题目描述

One day, Bessie decides to challenge Farmer John to a game of ‘Cow Checkers’. The game is played on an M*N (1 <= M <= 1,000,000; 1 <= N <= 1,000,000) checkerboard that initially contains a single checker piece on the checkboard square coordinates (X, Y) (0 <= X < M; 0 <= Y < N). The bottom leftmost square of the checkerboard has

coordinates (0, 0), and the top rightmost square has coordinates (M-1, N-1). Bessie always moves first, and then the two players alternate turns. Each turn comprises one of three types of moves:

1) Move the checker piece to any square on the same row to the left of its current position.

2) Move the checker piece to any square on the same column below its current position.

3) Move the checker piece to any spot k squares below and k squares to the left of the current square (where k is any positive integer such that this new spot is still on the checkerboard).

The first player unable to make a move (i.e., because the checker is at (0, 0)) loses. Given that Bessie always goes first, determine who will win the game if both play optimally.

Play and report the winner for T games (1 <= T <= 1,000) reading in a new X,Y starting value for each new game.

有一天,Bessie准备玩一个叫做奶牛跳棋的游戏,来挑战Farmer John。

这个游戏的棋盘大小为 M*N (1 <= M <= 1,000,000; 1 <= N <= 1,000,000) 。最初棋盘上只有一个棋子在(x,y),棋盘的左下角坐标是(0,0),右上角的坐标是(M-1,N-1)。

每次游戏Bessie都是第一个开始,之后两个人轮流。

每次下棋的时候都有三种走法:

1.向左走任意步

2.向下走任意步

3.向左走k步然后向下走k步(k可随便取值,只要不走出棋盘)

先走到(0,0)的人就算输。

游戏共有T次,每次都会给出一个新的坐标(x,y),请输出每一轮赢家的名字。
输入输出格式
输入格式:

  • Line 1: Two space-separated integers: M and N

  • Line 2: A single integer: T

  • Lines 3..T+2: Two space-separated integers: X and Y

第一行:M N

第二行:T

第三行到第T+2行:这一轮的X Y

输出格式:

  • Lines 1..T: Should contain either ‘Farmer John’ or ‘Bessie’ depending on who wins each game.

共T行,每一行输出那一轮的赢家

输入输出样例
输入样例#1:

3 3
1
1 1

输出样例#1:

Bessie

说明

Farmer John and Bessie are playing one game on a 3*3 checkerboard with the checker piece initially at (1, 1) (i.e. at the center of the board).

Bessie initially can only move the checker piece to either (1, 0) or (0, 1), or (0, 0). Bessie can immediately win by moving the piece to (0, 0).

起点在(1,1),一开始有三种选择(1,0),(0,1),(0,0)只要Bessie在开始时将棋子移到(1,0)或(0,1),就必胜无疑。

感谢@ 2014nhc 提供的翻译

解题思路

威佐夫博弈裸题。。。两堆石子,一堆n个,一堆m个

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath> using namespace std; int n,m,T; int main(){
scanf("%d%d%d",&n,&m,&T);
while(T--){
int x,y;
scanf("%d%d",&x,&y);
if(x>y) swap(x,y);
int tmp=floor((y-x)*(sqrt(5)+1)/2);
if(tmp==x) cout<<"Farmer John"<<endl;
else cout<<"Bessie"<<endl;
}
return 0;
}

LUOGU P3024 [USACO11OPEN]奶牛跳棋Cow Checkers的更多相关文章

  1. [USACO11OPEN]奶牛跳棋Cow Checkers(博弈论)

    题目描述 One day, Bessie decides to challenge Farmer John to a game of 'Cow Checkers'. The game is playe ...

  2. 3298: [USACO 2011Open]cow checkers

    3298: [USACO 2011Open]cow checkers Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 65  Solved: 26[Su ...

  3. BZOJ3298: [USACO 2011Open]cow checkers(佐威夫博弈)

    3298: [USACO 2011Open]cow checkers Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 195  Solved: 96[S ...

  4. bzoj 3298: [USACO 2011Open]cow checkers -- 数学

    3298: [USACO 2011Open]cow checkers Time Limit: 10 Sec  Memory Limit: 128 MB Description 一天,Besssie准备 ...

  5. 洛谷P1472 奶牛家谱 Cow Pedigrees

    P1472 奶牛家谱 Cow Pedigrees 102通过 193提交 题目提供者该用户不存在 标签USACO 难度普及+/提高 提交  讨论  题解 最新讨论 暂时没有讨论 题目描述 农民约翰准备 ...

  6. BZOJ3298[USACO 2011Open]cow checkers——威佐夫博弈

    题目描述 一天,Besssie准备和FJ挑战奶牛跳棋游戏.这个游戏上在一个M*N的棋盘上, 这个棋盘上在(x,y)(0<=x棋盘的左下角是(0,0)坐标,棋盘的右上角是坐标(M-1,N-1). ...

  7. BZOJ3298: [USACO 2011Open]cow checkers 威佐夫博弈

    Description 一天,Besssie准备和FJ挑战奶牛跳棋游戏.这个游戏上在一个M*N的棋盘上, 这个棋盘上在(x,y)(0<=x棋盘的左下角是(0,0)坐标,棋盘的右上角是坐标(M-1 ...

  8. Luogu P2419 [USACO08JAN]牛大赛Cow Contest

    题目背景 [Usaco2008 Jan] 题目描述 N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a p ...

  9. luogu P2915 [USACO08NOV]奶牛混合起来Mixed Up Cows

    题目描述 Each of Farmer John's N (4 <= N <= 16) cows has a unique serial number S_i (1 <= S_i & ...

随机推荐

  1. java_新特性未整理

    得到的.className method.isAnnotationPresent:判断是否有指定的注解,注解.class method.invoke:执行 */}

  2. UMP系统架构

  3. 05.Mybatis动态sql

    1.IF标签 需求:根据条件查询用户 在Mapper.xml中编写 <!-- 根据sex和username查询user --> <select id="findbySexa ...

  4. loj2509 hnoi2018排列

    题意:对于a数组,求它的一个合法排列的最大权值.合法排列:对于任意j,k,如果a[p[j]]=p[k],那么k<j. 权值:sigma(a[p[i]]*i).n<=50W. 标程: #in ...

  5. uoj60 怎样提高智商

    题意:你需要构造n个四项选择题.格式为:问在前i个问题中选了几个hi字母? 输出有最多正确答案的构造方案. 标程: #include<cstdio> using namespace std ...

  6. MySQL数据库CRUD命令用法

    数据库CRUD操作即添加(Create).读取(Read).更新(Update)和删除(Delete). 1. 添加操作也称插入操作,使用Insert语句,Insert语句可以用于几种情况: 插入完整 ...

  7. SPOJ - LOCKER

    SPOJ - LOCKERhttps://vjudge.net/problem/45908/origin暴力枚举2-102 23 34 2 25 2 36 3 37 2 2 38 2 3 39 3 3 ...

  8. thinkphp 快捷查询

    快捷查询方式是一种多字段查询的简化写法,可以进一步简化查询条件的写法,在多个字段之间用|分割表示OR查询,用&分割表示AND查询,可以实现下面的查询,例如: 大理石平台价格 一.不同字段相同的 ...

  9. 莫烦PyTorch学习笔记(五)——分类

    import torch from torch.autograd import Variable import torch.nn.functional as F import matplotlib.p ...

  10. IDEA启动springboot项目一直build

    启动main方法后,项目一直在不断的build,期间截了两张一闪而过的提示 我用的是Run Dashboard面板,不论是通过删除configuration,rebuild,删除IDEA缓存都没有效果 ...