A - Calendar Game
Time Limit:2000MS Memory Limit:65536KB 64bit
IO Format:%lld & %llu
Submit Status Practice ZOJ
1024
Appoint description:
System Crawler (2015-08-02)
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
Output for the Sample Input
YES
NO
NO
题目大意是：
两个家伙在区域赛前夕闲的无聊，然后玩一种无限纠结的游戏，随即给定一个日期，每次只能移动day OR month..........
而且如果下一个月没有当前day的话， 你就不能移动month，比如1月31日，你只能移动day 使其到2月1日，而不能移动月让其到达2月31日，原因你懂的！
嗯，现在Adam开始YY了要！需要你来找一个必胜策略！（到达2001.11.4日就不能移动，无法移动的孩纸败，超过2001.11.4的也算输！)。
思路：没什么好的方法，找规律吧；
假设我站在先手的状态
（注意： a -> b, 如果b全部是必胜态， 则a是必败态， 如果存在一个b是必败态，则a是必胜态。）
必败 必胜
11.4 11.3
11.2 11.1
10.31 10.30
10.29 10.28
…………
10.5 10.4
10.3 10.2
10.1
9.30（聪明的人会选择转移到 10月1日）
9.28
9.29（聪明的人会选择跳到10.29）
…………
9.4 9.3
9.2 9.1
8.31 8.30（聪明的人会选择转移到 8月31日）
8.29 8.28
..........
8.5 8.4
8.3 8.2
8.1 7.31
7.30 7.29
..........
7.4 7.3
7.2 7.1
6.30（聪明的人会选择跳到7.30）
6.29 6.28
..........
6.5 6.4
6.3 6.2
6.1 5.31
5.30 5.29
..........
5.4 5.3
5.2 5.1
4.30（聪明的人 会选择转移到5月30日）
4.29 4.28
..........
4.5 4.4
4.3 4.2
4.1 3.31
3.30 3.29
3.28 3.27
.........
3.4 3.3
3.2 3.1
2.29 2.28
2.27 2.26
..........
2.3 2.2
2.1 1.31
1.30 1.29
1.28 1.27
..........
1.4 1.3
1.2 1.1
12.31 12.30
12.29 12.28
..........
12.3 12.2
12.1 11.30（聪明的人会选择转移到12月1日）
11.29（聪明的人会选择转移到12.29）
11.28 11.27
11.4 11.3
.....................................................
不管是月份加一，还是日期加一，都改变了奇偶性，只有两个特殊日期9月30日，和11月30日例外。
那么目标日期是11月4日，为奇数。初始日期如果为偶数的话，先者必胜。
可以大致看到其实胜负和年份是没有关系的说，年份影响的只是2月是否存在第29天……而29是必败点，28为必胜……
看上图…得…必胜点 月份+日期 == 偶数 （除去两个例外）
特殊考虑……
转载请注明出处：寻找&星空の孩子
#include<stdio.h>
int main()
{
int T;
int yy,mm,dd;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&yy,&mm,&dd);
if((mm+dd)%2==0)
printf("YES\n");
else if(dd==30&&(mm==11||mm==9))
printf("YES\n");
else printf("NO\n");
}
return 0;
}
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