hdu 1079 Calendar Game(博弈找规律)

CalendarGame策略解析
本文介绍了一个名为CalendarGame的游戏,该游戏设定在1900年1月1日至2001年11月4日的时间范围内,由两名玩家Adam和Eve轮流进行。文章详细解释了游戏规则,并给出了一种快速判断初始日期下Adam是否拥有必胜策略的方法。

                                     Calendar Game

                                          Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                     Total Submission(s): 5242    Accepted Submission(s): 3193


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.
 

Input
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.
 

Output
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".
 

Sample Input
3 2001 11 3 2001 11 2 2001 10 3
 

Sample Output
YES NO NO
 
//我是从最后一个日期开始往前推的,先从(2001 12 2)开始一步步调整时间往前算,发现:我只要是月份加天数是偶数,那么就一定是Adam赢,但是出现了特例就是
//当月份是11或者是9时,如果此时的天数是30,那么也是Adam赢。
#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<vector>
#include<cmath>
using namespace std;

int main(){
    int t;
    int y, m, d;
    cin >> t;
    while(t--){
        scanf("%d%d%d", &y, &m, &d);
        if((m+d)%2==0 || (m==9 && d==30) || (m==11 && d==30))
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}

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