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…

Charm Bracelet Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 38909   Accepted: 16862 Description Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible fro…

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…

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…

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…

Brave Game Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 14395 Accepted Submission(s): 9749 Problem Description 十年前读大学的时候,中国每年都要从国外引进一些电影大片,其中有一部电影就叫<勇敢者的游戏>(英文名称:Zathura),一直到现在,我依然对于电影中的部分电脑特技…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 43168   Accepted: 20276 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh…

题意:给一个n和m,表示n个房间,m次操作,操作类型有2种,一种把求连续未租出的房间数有d个的最小的最左边的房间号,另一个操作时把从x到x+d-1的房间号收回. 建立线段树,值为1表示未租出,0为租出,线段树实现区间合并,必须记录区间左右端点的状态,这样才能实现区间合并,用lsum记录以左端点开始的最长连续可租房间数, rsum记录右端点开始的最长连续可组房间数,msum记录该区间的最长连续可租房间数,那么,区间合并时,根节点的更新 msum = max(rsum[rt<<1]+lsum[rt…