p.MsoNormal { margin: 0pt; margin-bottom: .0001pt; text-align: justify; font-family: Calibri; font-size: 10.5000pt } h1 { margin-top: 5.0000pt; margin-bottom: 5.0000pt; text-align: center; font-family: 宋体; color: rgb(26,92,200); font-weight: bold; fo…

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.…

Nim or not Nim? Problem Description Nim is a two-player mathematic game of strategy in which players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provide…

Nim or not Nim? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 858    Accepted Submission(s): 412 Problem Description Nim is a two-player mathematic game of strategy in which players take turns…

[HDU3032]Nim or not Nim?(博弈论) 题面 HDU 题解 \(Multi-SG\)模板题 #include<iostream> #include<cstdio> using namespace std; inline int read() { int x=0;bool t=false;char ch=getchar(); while((ch<'0'||ch>'9')&&ch!='-')ch=getchar(); if(ch=='-'…

Nim or not Nim? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 623    Accepted Submission(s): 288 Problem Description Nim is a two-player mathematic game of strategy in which players take turns…

Nim or not Nim? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1056    Accepted Submission(s): 523 Problem Description Nim is a two-player mathematic game of strategy in which players take turn…

Nim or not Nim? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 613    Accepted Submission(s): 282 Problem Description Nim is a two-player mathematic game of strategy in which players take turns…

Nim or not Nim? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2513    Accepted Submission(s): 1300 Problem Description Nim is a two-player mathematic game of strategy in which players take tur…

Nim or not Nim? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3759    Accepted Submission(s): 1937 Problem Description Nim is a two-player mathematic game of strategy in which players take tur…

NIM游戏,NIM游戏变形,威佐夫博弈以及巴什博奕总结 经典NIM游戏: 一共有N堆石子,编号1..n,第i堆中有个a[i]个石子. 每一次操作Alice和Bob可以从任意一堆石子中取出任意数量的石子,至少取一颗,至多取出这一堆剩下的所有石子. 两个人轮流行动,取走最后一个的人胜利.Alice为先手. 我们定义: P:表示当前局面下先手必败 N:表示当前局面下先手必胜 N,P状态的转移满足如下性质: 1.合法操作集合为空的局面为P 2.可以移动到P的局面为N,这个很好理解,以为只要能转换到P局面…

每次只能从取集合S中个数的物品,其他和普通Nim游戏相同 预处理出每种物品堆的sg值,然后直接xor一下,xor-sum>0即必胜 #include <set> #include <map> #include <cmath> #include <queue> #include <vector> #include <cstdio> #include <cstdlib> #include <cstring>…

加强版的NIM游戏,多了一个操作,可以将一堆石子分成两堆非空的. 数据范围太大,打出sg表后找规律. # include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # inc…

学习Nim语言 nim 语法上类似python ,是一门静态编译型语言,nim 使用空格缩进标示语句块的开始和结束, 喜欢python风格的程序员应该也会很容易适应和喜欢nim的风格. nim语言官方网站 nim-lang.org #最简单的nim 程序是这样的. echo "hi nim" nim语言中文教程下载:http://files.cnblogs.com/files/gayhub/%E5%AD%A6%E4%B9%A0Nim%E8%AF%AD%E8%A8%80.rar…

博弈的题目,打表找规律还是相当有用的一个技巧. 这个游戏在原始的Nim游戏基础上又新加了一个操作,就是游戏者可以将一堆分成两堆. 这个SG函数值是多少并不明显,还是用记忆化搜索的方式打个表,规律就相当显然了. #include <cstdio> #include <cstring> ; ]; ]; int mex(int v) { ) return sg[v]; memset(vis, false, sizeof(vis)); ; i < v; i++) vis[mex(i)…

 这题是Lasker’s Nim. Clearly the Sprague-Grundy function for the one-pile game satisfies g(0) = 0 and g(1) = 1. The followers of 2 are 0, 1 and (1,1), with respective Sprague-Grundy values of 0, 1, and 1⊕1 = 0. Hence, g(2) = 2. The followers of 3 are 0,…

意甲冠军:经典Nim游戏转换,给你n礧pi,每个堆栈有pi石头, Alice和Bob轮流石头,意一堆中拿走随意个石子,也能够将某一堆石子分成两个小堆 (每堆石子个数必须不能为0).先拿完者获胜 思路:求SG函数后找规律. SG函数定义及求法:点击打开链接 #include<cstdio> #include<stdlib.h> #include<string.h> #include<string> #include<map> #include<…

