HDU-1079——Calendar Game

- 日理万妓 2022-04-06 05:27 155阅读 0赞

#                                          Calendar Game #

### 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 ###

2001 11 3

2001 11 2

2001 10 3

### Sample Output ###

YES

### 题意描述： ###

两人轮流数日期，每次可以数到下一天，或者数到下个月的同日期。亚当开始先数，谁数到2001年11月4日谁就获胜了，问亚当是否能够获胜，获胜输出YES否则输出NO。

### 解题思路： ###

由时间从后往前推，可知2001 11 3 为必胜点2001 11 2为必败点，2001   10  4为必胜点2001 10 3为必败点……可推出当月和日加到一起为偶数时，先手必胜，当月和日为11  30或9  30时也是先手获胜。其他情况先手失败。

### 程序代码： ###

#include<stdio.h>
    int main()
    {
    	int t,year,month,day;
    	scanf("%d",&t);
    	while(t--)
        {
            scanf("%d%d%d",&year,&month,&day);
            if((month+day)%2==0)
                printf("YES\n");
            else if(day==30&&(month==11||month==9))
                printf("YES\n");
            else
                printf("NO\n");
        }
        return 0;
    }