HDU 1079

Calendar Game

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4503 Accepted Submission(s): 2735

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)

博弈论可真有意思，做了这么多博弈的题目，首先对于必败和必胜局面其实一定可以这么判断：
必胜局面：子局面中有一个必败局面就行（当然的了，你走到那个局面就好了）
必败局面：子局面中都是必胜局面（也很显然，意思就是不管你怎么走，对面都能赢，所以必败啊）
首先的想法是递推，略有些麻烦：

#include<bits/stdc++.h>
using namespace std;
int mon[15]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int sta[2155][20][50];
bool judge_rn(int y,int m)
{
    if(((y%4==0&&y%100)||y%400==0)&&m==2)
    return true;
    return false;
}
bool judge(int y,int m,int d)
{
    //首先考虑日期加1
    int dd=d+1,mm=m,yy=y;
    if(dd>(judge_rn(yy,mm)?mon[mm]+1:mon[mm]))
    {
        dd=1;
        mm++;
        if(mm>12) yy++;
    }
    if(sta[yy][mm][dd]==0) return true;
    //考虑月份加一
    dd=d,yy=y,mm=m+1;
    if(mm>12) {mm=1,yy++;}
    if((dd<=(judge_rn(yy,mm)?(mon[mm]+1):mon[mm]))&&(sta[yy][mm][dd]==0)) return true;
    return false;

}

void init(){
   memset(sta,-1,sizeof(sta));
   sta[2001][11][4]=0;
   for(int i=4;i<=30;i++)
    sta[2001][11][i]=0;
   for(int i=1;i<=31;i++)
    sta[2001][12][i]=0;
    sta[2001][11][2]=sta[2001][11][4]=0;
   for(int i=1;i<=31;i+=2)
    sta[2001][10][i]=0;
   for(int i=2001;i>=1900;i--)
   {
       for(int j=12;j>=1;j--)
       {
           for(int z=(judge_rn(i,j))?mon[j]+1:mon[j];z>=1;z--)
           {
               if(sta[i][j][z]!=-1) continue;
               sta[i][j][z]=judge(i,j,z)?1:0;
               //printf("sta[%d][%d][%d]=%d\n",i,j,z,sta[i][j][z]);
           }
       }
   }
   sta[2001][11][2]=sta[2001][11][4]=0;
   for(int i=1;i<=31;i+=2)
    sta[2001][10][i]=0;
}
int main()
{
   init();
   //cout<<judge_rn(2004,2)<<endl;
    int y,m,d;
    int t;
    cin>>t;
    while(t--)
    {
        cin>>y>>m>>d;
        if(sta[y][m][d]) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
}

然后发现一个结论：月份+天数为偶数则为必胜，为奇数则必败，然后9月30和11月30也是必胜。
其实可以把这个游戏看成在一个x,y坐标系中（x轴是按年份展开的月份，y轴是每个月天数）我们从某个起点出发，终点是2001年11月4号，每次x+1或者y+1，然后还有y+1>当月天数时相应的走法，会发现其实除了9月30和11月30外，总是无法改变（x+y）的奇偶性。所以结论成立。

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    int y,m,d;
    cin>>t;
    while(t--)
    {
        cin>>y>>m>>d;
        if((m+d)%2==0||(d==30&&(m==9||m==11)))
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
}