codeforces 768E Game of Stones

这是一个关于博弈论的游戏问题,玩家从n堆石头中按特定规则移除石头,每堆石头只能移除有限次。先手玩家(Sam)已经开始了游戏,后手玩家(Jon)想知道他是否有可能获胜。解决方案涉及到求解每堆石头的最大移除次数,并通过优化策略来确定胜利条件。如果Jon可以获胜,则输出'YES',否则输出'NO'。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

E. Game of Stones
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Sam has been teaching Jon the Game of Stones to sharpen his mind and help him devise a strategy to fight the white walkers. The rules of this game are quite simple:

The game starts with n piles of stones indexed from 1 to n. The i-th pile contains si stones.
The players make their moves alternatively. A move is considered as removal of some number of stones from a pile. Removal of 0 stones does not count as a move.
The player who is unable to make a move loses.
Now Jon believes that he is ready for battle, but Sam does not think so. To prove his argument, Sam suggested that they play a modified version of the game.

In this modified version, no move can be made more than once on a pile. For example, if 4 stones are removed from a pile, 4 stones cannot be removed from that pile again.

Sam sets up the

评论 2
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值