博弈论 ---- 1079 Calendar Game
2012-03-06 08:56
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[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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".
[align=left]Sample Input[/align]
3 2001 11 3 2001 11 2 2001 10 3
[align=left]Sample Output[/align]
YES NO NO
题意: 输入N,表示N组数据,之后输入N组数据,每组由年月日, 谁先到达2001.11.4日的为胜,超过的为败!Adam是先开始的人, Adam可能赢的情况
分析: 改变 月+1,日不变! 或者日+1,月不变,都能改变奇偶性,而11月4日为11+4是奇数,所以只要开始的时候
月+日 == 偶数 就可以使得Adam获胜,然而还有两个特殊时刻,11月30日, 虽然11+30 == 奇数, 但你要么移到12月1日 == 奇数, 或者移到12月30日 == 偶数!
虽然 9+30 == 39, 39 是奇数,然而你要么移到10月1日, 为奇数, 要么移到10月30日为偶数! 然而题目问的是 可能 赢
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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".
[align=left]Sample Input[/align]
3 2001 11 3 2001 11 2 2001 10 3
[align=left]Sample Output[/align]
YES NO NO
题意: 输入N,表示N组数据,之后输入N组数据,每组由年月日, 谁先到达2001.11.4日的为胜,超过的为败!Adam是先开始的人, Adam可能赢的情况
分析: 改变 月+1,日不变! 或者日+1,月不变,都能改变奇偶性,而11月4日为11+4是奇数,所以只要开始的时候
月+日 == 偶数 就可以使得Adam获胜,然而还有两个特殊时刻,11月30日, 虽然11+30 == 奇数, 但你要么移到12月1日 == 奇数, 或者移到12月30日 == 偶数!
虽然 9+30 == 39, 39 是奇数,然而你要么移到10月1日, 为奇数, 要么移到10月30日为偶数! 然而题目问的是 可能 赢
# include <iostream> using namespace std; int main() { int n, Y, M, D; cin>>n; while (n--) { cin>>Y>>M>>D; if( (D + M) % 2 == 0) cout<<"YES"<<endl; else if (M == 9 && D == 30) cout<<"YES"<<endl; else if (M == 11 && D == 30) cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0; }
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