传送门 题意: nim游戏,多了一种操作:将一堆分成两堆 Multi-SG游戏规定,在符合拓扑原则的前提下,一个单一游戏的后继可以为多个单一游戏. 仍然可以使用$SG$函数,分成多个游戏的后继$SG$值为多个游戏的异或和 然后本题规模很大,手动打一下表,发现$\mod 4=3$ 时$sg(x)=x+1$,$\mod 4=0$ 时$sg(x)=x-1$,其他不变 #include <iostream> #include <cstdio> #include <cstring>…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2508    Accepted Submission(s): 1297 Problem Description Nim is a two-player mathematic game of strategy in which players take turns removing obje…

题目链接 \(Description\) 有多堆石子, 每次可以将任意一堆拿走任意个或者将这一堆分成非空的两堆, 拿走最后一颗石子的人胜利.问谁会获得胜利. \(Solution\) Lasker's Nim游戏 具体见这 这个问题可以用SG函数来解决. 首先,操作(1)和Nim游戏没什么区别,对于一个石子数为k的点来说,后继可以为0-k-1. 而操作(2)实际上是把一个游戏分成了两个游戏,这两个游戏的和为两个子游戏的SG函数值的异或. 而求某一个点的SG函数要利用它的后继,它的后继就应该为 当…

翻译 你正在和你的朋友们玩以下这个Nim游戏:桌子上有一堆石头.每次你从中去掉1-3个.谁消除掉最后一个石头即为赢家.你在取出石头的第一轮. 你们中的每个人都有着聪明的头脑和绝佳的策略.写一个函数来确定对于给定的数字是否你能够赢得这场比赛. 比如,假设堆中有4个石头,那么你永远也无法赢得比赛:不管你移除了1.2或3个石头,最后一个石头都会被你的朋友所移除. 原文 You are playing the following Nim Game with your friend: There is a…

You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove th…

http://blog.csdn.net/qiankun1993/article/details/6765688 NIM 游戏 重点结论:对于一个Nim游戏的局面(a1,a2,...,an),它是P-position(先者有利)当且仅当a1^a2^...^an=0,其中^表示位异或(xor)运算. Nim游戏是博弈论中最经典的模型(之一?),它又有着十分简单的规则和无比优美的结论,由这个游戏开始了解博弈论恐怕是最合适不过了. Nim游戏是组合游戏(Combinatorial Games)的一种,…

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2016 Accepted Submission(s): 1048 Problem Description Nim is a two-player mathematic game of strategy in which players take turns removing objects f…

Nim is a two-player mathematic game of strategy in which players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap.…

题目链接 暴力出来,竟然眼花了以为sg(i) = i啊....看表要认真啊!!! #include <cstdio> #include <cstring> #include <iostream> using namespace std; #define LL __int64 ]; int sg(int x) { ],temp,i; ) return dp[x]; memset(flag,,sizeof(flag)); ;i <= x;i ++) { temp =…

题意: 有n堆石子,alice先取,每次可以选择拿走一堆石子中的1~x(该堆石子总数) , 也可以选择将这堆石子分成任意的两堆.alice与bob轮流取,取走最后一个石子的人胜利. 思路: 因为数的范围比较大,所以最好通过SG打表的结果找出规律在解. 打表代码 #include<cstdio> #include<cstring> ]; int find(int x) { ) return sg[x]; ]= {}; ; i<x; i++) { mex[find(i)]=;//…

题目链接 给出n堆石子, 每次可以取一堆中的任意x个(x>=1), 或者将一堆石子拆成两堆, 取到最后一堆的胜. 这个题需要打sg表找规律, 打表程序看代码. #include<bits/stdc++.h> using namespace std; #define pb(x) push_back(x) #define ll long long #define mk(x, y) make_pair(x, y) #define lson l, m, rt<<1 #define m…

题目链接 #include<iostream> #include<cstdio> using namespace std; int main() { ]; int sum,cnt; while(scanf("%d",&n)&&n) { sum=,cnt=; ;i<=n;i++) { scanf("%d",&k[i]); sum^=k[i]; } ) { printf("0\n"); c…

题意:有N堆石子,每堆有s[i]个,Alice和Bob两人轮流取石子,可以从一堆中取任意多的石子,也可以把一堆石子分成两小堆 Alice先取,问谁能获胜 思路:首先观察这道题的数据范围  1 ≤ N ≤ 10^6, 1 ≤ [Si] ≤ 2^31 - 1,很明显数据量太大,所以只能通过打表找规律 打表后发现,如果x%4==0 sg[x]=x-1 ;如果 x%4==3 sg[x]=x+1;如果 其他情况 sg[x]=x; 代码: 打表代码: #include <iostream> #includ…