HDU1944 S-NIM(多个NIM博弈)
The starting position has a number of heaps, all containing some, not necessarily equal, number of beads.
The players take turns chosing a heap and removing a positive number of beads from it.
The first player not able to make a move, loses.
Arthur and Caroll really enjoyed playing this simple game until they recently learned an easy way to always be able to find the best move:
Xor the number of beads in the heaps in the current position (i.e. if we have 2, 4 and 7 the xor-sum will be 1 as 2 xor 4 xor 7 = 1).
If the xor-sum is 0, too bad, you will lose.
Otherwise, move such that the xor-sum becomes 0. This is always possible.
It is quite easy to convince oneself that this works. Consider these facts:
The player that takes the last bead wins.
After the winning player's last move the xor-sum will be 0.
The xor-sum will change after every move.
Which means that if you make sure that the xor-sum always is 0 when you have made your move, your opponent will never be able to win, and, thus, you will win.
Understandibly it is no fun to play a game when both players know how to play perfectly (ignorance is bliss). Fourtunately, Arthur and Caroll soon came up with a similar game, S-Nim, that seemed to solve this problem. Each player is now only allowed to remove a number of beads in some predefined set S, e.g. if we have S =(2, 5) each player is only allowed to remove 2 or 5 beads. Now it is not always possible to make the xor-sum 0 and, thus, the strategy above is useless. Or is it?
your job is to write a program that determines if a position of S-Nim is a losing or a winning position. A position is a winning position if there is at least one move to a losing position. A position is a losing position if there are no moves to a losing position. This means, as expected, that a position with no legal moves is a losing position.
InputInput consists of a number of test cases.
For each test case: The rst line contains a number k (0 < k <= 100) describing the size of S, followed by k numbers si (0 < si <= 10000) describing S. The second line contains a number m (0 < m <= 100) describing the number of positions to evaluate. The next m lines each contain a number l (0 < l <= 100) describing the number of heaps and l numbers hi (0 <= hi <= 10000) describing the number of beads in the heaps.
The last test case is followed by a 0 on a line of its own.OutputFor each position:
If the described position is a winning position print a 'W'.
If the described position is a losing position print an 'L'.
Print a newline after each test case.Sample Input
2 2 5
3
2 5 12
3 2 4 7
4 2 3 7 12
5 1 2 3 4 5
3
2 5 12
3 2 4 7
4 2 3 7 12
0
Sample Output
LWW
WWL
题解:可以当成多个NIM博弈,最终答案等于每个NIMA博弈结果的异或;(注意求SG函数时,不要每次都把vis数组清空,用一个t标记即可,每次改变标记,否则会超时)
参考代码:
#include<bits/stdc++.h>
using namespace std;
#define clr(a,val) memset(a,val,sizeof a)
const int maxn=;
int num,f[maxn],ans;
int l,t,cas,SG[maxn],vis[maxn];
void GetSG(int x)
{
clr(SG,);
int t=;
for(int i=;i<=x;++i)
{
for(int j=;f[j]<=i&&j<=num;++j) vis[SG[i-f[j]]]=t;
for(int j=;j<=x;j++) {if(vis[j]!=t) {SG[i]=j;break;}}
++t;
}
} int main()
{
while(~scanf("%d",&num) && num)
{
for(int i=;i<=num;++i)scanf("%d",&f[i]);
sort(f+,f++num);
GetSG(maxn-);
scanf("%d",&cas);
while(cas--)
{
scanf("%d",&l);ans=;
for(int i=;i<=l;++i) scanf("%d",&t),ans^=SG[t];
if(!ans) printf("L");
else printf("W");
}
puts("");
}
return ;
}
HDU1944 S-NIM(多个NIM博弈)的更多相关文章
- NIM游戏,NIM游戏变形,威佐夫博弈以及巴什博奕总结
NIM游戏,NIM游戏变形,威佐夫博弈以及巴什博奕总结 经典NIM游戏: 一共有N堆石子,编号1..n,第i堆中有个a[i]个石子. 每一次操作Alice和Bob可以从任意一堆石子中取出任意数量的石子 ...
