S-Nim Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5637    Accepted Submission(s): 2414 Problem Description Arthur and his sister Caroll have been playing a game called Nim for some time now.…

S-Nim Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7262    Accepted Submission(s): 3074 Problem Description Arthur and his sister Caroll have been playing a game called Nim for some time now.…

S-Nim Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3077    Accepted Submission(s): 1361 Problem Description Arthur and his sister Caroll have been playing a game called Nim for some time now…

Fibonacci again and again Problem Description   任何一个大学生对菲波那契数列(Fibonacci numbers)应该都不会陌生,它是这样定义的:F(1)=1;F(2)=2;F(n)=F(n-1)+F(n-2)(n>=3);所以,1,2,3,5,8,13……就是菲波那契数列.在HDOJ上有不少相关的题目,比如1005 Fibonacci again就是曾经的浙江省赛题.今天,又一个关于Fibonacci的题目出现了,它是一个小游戏,定义如下:1. …

John Problem Description   Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to eat several M&Ms of the same color. Then his opponent has to make a turn. A…

S-Nim HDU 1536 博弈 sg函数 题意 首先输入K,表示一个集合的大小,之后输入集合,表示对于这对石子只能去这个集合中的元素的个数,之后输入 一个m表示接下来对于这个集合要进行m次询问,之后m行,每行输入一个n表示有n个堆,每堆有n1个石子,问这一行所表示的状态是赢还是输,如果赢输入W否则L. 解题思路 如果没有每次取石子个数的限制的话,那么仅仅需要把每堆石子的个数进行异或运算即可,如果结果不是1,那么先手赢,反之后手赢. 但是这里对每次取石子的个数进行了限制,每次只能从几个数中进行…

Nim or not Nim? Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 3032 Description Nim is a two-player mathematic game of strategy in which players take turns removing objects from distinct heaps.…

题意:每次可以选择n种操作,玩m次,问谁必胜.c堆,每堆数量告诉. 题意:sg—NIM系列博弈模板题 把每堆看成一个点,求该点的sg值,异或每堆sg值. 将多维转化成一维,性质与原始NIM博弈一样. // #pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <cstdio> #include <cstring> #include &l…

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…

题目来源:http://acm.hdu.edu.cn/showproblem.php?pid=1730 Nim博弈为:n堆石子,每个人可以在任意一堆中取任意数量的石子 n个数异或值为0就后手赢,否则先手赢 将这题转化成Nim游戏 可以在任意一行中移动任意距离,可以向左或右,但是仔细观察发现,其实只能接近对方棋子,如果你远离对方棋子,对方可以接近你相同距离 和nim相似的是,不能不移,所以两个棋子的距离差就是SG值 #include<cstdio> #include<iostream>…