ZJU1024-Calendar Game（模拟）
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
Output for the Sample Input
YES
NO
NO
思路分析：
比赛双方都想赢，都会以最佳的方式改变日期，我们采用模拟的方法，谁先把日期变成2001年11月4日时，谁就赢了；
（1）数据结构
每个月最多的天数：
int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}
1900年1月1日至2001年11月4日之间的所有日期：
int is[102][13][32];
当然存放结果时，1900年记为第0年。
该数组的每个单元，其值为下列三种情况之一：
1) 1, 亚当赢
2) 0, 亚当输
3) -1, 还没有计算
（2）判断输赢
采用模拟的方法进行递归搜索。
1)递归结束条件
如果亚当将日期变过头，亚当就输了：
if(y>101||(y==101&&(m>11||(m==11&&d>4)))) return 0;
如果亚当把日期变成2001年11月4日，亚当就赢了：
if(y==101&&m==11&&d==4) return 1;
2)递归搜索
如果这一天已经计算过，直接返回结果：
return is[y][m][d];
否则进行递归搜索，把这一天亚当的输赢计算出来。
首先改变月份，因为改变月份的速度比改变天数要快：
int win=0;
if(m!=12)//如果不是12月份
{ //下一个月有同一天或者闰年
if(d<=day[m+1]||(d==29&&m==1&&(y%4)==0&&y!=0)
if(judge(y,m+1,d)==1) win=1;
}
//是12月份，明年1月份的同一天
else if(judge(y+1,1,d)==1) win=1;
接着考虑改变日期：
if(win==0)//改变月份还没有获胜
{
if(d<day[m]||((y%4)==0&&y!=0&&m==2&&d==28))
win=judge(y,m,d+1);
else if(m!=12)
win=judge(y,m+1,1);
else
win=judge(y+1,1,1);
}
is[y][m][d]=win;
因为下一轮是夏娃，所以亚当的获胜正好与夏娃相反。