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
Source
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#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int sg[2009][20][40];
int to[2009][20][40];
bool judge(int y,int m,int d){
if(y==2000||y==1996||y==1992){
if(m==2&&d==29) return 1;
else if(m==1||m==3||m==5||m==7||m==8||m==10||m==12){
if(d==31) return 1;
else return 0;
}
else if(m==4||m==6||m==9||m==11){
if(d==30) return 1;
else return 0;
}
return 0;
}
else{
if(m==2&&d==28) return 1;
else if(m==1||m==3||m==5||m==7||m==8||m==10||m==12){
if(d==31) return 1;
else return 0;
}
else if(m==4||m==6||m==9||m==11){
if(d==30) return 1;
else return 0;
}
return 0;
}
}
bool ok(int y,int m,int d){
if(y>2001||(y==2001&&m==11&&d>4)||(y==2001&&m==12)) return 0;
if(y==2000||y==1996||y==1992){
if(m==2&&d<=29) return 1;
else if(m==1||m==3||m==5||m==7||m==8||m==10||m==12){
if(d<=31) return 1;
else return 0;
}
else if(m==4||m==6||m==9||m==11){
if(d<=30) return 1;
else return 0;
}
return 0;
}
else{
if(m==2&&d<=28) return 1;
else if(m==1||m==3||m==5||m==7||m==8||m==10||m==12){
if(d<=31) return 1;
else return 0;
}
else if(m==4||m==6||m==9||m==11){
if(d<=30) return 1;
else return 0;
}
return 0;
}
}
int dfs(int y,int m,int d){
// cout<<y<<" "<<m<<" "<<d<<endl;
if(to[y][m][d]!=-1) return to[y][m][d];
if(y==2001&&m==11&&d==4){
to[y][m][d]=0;
return 0;
}
if(judge(y,m,d)==1){
int yy=y,mm=m,dd=d;
mm++;dd=1;
if(mm==13){
yy++;
mm=1;
}
if(dfs(yy,mm,dd)==0){
to[y][m][d]=1;
return 1;
}
}
else{
if(dfs(y,m,d+1)==0){
to[y][m][d]=1;
return 1;
}
}
int yy=y,mm=m,dd=d;
if(mm+1==13){
mm=1;yy++;
}
else mm++;
if(ok(yy,mm,dd)==1){
if(dfs(yy,mm,dd)==0){
to[y][m][d]=1;
return 1;
}
}
to[y][m][d]=0;
return 0;
}
int main(){
memset(to,-1,sizeof(to));
int y,m,d;
int T;
scanf("%d",&T);
// for(int y=1990;y<=2001;y++){
// for(int m=1;m<=12;m++){
// for(int d=1;d<=31;d++){
// memset(to,-1,sizeof(to));
// printf("%d %d %d\n",y,m,d);
// }
// }
// }
while(T--)
{
scanf("%d%d%d",&y,&m,&d);
if(dfs(y,m,d)) printf("YES\n");
else printf("NO\n");
}
}
HDU 1079 Calendar Game（sg）
原文：https://www.cnblogs.com/ellery/p/11695504.html