- hdu 3032 Nim or not Nim? (SG函数博弈+打表找规律)
Nim or not Nim? Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Sub ...
- HDU 3032 Nim or not Nim?(博弈,SG打表找规律)
Nim or not Nim? Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)T ...
- HDU 3032 Nim or not Nim? (sg函数)
Nim or not Nim? Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)T ...
- HDU 5795 A Simple Nim(简单Nim)
p.MsoNormal { margin: 0pt; margin-bottom: .0001pt; text-align: justify; font-family: Calibri; font-s ...
- HDU 3032 Nim or not Nim? (sg函数求解)
Nim or not Nim? Problem Description Nim is a two-player mathematic game of strategy in which players ...
- Nim or not Nim? hdu3032 SG值打表找规律
Nim or not Nim? Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)T ...
- 【HDU3032】Nim or not Nim?(博弈论)
[HDU3032]Nim or not Nim?(博弈论) 题面 HDU 题解 \(Multi-SG\)模板题 #include<iostream> #include<cstdio& ...
- hdu 3032 Nim or not Nim? sg函数 难度:0
Nim or not Nim? Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)T ...
- HDU 3032 Nim or not Nim?(Multi_SG,打表找规律)
Nim or not Nim? Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)T ...
随机推荐
- 深入理解计算机系统 第九章 虚拟内存 Part1 第二遍
这次花了4小时40分钟,看了第 559~575 页,共 17 页 第一遍对应地址 https://www.cnblogs.com/stone94/p/10264044.html 注意:本章的练习题一定 ...
- python 豆瓣top250电影的爬取
我们先看一下豆瓣的robot.txt 然后我们查看top250的网页链接和源代码 通过对比不难发现网页间只是start数字发生了变化. 我们可以知道电影内容都存在ol标签下的 div class属性为 ...
- ReactJS的4行代码
Angular 2一个显著的变动是,把Angular 1的Promise pattern改成了Observer pattern,并且使用了ReactJS.这里有一篇值得一读的文章 要搞懂ReactJS ...
- 字体图标转base64
如果你在阿里矢量库下载了字体图标在项目引入无法显示时,可以把图标转成base64 在线转换的链接 https://transfonter.org/ css字体图标的制作
- 创建sql自定义的函数及商品分页sql存储过程
--商品筛选时判断品牌ID是否存在 --select dbo.isValite(94,94)create function isValite(@brandId int,@bId int)returns ...
- 勾股数专题-SCAU-1079 三角形-18203 神奇的勾股数(原创)
勾股数专题-SCAU-1079 三角形-18203 神奇的勾股数(原创) 大部分的勾股数的题目很多人都是用for来便利,然后判断是不是平方数什么什么的,这样做的时候要对变量类型和很多细节都是要掌握好的 ...
- 提高PHP性能效率的几个技巧!
如何提高效率问题,往往同样的功能,不一样的代码,出来的效率往往大不一样. ● 用单引号代替双引号来包含字符串,这样做会更快一些.因为PHP会在双引号包围的字符串中搜寻变量,单引号则不会,注意:只有ec ...
- Java多线程——对象及变量的并发访问
Java多线系列文章是Java多线程的详解介绍,对多线程还不熟悉的同学可以先去看一下我的这篇博客Java基础系列3:多线程超详细总结,这篇博客从宏观层面介绍了多线程的整体概况,接下来的几篇文章是对多线 ...
- Stream系列(七)distinct方法使用
EmployeeTestCase.java package com.example.demo; import lombok.Data; import lombok.ToString; import l ...
- 3 JAVA的基本变量类型
1. 数字 整数型 类型 字节数 范围 int 4 -2^31~ 2^31-1 short 2 -2^15~ 2^15 -1 long 8 -2^63 ~ 2^63 -1 byte 1 -2^ ...