HDU 1079 Calendar Game （日期博弈，真是日历都快被我走遍了...）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1079

题面：

Calendar Game

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3240    Accepted Submission(s): 1903

Problem Description

Adam and Eve enter this year’s ACM International Collegiate Programming Contest. Last night, they played the Calendar Game, in celebration of this contest. This game consists of the dates from January 1, 1900 to November 4, 2001, the contest day. The game starts 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.

 

Input

The input consists of T test cases. The number of test cases (T) is given in the first line of the input. Each test case is written in a line and corresponds to an initial date. The three integers in a line, YYYY MM DD, represent the date of the DD-th day of 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.

 

Output

Print exactly one line for each test case. The line should contain the answer "YES" or "NO" to the question of whether Adam has a winning strategy against Eve. Since we have T test cases, your program should output totally T lines of "YES" or "NO".

 

Sample Input

   
   
   
   

    
    
    
    3 
2001 11 3 
2001 11 2 
2001 10 3 
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    YES 
NO 
NO 
   
   
   
   

 

题目大意：

    给定一个初始日期，对于当前日期，A，B轮流操作。操作方式有两种，以天数为单位后移一天，或者以月为单位，后移一月，但如果后面那个月份没有对应的天，那么就不能进行月移操作。若刚好移到指定日期者获胜。

解题：

    抛开日期操作不说，这是一道很简单的博弈问题，但由于结合了日期操作，所以处理较繁琐。博弈策略是这样的，对于一个日期，如果能够通过日移操作或月移操作达到一个必败态，那么该状态就为胜态，否则就为必败态。也就是去检验后一个月，和后一天的状态。这道题的亮点在于，出了任何小差错，就可能导致不能AC。

提升版：

    ZJUT 1587

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
#define eps 1e-7
using namespace std;
struct date
{
	int y,m,d;
};
//日期状态
bool status[2016][13][32],flag,sign;
//每月天数
int month[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//日移操作
date next_day(int y,int m,int d)
{
	date tmp;
	//如果是12月份，可能会换年，需特殊处理
	if(m==12)
	{
		if(d==31)
		  tmp.y=y+1,tmp.m=1,tmp.d=1;
		else
		  tmp.y=y,tmp.m=12,tmp.d=d+1;
	}
	//如果不是2月，一般处理
	else if(m!=2)
	{
	   //看是不是该月最后一天，可能会换月，需特殊处理
       if(d==month[m])
		   tmp.y=y,tmp.m=m+1,tmp.d=1;
	   else
		   tmp.y=y,tmp.m=m,tmp.d=d+1;
	}
	//如果是2月，需要看是不是闰年
	else
	{
		if(y%400==0||(y%4==0&&y%100!=0))
		{
			if(d==29)
			  tmp.y=y,tmp.m=3,tmp.d=1;
			else
			  tmp.y=y,tmp.m=2,tmp.d=d+1;
		}
		else
		{
			if(d==28)
			  tmp.y=y,tmp.m=3,tmp.d=1;
			else
			  tmp.y=y,tmp.m=2,tmp.d=d+1;		
		}
	}
	return tmp;
}
//月移操作
date next_month(int y,int m,int d)
{
	date tmp;
	//flag为月移是否合法标志
	flag=true;
	//特判12月，可能会跨年
	if(m==12)
		tmp.y=y+1,tmp.m=1,tmp.d=d;
	//特判1、2月
	else if((m!=2)&&(m!=1))
	{
		//到达月末，换月
		if(d>month[m+1])
			flag=false;
		else
		  tmp.y=y,tmp.m=m+1,tmp.d=d;
	}
	//1月，因为后移到2月，要看2月是否在闰年
	else if(m==1)
	{
		if((y%4==0&&y%100!=0)||(y%400==0))
		{
			if(d>29)
				flag=false;
			else
			  tmp.y=y,tmp.m=2,tmp.d=d;
		}
		else
		{
			if(d>28)
				flag=false;
			else
			  tmp.y=y,tmp.m=2,tmp.d=d;
		}
	}
	//2月最多29天，对于3月都是可以的
	else
		tmp.y=y,tmp.m=m+1,tmp.d=d;
	return tmp;
}
//年移操作，此题不需要
date next_year(int y,int m,int d)
{
	//sign为年移是否合法标志
	sign=true;
	date tmp;
	//只有2月29号是特殊的
	if(m==2&&d==29)
		sign=false;
	else
	  tmp.y=y+1,tmp.m=m,tmp.d=d;
	return tmp;
}
int main()
{
	date tmp,tmp2;
	//标记11.1-11.4号
	for(int i=1;i<=4;i++)
	{
       if(i%2)
		   status[2001][11][i]=0;
	   else
		   status[2001][11][i]=1;
	}
	if(status[2001][11][1])
		status[2001][10][31]=0;
	else
		status[2001][10][31]=1;
	//10.5-10.30只能日移，不能月移
	for(int i=30;i>=5;i--)
	{
		tmp=next_day(2001,10,i);
		//如果后一天是必败态，那么当前就是胜态
		if(status[tmp.y][tmp.m][tmp.d])
			status[2001][10][i]=0;
		else
			status[2001][10][i]=1;
	}
	//10.1-10.4既可月移也可日移
	for(int i=4;i>=1;i--)
	{
        tmp=next_day(2001,10,i);
		tmp2=next_month(2001,10,i);
		//如果月移或日移到必败态，那么当前为胜态，否则为必败态
		if(status[tmp.y][tmp.m][tmp.d]||status[tmp2.y][tmp2.m][tmp2.d])
			status[2001][10][i]=0;
		else
			status[2001][10][i]=1;
	}
	//2001.1-2001.9既可月移又可日移
	for(int i=9;i>=1;i--)
	{
       for(int j=month[i];j>=1;j--)
	   {
		   tmp=next_day(2001,i,j);
		   if(status[tmp.y][tmp.m][tmp.d])
			   status[2001][i][j]=0;
		   else
		   {
			   tmp=next_month(2001,i,j);
			   if(flag)
			   {
				   if(status[tmp.y][tmp.m][tmp.d])
					   status[2001][i][j]=0;
				   else
					   status[2001][i][j]=1;
			   }
			   else
				   status[2001][i][j]=1;
		   }
	   }
	}
	//1900-2000，即可月移又可日移
	for(int i=2000;i>=1900;i--)
	{
		for(int j=12;j>=1;j--)
		{
			int k;
			if(j==2&&(i%4==0&&i%100!=0)||(i%400==0))
				k=29;
			else
				k=month[j];
			for(;k>=1;k--)
			{
				tmp=next_day(i,j,k);
				if(status[tmp.y][tmp.m][tmp.d])
					status[i][j][k]=0;
				else
				{
					tmp=next_month(i,j,k);
					if(flag)
					{
                       if(status[tmp.y][tmp.m][tmp.d])
						   status[i][j][k]=0;
					   else
						   status[i][j][k]=1;
					}
					else
						status[i][j][k]=1;
				}
			}
		}
	}
	//输入输出
	int yy,mm,dd,T;
	cin>>T;
	while(T--)
	{
		cin>>yy>>mm>>dd;
		if(status[yy][mm][dd])
			cout<<"NO\n";
		else
			cout<<"YES\n";
	}
	return 0;
}