hdu 1079 Calendar Game 博弈论

 
 

Calendar Game

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2320    Accepted Submission(s): 1315


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.

每次只能前进一天或者前进一个月,问是谁先到2001年11月4日。

明显是个博弈论问题,这里采用奇偶分析法。

11 + 4 = 15 为奇数。

很明显每次操作时都会是日月之和的奇偶性发生改变,所以很显然当为初始和为偶数是,先手肯定有必胜策略。

然而当初始和为奇数时有两个例外情况。那就是9月30和11月三十,这两个日期为起始日期时,先手只要前进一天,那么这时的和的奇偶性仍为奇数,先手仍有必胜策略。

那么为什么初始和为偶数时不考虑这两个日期是否会走到呢,那是因为先手可以保证不会过程中不会走到这两个日期。

从而题目已破

下面贴出代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <cctype>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <algorithm>

using namespace std;

int main()
{
    int n;
    scanf("%d", &n);
    while(n--)
    {
        int a,b,c;
        scanf("%d%d%d", &a, &b, &c);
        if((b + c) % 2 == 0)
            printf("YES\n");
        else if((b == 9 || b == 11) && c == 30)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}


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