HDU 1079 Calendar Game(規律博弈)
阿新 • • 發佈:2018-07-24
days == tps turn NPU november val cal and
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
Sample Output
題目鏈接:
https://cn.vjudge.net/problem/HDU-1079
題目:
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.
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
YES NO NO
解題思路:
已知兩個人的博弈方式只能是days+1或者month+1,可得days+month之和總是在奇數和偶數之間變換,所以目標是取到11+4為奇數,那就必須要面對偶數才能贏。所以當給出的日期是偶數時,先手必贏,奇數時,先手必輸,不過有兩個特例是,9.30和11.30,他們是奇數,但是他們的兩個後繼有一個是還是奇數,能造成先手還是面臨偶數,所以也行。
1 #include<cstdio> 2 3 int main() 4 { 5 int T; 6 scanf("%d", &T); 7int y, m, d; 8 while(T--){ 9 scanf("%d%d%d", &y, &m, &d); 10 if((m + d) % 2 == 0 || (d == 30) && (m == 9 || m == 11)) 11 printf("YES\n"); 12 else 13 printf("NO\n"); 14 } 15 return 0; 16 }
HDU 1079 Calendar Game(規律博弈)