<span style="color: green; font-family: Arial; font-size: 12px; background-color: rgb(255, 255, 255);">Daizhenyang's Coin
</span>
<span style="color: green; font-family: Arial; font-size: 12px; background-color: rgb(255, 255, 255);">Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)</span>

Total Submission(s): 320    Accepted Submission(s): 146

Problem Description
We know that Daizhenyang is chasing a girlfriend. As we all know, whenever you chase a beautiful girl, there'll always be an opponent, or a rival. In order to take one step ahead in this chasing process, Daizhenyang decided to prove
to the girl that he's better and more intelligent than any other chaser. So he arranged a simple game: Coin Flip Game. He invited the girl to be the judge.

In this game, n coins are set in a row, where n is smaller than 10^8. They took turns to flip coins, to flip one coin from head-up to tail-up or the other way around. Each turn, one can choose 1, 2 or 3 coins to flip, but the rightmost selected must be head-up
before flipping operation. If one cannot make such a flip, he lost.

As we all know, Daizhenyang is a very smart guy (He's famous for his 26 problems and Graph Theory Unified Theory-Network Flow does it all ). So he will always choose the optimal strategy to win the game. And it's a very very bad news for all the competitors.

But the girl did not want to see that happen so easily, because she's not sure about her feelings towards him. So she wants to make Daizhenyang lose this game. She knows Daizhenyang will be the first to play the game. Your task is to help her determine whether
her arrangement is a losable situation for Daizhenyang.

For simplicity, you are only told the position of head-up coins. And due to the girl's complicated emotions, the same coin may be described twice or more times. The other coins are tail-up, of course.

Coins are numbered from left to right, beginning with 0.
 
Input
Multiple test cases, for each test case, the first line contains only one integer n (0<=n<=100), representing the number of head-up coins. The second line has n integers a1, a2 … an (0<=ak<10^8) indicating the An-th coin is head up.
 
Output
Output a line for each test case, if it's a losable situation for Daizhenyang can, print "Yes", otherwise output "No" instead.
 
Sample Input
0
1
0
4
0 1 2 3
 
Sample Output
Yes
No
Yes
翻硬币的经典例子-MOCK-TURTLES
打表发现:
x: 0 1 2 3 4 5 6 7...
g(x): 1 2 4 7 8 11 13 14...
发现x化作2进制1的个数为奇数时。g(x)= 2*x 否则 g(x) = 2*x+1 SG和即是答案。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <set>
using namespace std;
int n;
set<int>st;
int main(){ while(~scanf("%d",&n)){
int ans = 0;
st.clear();
for(int i = 0; i < n; i++){
int t;
scanf("%d",&t);
if(st.count(t)==0){
if(t==0) ans ^= 1;
else{
int k = t,cnt = 0;
while(k){
k = k&(k-1);
cnt++;
}
if(cnt%2==0) ans ^= (2*t+1);
else ans ^= 2*t; }
st.insert(t);
} }
if(ans==0){
cout<<"Yes"<<endl;
}else{
cout<<"No"<<endl;
}
}
return 0;
}

HDU3537-Daizhenyang&#39;s Coin(博弈SG-打表)的更多相关文章

  1. 51nod_1714:B君的游戏(博弈 sg打表)

    题目链接:https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1714 nim游戏的一个变形,需要打出sg函数的表 #incl ...

  2. 51nod 1714:B君的游戏(博弈 sg打表)

    题目链接:https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1714 nim游戏的一个变形,需要打出sg函数的表 #incl ...

  3. 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 ...

  4. 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 ...

  5. HDU 3032 (SG打表找规律)

    题意: 有n堆石子,alice先取,每次可以选择拿走一堆石子中的1~x(该堆石子总数) ,也可以选择将这堆石子分成任意的两堆.alice与bob轮流取,取走最后一个石子的人胜利. 思路: 因为数的范围 ...

  6. 博弈SG

    先转一篇看得比较懂的,以后有时间自己再归纳下 转自:http://blog.csdn.net/logic_nut/article/details/4711489 博弈问题若你想仔细学习博弈论,我强烈推 ...

  7. BZOJ-1228 E&D 博弈SG+找啊找啊找规律

    讨厌博弈,找规律找半天还是错的.... 1228: [SDOI2009]E&D Time Limit: 10 Sec Memory Limit: 162 MB Submit: 666 Solv ...

  8. hdu 3032(博弈sg函数)

    题意:与原来基本的尼姆博弈不同的是,可以将一堆石子分成两堆石子也算一步操作,其它的都是一样的. 分析:由于石子的堆数和每一堆石子的数量都很大,所以肯定不能用搜索去求sg函数,现在我们只能通过找规律的办 ...

  9. HDU-4678 Mine 博弈SG函数

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4678 题意就不说了,太长了... 这个应该算简单博弈吧.先求联通分量,把空白区域边上的数字个数全部求出 ...

随机推荐

  1. javascript 交互取值

    var publicClassName; var classIdInMemory = { lastVal: publicClassName, set:function(x){ if(x != &quo ...

  2. html5的自定义data-*属性和jquery的data()方法的使用

    人们总喜欢往HTML标签上添加自定义属性来存储和操作数据.但这样做的问题是,你不知道将来会不会有其它脚本把你的自定义属性给重置掉,此外,你这样做也会导致html语法上不符合Html规范,以及一些其它副 ...

  3. html-图片button,抓包---Shinepans

    askLike.html <html> <meta http-equiv="content-type" content="text/html;chars ...

  4. SQL逆向工程

    话说有个现成的SQL数据库,四十来张表,每张表多的几十的字段,少的十几个字段.老板说为了下一步大家好好利用这个数据库中的数据,让你研究一下该数据库中的所有的表和字段之间的联系.这是一个什么性质的工作, ...

  5. S3C6410 纯粹的裸机启动,自己写的SD BOOT启动

    这几天晚上一直折腾S3C6410的裸机SD卡启动,不大想使用UBOOT,我是搞硬件的,对底层非常感兴趣,不喜欢已经写好的,因此自己一直在尝试,其实很早之前就试过SD卡启动,也就是ARM11上电后会把S ...

  6. DELPHI 通過窗口句柄或窗口标题得到进程句柄

    DELPHI 通過窗口句柄或窗口标题得到进程句柄2009年05月08日 星期五 10:15procedure TForm1.Button1Click(Sender: TObject);varhWind ...

  7. Complete Guide for Spring Boot Actuator

    You are here to learn about Spring Boot Actuator for collecting metrics about your production grade ...

  8. NYOJ 1066 CO-PRIME(数论)

    CO-PRIME 时间限制:1000 ms  |  内存限制:65535 KB 难度:3 描写叙述 This problem is so easy! Can you solve it? You are ...

  9. 无边无状态栏窗口(使用GetWindowLongPtr设置GWL_EXSTYLE)

    通过SetWindowLongPtr来设置窗口样式 var NewStyle: Integer; begin Application.Initialize; Application.MainFormO ...

  10. 《c陷阱与缺陷》笔记--注意边界值

    如果要自己实现一个获取绝对值的函数,应该都没有问题,我这边也自己写了一个: void myabs(int i){ if(i>=0){ printf("%d\n",i); }e ...