Calendar Game
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2071 Accepted Submission(s): 1185
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.
2001 11 3
2001 11 2
2001 10 3
NO
NO
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#pragma comment(linker,"/STACK:1024000000,1024000000")
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
int dp[2020][14][33];
int day[2][13]={{0,31,28,31,30,31,30,31,31,30,31,30,31},
{0,31,29,31,30,31,30,31,31,30,31,30,31}};
int ispri(int y,int m ,int d){
if(y%400==0||(y%100&&y%4==0))return 1;
return 0;
}
int dfs(int y,int m,int d){
if(dp[y][m][d]!=-1)return dp[y][m][d];
if(y==2001&&m==11&&d==4)return dp[y][m][d]=0;
if(y>2001)return dp[y][m][d]=1;
if(y==2001&&m>11)return dp[y][m][d]=1;
if(y==2001&&m==11&&d>4)return dp[y][m][d]=1;
int my,mm,md;
my=y;mm=m;md=d+1;
int temp=ispri(y,m,d);
if(md>day[temp][mm]){
mm++;md=1;
if(mm>12)mm=1,my++;
}
if(dfs(my,mm,md)==0)return dp[y][m][d]=1;
my=y;mm=m+1;md=d;
if(mm>12)mm=1,my++,temp=ispri(my,mm,md);
if(md<=day[temp][mm]&&dfs(my,mm,md)==0)return dp[y][m][d]=1;
return dp[y][m][d]=0;
}
int main()
{
int tcase,y,m,d;
scanf("%d",&tcase);
while(tcase--){
mem(dp,-1);
scanf("%d%d%d",&y,&m,&d);
if(dfs(y,m,d))printf("YES\n");
else printf("NO\n");
}
return 0;
}