HDOJ 1079 && POJ 1082 Calendar Game (博弈: 暴力枚举所有状态的P\N)

//1079 Calendar Game 博弈 暴力枚举所有状态的P\N，其实网上有非常简单的做法

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<stack>
using namespace std;

bool vis[105][15][35];
int mon[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

struct node{
	int y;
	int m;
	int d;
};

node GetNextDay(node p){
	int y = p.y;
	if(y%4==0&&y%100!=0||y%4==0&&y%400==0)	mon[2] = 29;
	else mon[2] = 28;
	++p.d;
	if(p.d > mon[p.m]){
		p.d = 1;
		++p.m;
		if(p.m > 12){
			p.m = 1;
			++p.y;
		}
	}
	return p;
}

node GetLastDay(node p){
	int y = p.y;
	if(y%4==0&&y%100!=0||y%4==0&&y%400==0)	mon[2] = 29;
	else mon[2] = 28;
	--p.d;
	if(p.d==0){
		--p.m;
		if(p.m == 0){
			p.d=31;
			p.m=12;
			--p.y;
		}
		else
			p.d = mon[p.m];
	}
	return p;
}

node GetSameDayInNextMon(node p){
	node q;
	q.y=q.m=q.d=-1;//不存在
	int y = p.y;
	if(y%4==0&&y%100!=0||y%4==0&&y%400==0)	mon[2] = 29;
	else mon[2] = 28;
	++p.m;
	if(p.m >12){
		p.m = 1;
		++p.y;
	}
	if(p.d<=mon[p.m])
		return p;
	return q;
}

node GetSameDayInLastMon(node p){
	node q;
	q.y=q.m=q.d=-1;//不存在
	int y = p.y;
	if(y%4==0&&y%100!=0||y%4==0&&y%400==0)	mon[2] = 29;
	else mon[2] = 28;
	--p.m;
	if(p.m == 1){
		p.m = 12;
		--p.y;
	}
	if(p.d<=mon[p.m])
		return p;
	return q;
}

bool check(node p){
	if(p.y > 2001 || p.y < 1900)
		return false;
	if(p.y == 2001 && p.m == 12)
		return false;
	if(p.y == 2001 && p.m == 11 && p.d > 4)
		return false;
	return true;
}

void init(){
	memset(vis,false,sizeof(vis));//P:false N:true
	node p,a,b;
	p.y=2001; p.m=11; p.d=4;
	while(1){
		p = GetLastDay(p);
		if(p.y == 1899)
			break;
		a = GetNextDay(p);
		b = GetSameDayInNextMon(p);
		if(check(a) && vis[a.y-1900][a.m][a.d] == false)
			vis[p.y-1900][p.m][p.d] = true;
		else if(check(b) && vis[b.y-1900][b.m][b.d] == false)
			vis[p.y-1900][p.m][p.d] = true;
	}
}
int main(){
	init();
	int T,y,m,d;
	scanf("%d",&T);
	while(T--){
		scanf("%d %d %d",&y,&m,&d);
		puts(vis[y-1900][m][d]?"YES":"NO");
	}
	return 0;
}