Accepted Submission(s): 1622 Problem Description 大学时光是浪漫的,女生是浪漫的,圣诞更是浪漫的,但是Rabbit和Grass这两个大学女生在今年的圣诞节却表现得一点都不浪漫:不去逛商场,不去逛…

Life Winner Bo Problem Description   Bo is a "Life Winner".He likes playing chessboard games with his girlfriend G. The size of the chessboard is N×M.The top left corner is numbered(1,1) and the lower right corner is numberd (N,M). For each game…

1.HDU 2509  2.题意:n堆苹果,两个人轮流,每次从一堆中取连续的多个,至少取一个,最后取光者败. 3.总结:Nim博弈的变形,还是不知道怎么分析,,,,看了大牛的博客. 传送门 首先给出结论:先手胜当且仅当(1)所有堆石子数都为1且游戏的SG值为0,(2)存在某堆石子数大于1且游戏的SG值不为0.证明:(1)若所有堆石子数都为1且SG值为0,则共有偶数堆石子,故先手胜.(2) i)只有一堆石子数大于1时,我们总可以对该堆石子操作,使操作后石子堆数为奇数且所有堆得石子数均为1 ii)有…

1.HDU 1907 2.题意:n堆糖,两人轮流,每次从任意一堆中至少取一个,最后取光者输. 3.总结:有点变形的Nim,还是不太明白,盗用一下学长的分析吧 传送门 分析:经典的Nim博弈的一点变形.设糖果数为1的叫孤独堆,糖果数大于1的叫充裕堆,设状态S0:a1^a2^..an!=0&&充裕堆=0,则先手必败(奇数个为1的堆,先手必败).S1:充裕堆=1,则先手必胜(若剩下的n-1个孤独堆个数为奇数个,那么将那个充裕堆全部拿掉,否则将那个充裕堆拿得只剩一个,这样的话先手必胜).T0:a1…

题目链接: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1072 题意: 中文题诶~ 思路: 博弈套路是有的, 找np局面, 然而要找还得靠yy, 之前做了一个bash模板, 自己还能yy出来, nim就有点变态了, 看题解才想到, 至于这道题, 诶, 找到了一半规律, 还是没发现那个黄金比关系, 还是yy能力不行~ 这里给出一个不错的题解: http://blog.csdn.net/y990041769/artic…