POJ1704 Georgia and Bob Nim阶梯博弈

多GPU时代的虚拟内存革命:CUDA VMM API的跨设备协同设计哲学 本文深入探讨了NVIDIA CUDA VMM API在多GPU系统中的虚拟内存管理革命,解决了传统Peer Access的显存碎片化和同步开销问题。通过地址-内存解耦、按需映射和精细权限控制,VMM API为分布式训练提供了高效的内存协同方案,显著提升大模型训练性能。 阅读详情

我们将模型转化

如果我们看成这个棋子能向左移动多少距离的话,那么序列就成为了一组非负的序列

如果我们在这组序列中考虑某一个棋子向左移动一步,那么就会发现,这枚棋子向左能移动的距离-1,而右边i+1那枚棋子能移动的距离+1,那么就转化成了一个Nim的模型

我们可以把把某一颗石头向i+1位置移动一个,或者是把N位置的棋子丢弃一堆,不能操作就输

显然Nim博弈

Problem: 1704		User: BPM136
Memory: 704K		Time: 16MS
Language: G++		Result: Accepted

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#define LL long long
#define DB double 
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
#define efo(i,x) for(int i=last[x];i!=0;i=e[i].next)
using namespace std;
inline LL read()
{
	LL d=0,f=1;char s=getchar();
	while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
	while(s>='0'&&s<='9'){d=d*10+s-'0';s=getchar();}
	return d*f;
}
#define N 1005
int a[N];
int s[N];
int n,m;


int main()
{
	int T=read();
	while(T--){
		n=read();
		fo(i,1,n)a[i]=read();a[0]=0;
		sort(a+1,a+n+1);
		memset(s,0,sizeof(s));
		for(int i=n,j=1;i>=1;i--,j++){
			s[j]=a[i]-a[i-1]-1;
		}
		int k=0;
		for(int i=1;i<=n;i+=2){
			k^=s[i];
		}
		if(!k){
			cout<<"Bob will win"<<endl;
		}else{
			cout<<"Georgia will win"<<endl;
		}
	}
	return 0;
}


YOLO11全网首发优化:卷积魔改创新 | AAAI 2025 | 一种新颖的风车形卷积(PConv)符合微弱小目标的像素高斯空间分布,增强特征提取,显著增加接受野 我们提出了一种新颖的风车形卷积(PConv)作为骨干网络下层标准卷积的替代品。PConv能更好地符合微弱小目标的像素高斯空间分布,增强特征提取,显著增加接受野,且只引入最小的参数增加。 阅读详情

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POJ1704Georgia and Bob

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Fsss 790

poj 1704 Georgia and Bob 阶梯博弈

这还是一道博弈的题目,但是我没有思路,。开始我想的是把它转化为取石子的问题,直接就是将每一个放的棋子的位置然后再看它到自己最终的位置的长度,就将其抽象成第几堆的棋子的个数,后来发现这是错误的,也还是我对昨天的nim的理解太浅显了。。。对于奇异局势的理解不够好。其实对于每一个博弈的问题我最开始需要找到的是它的奇异局势,而不是最终的状态。这个题目的求解是这样的。就是两两组合,当两者相邻的时候就是奇异局

u013087645的专栏 856

poj1704 Georgia and Bob(阶梯博弈)

题目 有N个棋子,第i个在x轴上的ai位置, 如果棋子左边有空,可以不越过其它棋子向左移动, 特别地,最左边的棋子可以移动到1,然后就动不了了 不能移动的输,对于给定局面,问谁必胜 思路来源 https://blog.youkuaiyun.com/qkoqhh/article/details/80434749 题解 Nim游戏的变种 直观的来看, ①有一个棋子的时候,如果在1处则先手败,否...

小衣同学的博客 308

Georgia and Bob(POJ - 1704)

Georgia and Bob 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, ..., and place N chessmen on different grids,...

the blog of Lee 424

蓝书(算法竞赛进阶指南)刷题记录——POJ1704 Georgia and Bob(博弈论)

题目:POJ1704. 题目大意:给定1∗n1*n1∗n的棋盘,mmm颗棋子的坐标为a1,a2,...,ama_1,a_2,...,a_ma1​,a2​,...,am​.现在两人轮流操作,每次操作可以把一颗棋子往左移,但不能超过它左边的棋子或走出棋盘(初始坐标111),无法操作的人输,问是否先手必胜. 这貌似是博弈论中经典的模型,叫阶梯型博弈. 考虑这个模型是否可以转化为NIM博弈.考虑只有两颗棋...

hzk_cpp的博客 391

POJ 1704 Georgia and Bob(阶梯博弈)

POJ - 1704 博弈还是神奇啊,做题的时候又发现了什么了不得的东西,这道题一开始只是有点思路,做不出来,看了人家的题解才知道是阶梯博弈,这个博客讲的很好啊:阶梯博弈传送门。。。所谓阶梯博弈核心思想,我感觉也就是把非nim博弈的规则带来的影响通过下一步可以消除,从而实现nim规则的保留。。思路:将两个相邻石子看成一组,每一组中前一个和后一个的距离等效成nim中石子个数(相邻两组之间距离无影响)...

STOP 630

poj1704:Georgia and Bob(阶梯NIM)

传送门题意: 从左到右有一排石子,给出石子所在的位置。规定每个石子只能向左移动,且不能跨过前面的石子。最左边的石子最多只能移动到1位置。每次选择一个石子按规则向左移动,问先手是否能赢。题解: 把相邻石子之间的空位看做一个阶梯就转化为阶梯NIM了。关于阶梯NIM,显然奇数位没有用,对偶数位异或即可。#include<iostream> #include<cstdio> #include<cstdl

DZYO的博客 308

poj1704 Georgia and Bob(阶梯博弈)

Georgia and Bob Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9291   Accepted: 3021 Description Georgia and Bob decide to play a self-invented game. They

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