POJ 1082 Calendar Game

http://poj.org/problem?id=1082

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 file. 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

翻黄历：Adam和Eve玩游戏，在1900年的1月1号到2001年的11月4号之间随机选一个日期，两人轮流增加日期，Adam先手。规定只能往此日期的下一天移动或者下个月的这一天移动（如果下个月没有这一天，则不能移动）。最终谁先移动到2001年的11月4号，谁就获胜。现给定日期，判断Adam是否有取胜策略。

如果抽到最后一天肯定是必败的，那就用这个必败态往前面递推。状态按照日期先后排列，每个current状态取决于两个next状态。只要两个next状态中有一个为必败态，则Adam可以选择移动到该next状态，让Eve落败；否则Adam没有选择，只能移动到成功态，将胜利的果实拱手让给Eve。

那么只需将这102年的日期都算出来，每个日期的状态都预先算好，就可以应对题目的多case了。

const int MAX_D = 365 * 105;
const int ds[] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

bool is_leap_year(int year)
{
    return year % 400 == 0 || (year % 100 != 0 && year % 4 == 0);
}

struct date
{
    int y, m, d;
    date() { }
    date(int y, int m, int d) : y(y), m(m), d(d) { }
    bool operator<(const date &other) const
    {
        return y != other.y ? y < other.y : m != other.m ? m < other.m : d < other.d;
    }
};

int D;
date days[MAX_D];
bool win[MAX_D];

void init_calendar()
{
    for (int y = 1900; y <= 2001; y++)
    {
        for (int m = 0; m < 12; m++)
        {
            for (int d = 0, ed = ds[m] + ((m == 1 && is_leap_year(y)) ? 1 : 0); d < ed; d++)
            {
                days[D++] = date(y, m, d);
            }
        }
    }
}

void solve()
{
    memset(win, true, sizeof(win));
    int start = lower_bound(days, days + D, date(2001, 10, 3)) - days;
    win[start] = false;
    for (int i = start - 1; i >= 0; --i)
    {
        win[i] = false;
        if (!win[i + 1])        // 如果两个next状态中有一个必败状态
        {
            win[i] = true;      // 则current状态为必胜状态
            continue;
        }
        else
        {
            date nxt = days[i];
            nxt.m++;
            if (nxt.m == 12)
            {
                nxt.y++;
                nxt.m = 0;
            }
            if (binary_search(days, days + D, nxt) &&
                    !win[lower_bound(days, days + D, nxt) - days])// 如果两个next状态中有一个必败状态
            {
                win[i] = true;  // 则current状态为必胜状态
            }
        }
    }
}

///////////////////////////SubMain//////////////////////////////////
int main(int argc, char *argv[])
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    init_calendar();
    solve();
    int T;
    scanf("%d", &T);
    while (T--)
    {
        int y, m, d;
        scanf("%d%d%d", &y, &m, &d);
        m--;
        d--;
        puts(win[lower_bound(days, days + D, date(y, m, d)) - days] ? "YES" : "NO");
    }
#ifndef ONLINE_JUDGE
    fclose(stdin);
#endif
    return 0;
}

http://www.acmerblog.com/POJ-1082-Calendar-Game-blog-258.html