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

3 
2001 11 3 
2001 11 2 
2001 10 3 

 

Sample Output

YES 
NO 
NO 

 

Source

Asia 2001, Taejon (South Korea) 

大神的解释很深刻，下面将大牛的分析链接粘贴如下

http://acm.hdu.edu.cn/discuss/problem/post/reply.php?postid=17033&messageid=1&deep=0

//个人总结下来就是要有两个预备知识点：
// 一、 sum=m+d，sum为偶数的时候，任意操作都能使sum 变为 奇数状态。
// 二、 奇数状态下 只有两个日子比较特殊，能够在进行一次操作之后，仍保持奇数状态。
//  题目分析：因为最后的日子是15 为奇数，所以为必败态。则由预备知识点一得，偶数
//  可以一步变成奇数，所以偶数点都为必胜点。又因为两个特殊点也能够一步变成奇数点，
//  所以也为必胜点。
//  演示过程如下（倒推）：
//  必败点（奇数）<-- 必胜点（偶数，特殊日子），所以在偶数日子の时候能够变为奇数
//  日子，即此时存在可能到达最终的奇数日子。奇数日子除了特殊日子，只能变为偶数点。
//  所以其他的奇数日子，不具备获胜的可能，即为必败点！ 

#include<stdio.h>
int main()
{
    int y, m,d,t;
    scanf("%d",&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;
}