HDU1079 Calender Game
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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.
InputThe 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.
OutputPrint 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
题解:任何一部移动都会改变日期的奇偶性(除了9.30和11.30);
参考代码:
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 "); 12 else printf("NO "); 13 } 14 return 0; 15 }
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