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)

题目大意：

给定日期。轮流来，能够在日期的月上加1，或者在天数上加1 ，假设约数上加1无效，自己主动转化为在天数上加1，轮流来，问先手是否赢？

解题思路：

这非常明显是道博弈题，对于SG的性质定义

必胜态记为P。用数值0表示。当且仅当其后继都是 N，也就是SG()>0

必输态记为N，用数值1表示，当且仅当其后继存在P，也就是SG()=0

对于这题，全然不是必需这样用SG去推理。能够结合DP，用记忆化搜索划分为子问题，每一步取对自己最优的。

解题代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int dp[2100][15][40][2];
const int day[]={0,31,28,31,30,31,30,31,31,30,31,30,31};

int getday(int y,int m){
    if(m!=2) return day[m];
    else{
        if( ( y%4==0 && y%100!=0 ) || y%400==0 ) return day[m]+1;
        else return day[m];
    }
}

bool valid(int y,int m,int d){
    if(m>12){
        m=1;
        y++;
    }
    if( y>2001 || ( y==2001 && m>11 ) || (y==2001 && m==11 && d>4) || d>getday(y,m) ) return false;
    else return true;
}

int DP(int y,int m,int d,int f){
    if(m>12){
        m=1;
        y++;
    }
    if(getday(y,m)<d){
        d=1;
        m++;
        if(m>12){
            m=1;
            y++;
        }
    }

    if(y>=2001 && m>=11 && d>=4) return 1-f;
    if(dp[y][m][d][f]!=-1) return dp[y][m][d][f];

    int ans;
    if(f==0){
        ans=1;
        if( valid(y,m+1,d) && DP(y,m+1,d,1-f)<ans ) ans=DP(y,m+1,d,1-f);
        if( DP(y,m,d+1,1-f)<ans ) ans=DP(y,m,d+1,1-f);
    }else{
        ans=0;
        if( valid(y,m+1,d) && DP(y,m+1,d,1-f)>ans ) ans=DP(y,m+1,d,1-f);
        if( DP(y,m,d+1,1-f)>ans ) ans=DP(y,m,d+1,1-f);
    }
    return dp[y][m][d][f]=ans;
}

int main(){
    memset(dp,-1,sizeof(dp));
    int t,y,m,d;
    cin>>t;
    while(t-- >0){
        cin>>y>>m>>d;
        if ( DP(y,m,d,0) ) cout<<"NO"<<endl;
        else cout<<"YES"<<endl;
    }
    return 0;
}

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