poj 1079 Calendar Game(博弈论 SG)
Calendar Game
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2519 Accepted Submission(s): 1438
by randomly choosing a date from this interval. Then, the players, Adam and Eve, make moves in their turn with Adam moving first: Adam, Eve, Adam, Eve, etc. There is only one rule for moves and it is simple: from a current date, a player in his/her turn can
move either to the next calendar date or the same day of the next month. When the next month does not have the same day, the player moves only to the next calendar date. For example, from December 19, 1924, you can move either to December 20, 1924, the next
calendar date, or January 19, 1925, the same day of the next month. From January 31 2001, however, you can move only to February 1, 2001, because February 31, 2001 is invalid.
A player wins the game when he/she exactly reaches the date of November 4, 2001. If a player moves to a date after November 4, 2001, he/she looses the game.
Write a program that decides whether, given an initial date, Adam, the first mover, has a winning strategy.
For this game, you need to identify leap years, where February has 29 days. In the Gregorian calendar, leap years occur in years exactly divisible by four. So, 1993, 1994, and 1995 are not leap years, while 1992 and 1996 are leap years. Additionally, the years
ending with 00 are leap years only if they are divisible by 400. So, 1700, 1800, 1900, 2100, and 2200 are not leap years, while 1600, 2000, and 2400 are leap years.
MM-th month in the year of YYYY. Remember that initial dates are randomly chosen from the interval between January 1, 1900 and November 4, 2001.
3
2001 11 3
2001 11 2
2001 10 3
YES
NO
NO
#include <iostream>
#include <cstdio>
using namespace std; int PN[2010][20][50] , year = 2001 , month = 11 , day = 4;
int mp[13] = {0 , 31 , 30 , 31 , 30 , 31 , 30 , 31 , 31 , 30 , 31 , 30 , 31}; void sub_day(){
if(day == 0){
month--;
if(month == 0){
month = 12;
day = 31;
year--;
}else{
if(month == 2){
if(year%4==0&&year%100!=0||year%400==0) day = 29;
else day = 28;
}else{
day = mp[month];
}
}
}
} bool add_month(int &Y , int &M , int &D){
if(M > 12){
Y++;
M = 1;
return true;
}else{
if(M == 2){
if(Y%4==0&&Y%100!=0||Y%400==0){
if(D > 29) return false;
return true;
}else{
if(D > 28) return false;
return true;
}
}else{
if(mp[M] < D) return false;
return true;
}
}
} int SG(){
int Y = year , M = month+1 , D = day;
if(add_month(Y , M , D)){
if(PN[Y][M][D] == 1) return 0;
}
return 1;
} void initial(){
PN[year][month][day] = 1;
int ty = year , tm = month , td = day;
day--;
while(year >= 1900){
if(PN[ty][tm][td] == 1) PN[year][month][day] = 0;
else PN[year][month][day] = SG();
ty = year , tm = month , td = day;
day--;
sub_day();
}
} int main(){
initial();
int YYYY ,MM, DD ,T;
scanf("%d" , &T);
while(T--){
scanf("%d%d%d" , &YYYY ,&MM, &DD);
if(PN[YYYY][MM][DD]) printf("NO\n");
else printf("YES\n");
}
return 0;
}
poj 1079 Calendar Game(博弈论 SG)的更多相关文章
- POJ 2960 S-Nim 博弈论 sg函数
http://poj.org/problem?id=2960 sg函数几乎是模板题. 调试代码的最大障碍仍然是手残在循环里打错变量名,是时候换个hydra产的机械臂了[超想要.jpg] #includ ...
- HDU 1847 Good Luck in CET-4 Everybody! (博弈论sg)
Good Luck in CET-4 Everybody! Problem Description 大学英语四级考试就要来临了,你是不是在紧张的复习?或许紧张得连短学期的ACM都没工夫练习了.反正我知 ...
- HDU.1847 Good Luck in CET-4 Everybody! ( 博弈论 SG分析)
HDU.1847 Good Luck in CET-4 Everybody! ( 博弈论 SG分析) 题意分析 简单的SG分析 题意分析 简单的nim 博弈 博弈论快速入门 代码总览 //#inclu ...
- 【bzoj3576】[Hnoi2014]江南乐 博弈论+SG定理+数学
题目描述 两人进行 $T$ 轮游戏,给定参数 $F$ ,每轮给出 $N$ 堆石子,先手和后手轮流选择石子数大于等于 $F$ 的一堆,将其分成任意(大于1)堆,使得这些堆中石子数最多的和最少的相差不超过 ...
- HDU 1079 Calendar Game (博弈论-sg)
版权声明:欢迎关注我的博客,本文为博主[炒饭君]原创文章.未经博主同意不得转载 https://blog.csdn.net/a1061747415/article/details/32336485 C ...
- hdu 1079 Calendar Game sg函数 难度:0
Calendar Game Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Tot ...
- POJ 2425 A Chess Game 博弈论 sg函数
http://poj.org/problem?id=2425 典型的sg函数,建图搜sg函数预处理之后直接求每次游戏的异或和.仍然是因为看不懂题目卡了好久. 这道题大概有两个坑, 1.是搜索的时候vi ...
- (博弈论)hdoj 1079 Calendar Game
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1079 题解:题目大意,两个人Adam和Eve一块儿玩游戏,游戏规则是从1900年1月1日到2001年1 ...
- poj 3710 Christmas Game【博弈论+SG】
也就是转换到树形删边游戏,详见 https://wenku.baidu.com/view/25540742a8956bec0975e3a8.html #include<iostream> ...
随机推荐
- 如何防止Android反编译
转自: http://my.eoe.cn/sandking/archive/19772.html http://www.cnblogs.com/zdz8207/archive/2012/01/28/d ...
- poj 1981(单位圆覆盖最多点问题模板)
Circle and Points Time Limit: 5000MS Memory Limit: 30000K Total Submissions: 7327 Accepted: 2651 ...
- C#获取本机的外网IP
/// <summary> /// 功能:获取本地的外网IP地址 /// 作者:黄海 /// 时间:2016-07-22 /// </summary> /// <retu ...
- Hmailserver搭建邮件服务器
Hmailserver安装很简单不需要多说,自己去百度 Hmailserver配置: 输入安装时设置的密码登录Hmailserver 添加域名,如:域名是:mail.abc.com这里添加的时候应该填 ...
- PHP防止表单重复提交的解决方法
PHP+SESSION防止表单重复提交 index.php 当前表单页面is_submit设为0 SESSION_START(); $_SESSION['is_submit'] = 0; <fo ...
- Delphi中获取文件大小
大概有这些方法可以获得文件大小FileSizeByName(需要引用IdGlobal单元)GetFileSizeFileSize(不能获得正在使用的文件大小)FileSeekTFileStream.S ...
- 原来是adblock惹的祸
一个在本地开发好的网站,放到服务器就不行了.花了好几个小时的时间,最后试着把adblock关了,然后正常了.
- POJ 2226.Muddy Fields-二分图最大匹配(最小点覆盖)
Muddy Fields Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 12565 Accepted: 4651 Des ...
- WORDPRESS改为https部署
1.确保你已经正确开启了httpd 2.4.6的https配置,并且配置了该网站的虚拟主机,如下所示: <VirtualHost *:443> DocumentRoot "/ap ...
- Union与UnionAll
UNION指令的目的是将两个SQL语句的结果合并起来.从这个角度来看, 我们会产生这样的感觉,UNION跟JOIN似乎有些许类似,因为这两个指令都可以由多个表格中撷取资料. UNION的一个限制是两个 ...