A Simple Nim

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description
Two players take turns picking candies from n heaps,the player who picks the last one will win the game.On each turn they can pick any number of candies which come from the same heap(picking no candy is not allowed).To make the game more interesting,players can separate one heap into three smaller heaps(no empty heaps)instead of the picking operation.Please find out which player will win the game if each of them never make mistakes.
 
Input
Intput contains multiple test cases. The first line is an integer 1≤T≤100, the number of test cases. Each case begins with an integer n, indicating the number of the heaps, the next line contains N integers s[0],s[1],....,s[n−1], representing heaps with s[0],s[1],...,s[n−1] objects respectively.(1≤n≤106,1≤s[i]≤109)
 
Output
For each test case,output a line whick contains either"First player wins."or"Second player wins".
 
Sample Input
2
2
4 4
3
1 2 4
 
Sample Output
Second player wins.
First player wins.
 
Author
UESTC
 
Source

题意:跟普通的nim不一样的是,一堆可以分成三小堆;

思路:sg函数打表找规律;

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<vector>
using namespace std;
typedef long long ll;
#define PI acos(-1.0)
const int N=1e3+,maxm=1e5+,inf=0x3f3f3f3f,mod=1e9+;
const ll INF=1e18+; int sg[N],mex[N];
void initsg()
{
for(int i=;i<=;i++)
{
memset(mex,,sizeof(mex));
for(int j=;j<=i-;j++)
{
for(int k=;k+j<i;k++)
{
int l=i-j-k;
mex[sg[l]^sg[j]^sg[k]]=;
}
}
for(int j=;j<=i;j++)
mex[sg[j]]=;
for(int j=;;j++)
if(!mex[j])
{
sg[i]=j;
break;
}
}
for(int i=;i<=;i++)
if(sg[i]!=i)cout<<sg[i]<<" "<<i<<endl;
}
int SG(int x)
{
if(x%==)return x+;
if(x%==)return x-;
return x;
}
int main()
{
//initsg();
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
int ans=;
for(int i=;i<=n;i++)
{
int x;
scanf("%d",&x);
ans^=SG(x);
}
if(ans)printf("First player wins.\n");
else printf("Second player wins.\n");
}
return ;
}

hdu 5795 A Simple Nim 博弈sg函数的更多相关文章

  1. HDU 5795 A Simple Nim (博弈) ---2016杭电多校联合第六场

    A Simple Nim Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Tota ...

  2. HDU 5795 A Simple Nim(SG打表找规律)

    SG打表找规律 HDU 5795 题目连接 #include<iostream> #include<cstdio> #include<cmath> #include ...

  3. HDU 5795 A Simple Nim(简单Nim)

    p.MsoNormal { margin: 0pt; margin-bottom: .0001pt; text-align: justify; font-family: Calibri; font-s ...

  4. HDU 5795 A Simple Nim 打表求SG函数的规律

    A Simple Nim Problem Description   Two players take turns picking candies from n heaps,the player wh ...

  5. HDU 5795 A Simple Nim (博弈 打表找规律)

    A Simple Nim 题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5795 Description Two players take turns ...

  6. HDU 5795 A Simple Nim

    打表找SG函数规律. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> ...

  7. HDU 5795 A Simple Nim ——(Nim博弈 + 打表)

    题意:在nim游戏的规则上再增加了一条,即可以将任意一堆分为三堆都不为0的子堆也视为一次操作. 分析:打表找sg值的规律即可. 感想:又学会了一种新的方法,以后看到sg值找不出规律的,就打表即可~ 打 ...

  8. POJ 2311 Cutting Game(Nim博弈-sg函数/记忆化搜索)

    Cutting Game 题意: 有一张被分成 w*h 的格子的长方形纸张,两人轮流沿着格子的边界水平或垂直切割,将纸张分割成两部分.切割了n次之后就得到了n+1张纸,每次都可以选择切得的某一张纸再进 ...

  9. S-Nim HDU 1536 博弈 sg函数

    S-Nim HDU 1536 博弈 sg函数 题意 首先输入K,表示一个集合的大小,之后输入集合,表示对于这对石子只能去这个集合中的元素的个数,之后输入 一个m表示接下来对于这个集合要进行m次询问,之 ...

随机推荐

  1. linux系统Centos环境下搭建SVN服务器及权限配置

    linux系统Centos环境下如何搭建SVN服务器以及svnserve.conf.authz.passwd配置文件详细介绍   至于svn的概念,这里就不做详细阐述了,可以自行百度.简单来讲就是一个 ...

  2. selenium webdriver 实现Canvas画布自动化测试

    https://blog.csdn.net/xiaoguanyusb/article/details/80324210 由借鉴意义, 转过来 canvas 是一个画布,定位元素时只能定位到画布上,如下 ...

  3. JDK源码之LinkedHashSet

    LinkedHashSet是HashSet和LinkList结合产生的集合,集合中的元素互不相同,且元素采用双向链表进行连接. 1.定义 LinkedHashSet继承了HashSet并且实现了Set ...

  4. Fiddler(三)Fiddler 报错creation of the root certificate was not successful

    打开CMD,找到Fiddler所在目录,我这里是把汉化版的解压在了桌面,所以通过CMD进入桌面下的Fiddler文件夹. 执行命令 makecert.exe -r -ss my -n -h -cy a ...

  5. Django XSS攻击

    Django XSS攻击 XSS(cross-site scripting跨域脚本攻击)攻击是最常见的web攻击,其特点是“跨域”和“客户端执行”,XSS攻击分为三种: Reflected XSS(基 ...

  6. Sorl搜索技术

    在一些大型门户网站.电子商务网站等都需要站内搜索功能,使用传统的数据库查询方式实现搜索无法满足一高级的搜索需求,比如:搜索速度要快.搜索结果按相关度排序.搜索内容格式不固定等,这里就需要使用全文检索技 ...

  7. RTP协议全解析(H264码流和PS流)

    转自:http://blog.csdn.net/chen495810242/article/details/39207305 写在前面:RTP的解析,网上找了很多资料,但是都不全,所以我力图整理出一个 ...

  8. httpclient新旧版本分割点4.3

    从这个版本开始,httpclient的api发生了一次重大调整.主要包括如下:Release 4.3 Final ------------------- This is the first stabl ...

  9. Hibernate properties文件

    ###################### ### Query Language ### ###################### ## define query language consta ...

  10. Codeforces Round #439 (Div. 2) Problem B (Codeforces 869B)

    Even if the world is full of counterfeits, I still regard it as wonderful. Pile up herbs and incense ...