hdu1079
来源:互联网 发布:淘宝网宝贝主图尺寸 编辑:程序博客网 时间:2024/05/29 11:18
Calendar Game
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4583 Accepted Submission(s): 2786
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.
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思路:可以发现无论是改变日期还是改变月份,奇偶性都会发生变化(除了9月30日和11月30日),因为11月4日是奇数,如果为偶数先手一定赢。代码:#include <map>#include <set>#include <queue>#include <vector>#include <stack>#include <cmath>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;#define long long int ll#define ma(a) memset(a,0,sizeof(a))int SG[10005],f[105];bool S[10005];int main(){ int T; scanf("%d",&T); while(T--) { int x,y,z; scanf("%d %d %d",&x,&y,&z); if((y+z)%2==0||y==9&&z==30||y==11&&z==30) { cout<<"YES"<<endl; } else cout<<"NO"<<endl; } return 0;}
阅读全文
1 0
- Hdu1079
- hdu1079
- hdu1079 Calendar Game
- hdu1079 博弈 奇偶规律
- hdu1079 Calendar Game
- hdu1079 Calendar Game(博弈)
- hdu1079博弈(规律)
- hdu1079 Calendar Game
- hdu1079 找规律/博弈
- HDU1079 POJ1082 Calendar Game【博弈】
- HDU1079 Calendar Game(博弈)
- HDU1079 Calendar Game SG函数
- [POJ1082]Calendar Game & HDU1079 Calendar Game
- 【HDU1079】Calendar Game(博弈,PN状态枚举)
- 博弈小结:HDU1907 HDU 1760 HDU1079
- 【HDU1079】Calendar Game(博弈论,博弈找规律)
- HDU1079 Calendar Game(找规律+特殊点判断博弈)
- 23种设计模式之——装饰模式
- oracle 12c 中scott账户与表问题
- HDU 6034 Balala Power!(贪心)
- 螺旋二维数组
- 排序算法大集成
- hdu1079
- [第五季]html格式化标签
- 数论-HDU1852
- 上楼梯问题
- 设计模式——单例模式
- c++实现单例类(懒汉与饿汉)
- kindeditor编辑器提示:PHP Warning: File upload error
- 常用的IntelliJ IDEA 快捷键
- 顺序表---模板