Calendar Game
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6052 Accepted Submission(s): 3618
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
2001 11 3
2001 11 2
2001 10 3
Sample Output
YES
NO
NO
NO
NO
Source
题意：
从输入的日期开始，可以向后跳，一共两种跳跃方式，1.进入后一天，2，进入下一月的同一天，如果下个月没有这个日期，那么这个方法不可用。
跳跃到2001年11月4号即为获胜。
思路：
设定2001年11月4日为先手必败。
从2001年11月4号，枚举到1990年1月1日，判断当前日期可以向后跳的日期是否为先手必败态，如果有任意一种状态是先手必败态，则当前是先手必胜态，否则当前是先手必败态。如果存在第二种跳跃不可用的情况，则当做必胜态处理。
代码：
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = ;
const int inf = 2.1e9;
const ll Inf = ;
const int mod = ;
const double eps = 1e-;
const double pi = acos(-);
int date[][][];
bool judge(int x)
{
if(x%==&&x%!=){return false;}
else return x%==;
}
int dd[]={,,,,,,,,,,,,}; bool check1(int y,int m,int d)
{
m++;
if(m==){y++;m=;}
if(judge(y)){dd[]++;}
if(d>dd[m]){
if(judge(y)){dd[]--;}
return true;
}
if(judge(y)){dd[]--;}
return date[y][m][d];
} bool check2(int y,int m,int d)
{
if(judge(y)){dd[]++;}
d++;
if(d>dd[m]){
m++;
if(m==){
y++;
m=;
}
d=;
}
if(judge(y)){dd[]--;} return date[y][m][d];
} void solve(int y,int m,int d)
{
bool flag1=check1(y,m,d);
bool flag2=check2(y,m,d);
if(!flag1||!flag2){date[y][m][d]=true;}
else date[y][m][d]=false;
} int main()
{
// ios::sync_with_stdio(false);
// freopen("in.txt","r",stdin); for(int i=;i<=;i++){
for(int j=;j<=;j++){
for(int k=;k<=;k++){
date[i][j][k]=true;
}
}
} int y,m,d;
y=,m=,d=;
date[y][m][d]=false;
while(true){
if(y==&&m==&&d==){break;}
d--;
if(d==){
m--;
if(m==){
y--;
m=;
}
d=dd[m];
if(judge(y)&&m==){d++;}
}
solve(y,m,d);
} int T;
scanf("%d",&T);
while(T--){
scanf("%d%d%d",&y,&m,&d);
if(date[y][m][d]){printf("YES\n");}
else printf("NO\n");
} return ;
}
抱着比WA的心态交了一发，没想到居然A了，哈哈哈哈或或。