【Georgia and Bob POJ - 1704 阶梯Nim】的更多相关文章

Georgia and Bob POJ - 1704 阶梯Nim

$ \color{#0066ff}{ 题目描述 }$ 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, as shown in the following figure for exam…

Georgia and Bob(POJ 1704)

原题如下: Georgia and Bob Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12712   Accepted: 4262 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,…

POJ 1704 Staircase Nim 阶梯博弈

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int main() { int t,n,ans; int i,j,js; ]; //int b[1100]; scanf("%d",&t); while(t--) { memset(a,,sizeof(a)); //memset(b,0,sizeof(b)); ans=; js=; sc…

poj 1704 阶梯博弈

转自http://blog.sina.com.cn/s/blog_63e4cf2f0100tq4i.html 今天在POJ做了一道博弈题..进而了解到了阶梯博弈...下面阐述一下我对于阶梯博弈的理解..  首先是对阶梯博弈的阐述...博弈在一列阶梯上进行...每个阶梯上放着自然数个点..两个人进行阶梯博弈...每一步则是将一个集体上的若干个点( >=1 )移到前面去..最后没有点可以移动的人输.. 如这就是一个阶梯博弈的初始状态 2 1 3 2 4 ... 只能把后面的点往前面放...如何来分析…

poj 1704 Georgia and Bob(阶梯博弈)

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, ...…

POJ 1704 Georgia and Bob [阶梯Nim]

题意: 每次可以向左移动一个棋子任意步,不能跨过棋子 很巧妙的转化,把棋子间的空隙看成石子堆 然后裸阶梯Nim #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; typedef long long ll; ; inline int read(){ ,f=; ;c=ge…

POJ 1704 Georgia and Bob(阶梯Nim博弈)

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 11357   Accepted: 3749 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, ..., and place N c…

POJ 1704 Georgia and Bob（阶梯博弈+证明）

POJ 1704 题目链接 关于阶梯博弈有如下定理: 将所有奇数阶梯看作n堆石头,做Nim,将石头从奇数堆移动到偶数堆看作取走石头,同样地,异或值不为0(利己态)时,先手必胜. 定理证明看此博:http://blog.csdn.net/kk303/article/details/6692506 以下是POJ 1704的AC代码: //棋子只能往左走(最左有界线),可以走任意多格(>=1) //而且棋子不能越过在它前面的棋子(它左边的棋子) //每个格最多放一个棋子,说明棋子也不能走到另一个棋子所…

poj 1704 Georgia and Bob（阶梯博弈）

Georgia and Bob Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8656   Accepted: 2751 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, ...…

【POJ】1704 Georgia and Bob（Staircase Nim）

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, ..., and place N chessmen on different grids, as shown in the following figure for example: Georgia an…