Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3935 Accepted Submission(s): 2374
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
題意：從1900.1.1到2001.11.04中輸入一個日期，兩個人輪流移動日期，每次移動可以到當前日期的下一天或者下一個月的當前天，如果不存在，只能移動到下一天，比如，1924.12.19可以移動到1924.12.20或者1925.1.19，
但是2001.1.31只能移動到2001.2.1,因爲2001.2.31是不存在的。。最後誰先移到2001.11.04誰贏。。
思路：假設2001年11月3號是先手的必勝態，則11.02是必輸態，因爲沒有2002.11.02,所以只能移動到下一天，於是使對方達到了必勝態（11.03），那麼2001.11.02是先手的必敗態，依次往後推，每一次，都要看如果往後走一天或者到下個月相同一天，如果這兩種選擇都使得對方到達必勝態，那麼當前點就是必輸態，否則如果其中一種選擇使得對方到達必輸態，那麼當前點就是必勝態，，，
最後發現，當月份和天數互爲奇偶時，一定是必輸態，，除了9月30號與11月30號，，
以下AC代碼：
#include<stdio.h>
int main()
{
int t;
int y,m,d;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&y,&m,&d);
if((d+m)%2==0)
printf("YES\n");
else if(m==9 && d==30)
printf("YES\n");
else if(m==11 && d==30)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}