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>…

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…

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…

传送门 #include<iostream> #include<cstdio> #include<cstring> using namespace std; int main() { int n; while(~scanf("%d",&n)) { ; ,g=; ;i<n;i++) { scanf("%d",&a); ) c++; ) g++; res^=a; } if(!res) { )//T2 printf…

传送门 #include<iostream> #include<cstdio> #include<cstring> using namespace std; int main() { int t; scanf("%d",&t); while(t--) { ; ,g=;//¹Âµ¥¶Ñ¡¢³äÔ£¶Ñ scanf("%d",&n); ;i<n;i++) { scanf("%d",&a…

好久之前看的sg函数了 好像就记住一个nim博弈qwq 第一次啊看的时候很迷,现在感觉可以了qwq 首先我们来看一个其他的游戏.(以下游戏只有两个人参与,且足够聪明) 两个人在一张圆形的桌子上放等大的盘子,最后一个无法放盘子的人输掉比赛 很显然,先手必胜. 为什么? 第一个人可以将盘子放在桌子的中心. 然后只要第二个人可以放置盘子,我们就在其中心对称的位置上放盘子. 如此,只要后手可以放,我先手就一定能放 可以看出,有时候如果处于先手必胜的状态时,模仿对手的策略不妨是个好方法.这可以保证如果游戏…

Georgia and Bob Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9363   Accepted: 3055 Description Georgia and Bob decide to play a self-invented game. They draw a row of grids on paper, number the grids from left to right by 1, 2, 3, ...…