C - Crazy Calendar

Time Limit:4000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Description

2011 was a crazy year. Many people all over the world proposed on 11-11-11, married on 11-11-11, some even went through surgery only to have 11-11-11 as their child's birth date. How crazy people can be! Don't they see there is a "20" hidden? Then what to do? A very elegant solution came from ARR, a very famous and funny character - why do we need to follow Christian (or some calls it Gregorian) calendar? Why don't we start our own calendar on the day of marriage? And those who like to celebrate their marriage ceremony too frequent, why don't they declare only 1 day per year. In that fashion they can celebrate their anniversary every day. And may be one minute a year or a second or ... Uh.. getting complex. Let's back to the title. From now, we start to have a new calendar system, "Kisu Pari Na". And we hope to update this calendar on every national contest.

The purpose of this calendar is - we all will try our best to learn something new in every year. For this first year let's learn some combinatory. It reminds me of my first year in college. I faced this problem but could not solve this then. But see how easy it is:

Say you start from upper left cell and want to go to lower right cell. The only restriction is you can only move downward or rightward. How many ways are there? How to solve it? Not that difficult. You have to go two times Down and three times Right (whichever way you try) to reach the goal from the starting cell, right? So the answer is number of ways you can arrange two D (represents Down) and three R (represent Right). 2 same characters and 3 same characters, total 5 characters. So it is:

Or =D+RCR. Easy isn't it?

Ok enough with learning. Now back to problem, given a grid and at each cell there are some coins. Inky and Pinky are playing a game getting inspiration from the above problem. At each turn, a player chooses a non empty cell and then removes one or more coins from that cell and put them to the cell exactly right of it or exactly beneath it. A player can't divide the coins and put one part to right and others to down. Note that, for the cells at the right column the player can't move it to more right, and same for the bottom-most row. So a player can't move coins from the lower right cell. The game will finish when no moves are available and the player who moved last will win. Now inky being very modest asked Pinky to move first. Can you say if Pinky will win if both play perfectly?

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing two integers R C (1 ≤ R * C ≤ 50000), where R denotes the number of rows and C denotes the number of columns of the grid respectively. Each of the next R lines contains C space separated integers denoting the grid. These integers lie in the range [0, 109].

Output

For every test case, output case number followed by "win" if Pinky can win or "lose".

Sample Input

1

2 2

1 1

1 1

Sample Output

Case 1: lose

题意:给定n*m的一个方格,每个方格含x个石子,Pinky 先手,每次可以取任意数目石子向右或者下移动,谁不能移动谁输。

题解:如果该方格中的一个点的坐标与n+m的奇偶性相同,则不需要考虑该点。因为若该点与n+m奇偶性相同,设该点坐标为(i,j)(0<=i<=n,0<=j<=m)则(n+m)-(i+j),即到达终点的步数为偶数,也就是说先手怎么移动后手就怎么移动,先手右后手就下,先手下后手就右,也就是说等于没移动。所以我们只需要考虑与n+m不同的,也就是n+m和i+j一个奇数一个偶数,也就是说,这些点距离终点只需要也只能移动一次,每次可以从一个方格中取任意个石子移动,这就转化为了尼姆博弈中的取石子问题,因为尼姆博弈中可以随意取石子。异或为0为必败态一定会转化为必胜态,异或为1为必胜态可以转化为必败态,具体详见下最基础的尼姆博弈--取石子问题,就ok了。

#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int t,cas=;
cin>>t;
while(t--)
{
int r,c,data,ans=;
cin>>r>>c;
for(int i=; i<r; i++)
{
for(int j=; j<c; j++)
{
cin>>data;
int a=((r+c)-(i+j))%;
if(a)
ans^=data;
}
}
if(ans)
printf("Case %d: win\n",cas++);
else
printf("Case %d: lose\n",cas++);
}
return ;
}

Light OJ 1393 Crazy Calendar (尼姆博弈)的更多相关文章

  1. light oj 1393 - Crazy Calendar 博弈论

    思路:当移到右下角时,就不能移动了.所以与右下角的奇偶性相同的位置,都不能直接到达,先手必败! 只需考虑与右下角奇偶不同的位置,可以看成NIM博弈.最后NIM和不为0的胜,否者败!! 代码如下: #i ...

  2. Light OJ 1253 Misere Nim (尼姆博弈(2))

    LightOJ1253 :Misere Nim 时间限制:1000MS    内存限制:32768KByte   64位IO格式:%lld & %llu 描述 Alice and Bob ar ...

  3. hdu----(1849)Rabbit and Grass(简单的尼姆博弈)

    Rabbit and Grass Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...

  4. hdu 1849(Rabbit and Grass) 尼姆博弈

    Rabbit and Grass Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...

  5. Being a Good Boy in Spring Festival 尼姆博弈

    Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Descr ...

  6. HDU 4315 Climbing the Hill (阶梯博弈转尼姆博弈)

    Climbing the Hill Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u Su ...

  7. LightOJ 1247 Matrix Game (尼姆博弈)

    A - Matrix Game Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu Submi ...

  8. hdu-------(1848)Fibonacci again and again(sg函数版的尼姆博弈)

    Fibonacci again and again Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Jav ...

  9. BestCoder Round #65 hdu5591(尼姆博弈)

    ZYB's Game Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total ...

随机推荐

  1. BZOJ-1305 dance跳舞 建图+最大流+二分判定

    跟随YveH的脚步又做了道网络流...%%% 1305: [CQOI2009]dance跳舞 Time Limit: 5 Sec Memory Limit: 162 MB Submit: 2119 S ...

  2. window自动切换ip的脚本

    因为总要切换ip,所以百度了一下脚本 如下http://jingyan.baidu.com/article/d2b1d1029d21b95c7e37d4fa.html 动态ip netsh inter ...

  3. Ubuntu系统启动过程详解

    作者:杨硕,华清远见嵌入式学院讲师. 一. Ubuntu的启动流程 ubuntu的启动流程和我们熟知的RedHat的启动方式有所区别. RedHat的启动过程如下图: 这是我们熟知的linux启动流程 ...

  4. ios 图片尺寸

  5. WPF 注册全局快捷键

    .NET技术交流群 199281001 .欢迎加入. using System; using System.Collections.Generic; using System.Linq; using ...

  6. redhat RHEL 5.5 下载地址

    redhat RHEL 5.5 下载地址 RHEL 5 update 5 已经release许久了, redhat RHEL 5.5 下载地址: RHEL 5 安装 序列号 rhel-server-5 ...

  7. C#原始类型扩展方法—this参数修饰符

    扩展方法使您能够向现有类型“添加”方法,而无需创建新的派生类型.重新编译或以其他方式修改原始类型.扩展方法是一种特殊的静态方法,但可以像扩展类型上的实例方法一样进行调用.对于用 C# 和 Visual ...

  8. 《C++代码设计与重用》 书评

    作者:唐风 主页:www.cnblogs.com/liyiwen   前几个星期买了,一直没有直接细翻,买的时候看了背面的两个推荐,一个是孟岩,一个是Scott Meyers(Effective C+ ...

  9. LDA(Linear discriminate analysis)线性判别分析

    LDA 线性判别分析与Fisher算法完全不同 LDA是基于最小错误贝叶斯决策规则的. 在EMG肌电信号分析中,... 未完待续:.....

  10. iOS设备屏幕像素总览

    本文永久地址为http://www.cnblogs.com/ChenYilong/p/4011728.html ,转载请注明出处.   本文永久地址为http://www.cnblogs.com/Ch ...