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. 5、CentOS 6.5系统安装配置Nginx-1.2.7+PHP-5.3.22环境

    一,操作系统 以最小服务器形式安装系统,并添加开发工具库,便于后期编译使用. 此处基本都是下一步,下一步,不再废话. 安装完成,进入系统,调通网络,关闭防火墙或打开相应的WEB端口. 以下安装操作默认 ...

  2. 前端框架VUE----nodejs中npm的使用

    NPM是什么? 简单的说,npm就是JavaScript的包管理工具.类似Java语法中的maven,gradle,python中的pip. 安装 傻瓜式的安装. 第一步:打开https://node ...

  3. How many zero's and how many digits ? UVA - 10061

    Given a decimal integer number you will have to find out how many trailing zeros will be there in it ...

  4. Android NDK MediaCodec在ijkplayer中的实践

    https://www.jianshu.com/p/41d3147a5e07 从API 21(Android 5.0)开始Android提供C层的NDK MediaCodec的接口. Java Med ...

  5. 控制层和ajax用法的详解

    商城项目第二天复习的内容 package cn.tedu.store.entity; public class ResponseResult<T> { public static fina ...

  6. python-selenium,关于页面滑动的操作

    //移动到元素element对象的“顶端”与当前窗口的“顶部”对齐 ((JavascriptExecutor) driver).executeScript("arguments[0].scr ...

  7. bzoj2152 / P2634 [国家集训队]聪聪可可(点分治)

    P2634 [国家集训队]聪聪可可 淀粉质点分治板子 边权直接 mod 3 直接点分治统计出所有的符合条件的点对再和总方案数约分 至于约分.....gcd搞搞就好辣 #include<iostr ...

  8. tcp编程 示例

    #include <stdio.h> #include <sys/types.h> #include <sys/socket.h> #include <net ...

  9. Improving your submission -- Kaggle Competitions

    1: Improving Our Features In the last mission, we made our first submission to Titanic: Machine Lear ...

  10. DOM元素加载之前执行的jQuery代码

    <script type="text/javascript"> (function() { alert("DOM还没加载哦!"); })(jQuer ...