简单博弈问题(巴什博弈-Bash Game) 巴什博弈:只有一堆n个物品,两个人轮流从这对物品中取物,规定每次至少取一个,最多取m个,最后取光着得胜. 很容易想到当n%(m+1)!＝0时,先取者必胜,第一次先拿走n%(m+1)个,以后每个回合都保持两人拿走的物品总和为m+1即可. 个,谁能报到100者胜. 此题可以把每堆石头的取法看作是一个BashGame,这样只需将每组石头按照BashGame的取法判断,然后将n堆石头做异或, 如果异或的结果不为0,则老师获胜,否则Agrael取胜. 代码如下…

典型的尼姆博弈,在n对石子中,告诉你每堆的数目和每次从该堆最多可以取的数目,求最终谁将其取完. 题解:SG(i)=mi%(li+1),求异或值即可. #include <cstdio> int main(){ int T,i,n,SG,m,l; scanf("%d",&T); while(T--){ scanf("%d",&n); SG=0; for(i=1;i<=n;i++)scanf("%d%d",&…

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HDU 5794 - A Simple Chess题意: 马(象棋)初始位置在(1,1), 现在要走到(n,m), 问有几种走法 棋盘上有r个障碍物, 该位置不能走, 并规定只能走右下方 数据范围: ( 1 ≤ n,m ≤10^18, 0 ≤ r ≤100 )    分析: 分析不存在障碍物时的走法规律:                      (1,1)                  1              (2,3) (3,2)                      1  …

HDU 4974 A simple water problem pid=4974" target="_blank" style="">题目链接 签到题,非常easy贪心得到答案是(sum + 1) / 2和ai最大值的最大值 代码: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N =…

pid=4972" target="_blank" style="">题目链接:hdu 4972 A simple dynamic programming problem 题目大意:两支球队进行篮球比赛,每进一次球后更新比分牌,比分牌的计数方法是记录两队比分差的绝对值,每次进球的分可能是1,2,3分. 给定比赛中的计分情况.问说最后比分有多少种情况. 解题思路:分类讨论: 相邻计分为1-2或者2-1的时候,会相应有两种的的分情况 相邻计分之差大于3或…

A Simple Game Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)Total Submission(s): 1487    Accepted Submission(s): 939 Problem Description Agrael likes play a simple game with his friend Animal during the classes. I…

这是由n个巴什博奕的游戏合成的组合游戏. 对于一个有m个石子,每次至多取l个的巴什博奕,这个状态的SG函数值为m % (l + 1). 然后根据SG定理,合成游戏的SG函数就是各个子游戏SG函数值的异或和. #include <cstdio> int main() { int T, n; scanf("%d", &T); while(T--) { scanf("%d", &n); , m, l; ); } printf("%s\…

A Simple Chess 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5794 Description There is a n×m board, a chess want to go to the position (n,m) from the position (1,1). The chess is able to go to position (x2,y2) from the position (x1,y1), only and if…

A Simple Nim 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5795 Description Two players take turns picking candies from n heaps,the player who picks the last one will win the game.On each turn they can pick any number of candies which come from the…