HDOJ 1079.Calendar Game 找规律博弈

Calendar Game

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4793    Accepted Submission(s): 2922

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

找规律，不管是月份加一，还是日期加一，都改变了奇偶性。

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。如果是奇数，必败。

有两个特殊奇数日期9月30日，和11月30日例外（不管该年是否为润年，2月28\ 29同样一步都会发生正常奇偶变化）。

考虑特殊是日期，两个特殊日期本来为奇数，可以做到移动一步还是奇数，那么必胜点与必败点发生变化。

那么会不会在中途经过这两个日期呢。

如果本来为偶数，如果经过特殊日期就会改变奇偶，从必胜到必败。作为先手，不会主动进入特殊日期，而后者不可能从奇数依旧到达特殊日期的奇数。

如果本来为奇数，同样先手想赢，但是不可能进入特殊日期。保持奇偶性交替变化。

这样一来只可能是初始为特殊日期，否则中途不可能出现特殊日期

 代码：

[cpp]  view plain  copy

1. #include<iostream>  
2. #include<string.h>  
3. using namespace std;  
4. int main()  
5. {  
6.     int y,m,d , t;  
7.     scanf("%d",&t);  
8.     while(t--)  
9.     {  
10.          scanf("%d%d%d",&y,&m,&d);  
11.          if((m+d)%2==0 || m==9 && d==30 || m==11 && d==30)  printf("YES\n");  
12.          else printf("NO\n");  
13.     }  
14.     return 0;  
15. }