Calendar Game
Time limit:5000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 3628 Accepted Submission (s): 2163
Problem Descriptionadam 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, and 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's simple:from a current date, a player in his/her turn can move either to the NE XT Calendar date or the same day of the next month. When the next month does not has the same day, the player moves is only to the next calendar date. For example, from December, 1924, your can move either to December, 1924, the next calendar date, or January 19, 1925 , the same day of the next month. From January-2001, however, you can move only to February 1, 2001, because February, 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 This decides whether, given an initial date, Adam, the first mover, have a winning strategy.
For the This game, you need to identify leap years, where February have. In the Gregorian calendar, leap years occur in years exactly divisible by four. So, 1993, 1994, and 1995 is not a leap years, while 1992 and 1996 is leap years. Additionally, the years ending with XX is leap years only if they is divisible by 400. So, 1700, 1800, 1900, 2100, and 2200 is not a leap years, while, and 2400 is leap years.
Inputthe input consists of T test cases. The number of test cases (T) is given on the first line of the input. Each test case was 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 is randomly chosen from the interval between January 1, 1900 and November 4, 2001.
Outputprint exactly one line for each test case. The line should contain the answer ' YES ' or ' NO ' to the question of whether Adam had a winning strategy against Eve. Since we have t-Test cases, your program should output totally T lines of "YES" or "NO".
Sample INPUT3 2001 11 3 2001 11 2 2001 10 3
Sample Outputyes No no I want to say this question is a divine question (at present this level looks) ... There should be other ideas, but this is the solution to the idea that I did not expect ... Idea: To consider month and day as a whole sum=month+day, according to the rules of the topic, you can jump to the next day of the current date, the latter jumps to the next month corresponding to the current day, that is, month+1 or day+1, then the parity of sum has changed, November 4 The corresponding sum is odd, then to win, it always throws the odd number to the latter, if it is a normal date (not every month boundary) sum is odd, then the throw must be even, sum is even, throw must be odd. Boundary words: (1.31), (2.1) (2.28), (3.28) | | (3.1) Common year (2.29), (3.29) | | (3.1) Leap year (3.31), (4.1) (4.30), (5.1) | | (5.30) (5.31), (6.1) (6.30), (7.1) | | (7.30) (7.31), (8.1) | | (8.31) (8.31), (9.1) (9.30), (10.1) | | (10.30) (10.31) (11.1)(11.30), (12.1) | | (12.30)(12.31), (1.1) | | (1.31) where odd numbers can be thrown with odd numbers (9.30) and (11.30), (9.30) precursors are (9.29) and (8.30), (11.30) precursors are (10.30) and (11.29), that is to say that these two dates can be bypassed, Then there is the current position on both dates that the person must be able to win. In summary, if the former initial position of sum is even, the former must be able to win, always let the latter go when the position in the odd number, go after the even, if the initial position is (9.30) or (11.30), he can also throw an odd number.
#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespacestd;intMain () {intN; scanf ("%d",&N); while(n--) { intYear,mon,day; scanf ("%d%d%d",&year,&mon,&Day ); if((mon+day)%2==0|| ((mon==9|| mon== One) &&day== -)) printf ("yes\n"); Elseprintf ("no\n"); } return 0;}
Hdu_1079_ thinking problem
