Matches Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7567   Accepted: 4327 Description Here is a simple game. In this game, there are several piles of matches and two players. The two player play in turn. In each turn, one can ch…

这道题也是一个博弈论 根据一个性质 对于\( Nim \)游戏,即双方可以任取石子的游戏,\( SG(x) = x \) 所以直接读入后异或起来输出就好了 代码 #include <cstdio> #include <algorithm> #include <cstring> using namespace std; int m; int main(){ while(scanf("%d",&m)!=EOF){ ,mid; ;i<=m;i…

Description Here is a simple game. In this game, there are several piles of matches and two players. The two player play in turn. In each turn, one can choose a pile and take away arbitrary number of matches from the pile (Of course the number of mat…

题目链接: https://cn.vjudge.net/problem/POJ-2234 题目描述: Here is a simple game. In this game, there are several piles of matches and two players. The two player play in turn. In each turn, one can choose a pile and take away arbitrary number of matches fro…

传送门 #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int main() { int n; while(~scanf("%d",&n)) { ; ;i<n;i++) { scanf("%d",&a); res^=a; } if(!res)//T prin…

题目大意:尼姆博弈,判断是否先手必胜. 题目思路: 尼姆博弈:有n堆各a[]个物品,两个人轮流从某一堆取任意多的物品,规定每次至少取一个,多者不限,最后取光者得胜. 获胜规则:ans=(a[1]^a[2] --^a[n]),若ans==0则后手必胜,否则先手必胜. #include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<stdio.h>…

http://poj.org/problem?id=2234 博弈论真是博大精深orz 首先我们仔细分析很容易分析出来,当只有一堆的时候,先手必胜:两堆并且相同的时候,先手必败,反之必胜. 根据博弈论的知识(论文 张一飞:<由感性认识到理性认识——透析一类搏弈游戏的解答过程>) 局面可以分解,且结果可以合并. 局面均是先手 当子局面是 胜 和 败,那么局面则为胜 当子局面是 败 和 胜,那么局面则为胜 当子局面是 败 和 败,那么局面则为败 当子局面为 胜 和 胜,那么局面为不确定 而这些性质…

思路:首先用Tarjan算法找出树中的环,环为奇数变为边,为偶数变为点. 之后用博弈论的知识:某点的SG值等于子节点+1后的异或和. 代码如下: #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<vector> #include<cstring> using namespace std; int ans; vector<…

[题目链接] http://poj.org/problem?id=2975 [题目大意] 问在传统的nim游戏中先手必胜策略的数量 [题解] 设sg=a1^a1^a3^a4^………^an,当sg为0时为必败态, 因此先手只需改变一个aj,让其减少m,使得sg^aj^(aj-m)=0即可让对手处于必败态, 即先手必胜策略,因为异或为0的两个数相同,所以sg^aj=aj-m, 即m=aj-sg^aj,因为m大于0,所以aj>sg^aj,至此我们就得到了必胜策略的重要条件 [代码] #include…

[题目链接] http://poj.org/problem?id=2068 [题目大意] 给出两队人,交叉放置围成一圈,每个人能取的石子数有个上限,各不相同 轮流取石头,取到最后一块石头的队伍算输,问哪个队伍能赢 [题解] 用dp[i][j]记录第i个人取石头时候还有j个石头的状态, 显然j==0时候为必胜态,我们对每个状态搜索后继状态,如果能导向必败态则为必胜态, 否则必败,记忆化搜索即可. [代码] #include <cstdio> #include <cstring> us…