【hdu 3863】No Gambling
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65568/32768 K (Java/Others)
Total Submission(s): 1716 Accepted Submission(s): 1243
Problem Description
One day, Flyvan introduced a new game to his two friends, Oregon Maple and Grape Skin. The game is quite simple. Given an N-sized grids, like the figure A shown below (as N = 4). The blue points are the places the first player can choose, and the red points are the places the second player can choose.
In the game, the two players take turns to choose two points to get connected by a stick. The two chosen points’ distance should be exactly one-unit length. The first player’s goal is to create a ‘bridge’ that connects a most left point and a most right point. The second player’s goal is to create a ‘bridge’ that connects a most top point and a most bottom point. Figure B shows a possible result (the first player won). In addition, the stick shouldn’t get crossed.
Now Flyvan will give the number N, and his two friends will play the game. Both of the two players will choose the best strategy. You can bet on one player, and if he wins the game, you’ll get twice money you bet~
Since you are a talented programmer, you surely won’t just do gambling. Please write a program to find out the player who you should bet on. As Oregon Maple is elder, he will always play first.
Input
Each line of the input is an integer N (2 <= N <= 270000), which indicated the number Flyvan chose. The end-of-file is denoted by a single line containing the number -1.
Output
If you think the first player will win, please output “I bet on Oregon Maple~”, else please output “I bet on Grape Skin~”.
Sample Input
2
-1
Sample Output
I bet on Oregon Maple~
【题目链接】:http://acm.hdu.edu.cn/showproblem.php?pid=3863
【题解】
画了N=3和N=4的情况;
(奇数和偶数的棋盘都是一样的);
(都是相同的模式:一个横着放一个竖着放);
模拟了一下发现红色的线堵不住蓝色的线。
不信你自己试试。。
所以都输出Maple就好。。(先手赢);
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
int n;
int main()
{
//freopen("F:\\rush.txt","r",stdin);
while (~scanf("%d",&n))
{
if (n==-1)
break;
puts("I bet on Oregon Maple~");
}
return 0;
}
【hdu 3863】No Gambling的更多相关文章
- 【数位dp】【HDU 3555】【HDU 2089】数位DP入门题
[HDU 3555]原题直通车: 代码: // 31MS 900K 909 B G++ #include<iostream> #include<cstdio> #includ ...
- 【HDU 5647】DZY Loves Connecting(树DP)
pid=5647">[HDU 5647]DZY Loves Connecting(树DP) DZY Loves Connecting Time Limit: 4000/2000 MS ...
- -【线性基】【BZOJ 2460】【BZOJ 2115】【HDU 3949】
[把三道我做过的线性基题目放在一起总结一下,代码都挺简单,主要就是贪心思想和异或的高斯消元] [然后把网上的讲解归纳一下] 1.线性基: 若干数的线性基是一组数a1,a2,a3...an,其中ax的最 ...
- 【HDU 2196】 Computer(树的直径)
[HDU 2196] Computer(树的直径) 题链http://acm.hdu.edu.cn/showproblem.php?pid=2196 这题可以用树形DP解决,自然也可以用最直观的方法解 ...
- 【HDU 2196】 Computer (树形DP)
[HDU 2196] Computer 题链http://acm.hdu.edu.cn/showproblem.php?pid=2196 刘汝佳<算法竞赛入门经典>P282页留下了这个问题 ...
- 【HDU 5145】 NPY and girls(组合+莫队)
pid=5145">[HDU 5145] NPY and girls(组合+莫队) NPY and girls Time Limit: 8000/4000 MS (Java/Other ...
- 【hdu 1043】Eight
[题目链接]:http://acm.hdu.edu.cn/showproblem.php?pid=1043 [题意] 会给你很多组数据; 让你输出这组数据到目标状态的具体步骤; [题解] 从12345 ...
- 【HDU 3068】 最长回文
[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=3068 [算法] Manacher算法求最长回文子串 [代码] #include<bits/s ...
- 【HDU 4699】 Editor
[题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=4699 [算法] 维护两个栈,一个栈放光标之前的数,另外一个放光标之后的数 在维护栈的同时求最大前缀 ...
随机推荐
- 5.9 enum--支持枚举类型
enum模块提供了枚举类型的支持.枚举类型是由一个名称和一个统一值来组成.值是常量的值.它们之间能够通过名称进行比較和引用,还能够迭代訪问. 5.9.1 模块内容 本模块主要定义了两种枚举类型:Enu ...
- 【JavaScript】--JavaScript总结一览无余
对于 北风网李炎恢老师的JavaScript的视频也真的是醉了.视频整体来说结构清晰.内容比較简单.JS是一种灵活,开放的语言,语法规则并没有那么的死板.非常easy让人接受. JS的基础部分跟C#类 ...
- .v 和 .sdf
DC输出的.v(网表?)和.sdf(储存的是延时的信息) 用于后仿真
- C# 报表
报表技术 1.OWC控件的使用 OWC是office web Components 是组件对象模型(COM)控件的集合,可用于将电子表格,图表和数据库发布到网站上. 在Office2003以后的版本中 ...
- LuceneIndexFileDeleter会保留初始的commit
给实时索引加入了merge策略,持续更新时发现有做merge,但索引文件夹中的段数远远大于RealTimeIndexWriter中的段数,就是有些merge的段应该删除,文件夹中没有删除.而关闭sea ...
- LeetCode Algorithm 02_Add Two Numbers
You are given two linked lists representing two non-negative numbers. The digits are stored in rever ...
- Altium Designer导入pcb原件之后都是绿的
- Testfan软件测试社区
1. http://ask.testfan.cn/article/902 Appium 服务端安装-windows2. http://ask.testfan.cn/article/1078 最新 ...
- mahout測试朴素贝叶斯分类样例
对于这个測试建议大家先理解原理,这里我画了例如以下的示意图 接下来就依照例如以下的细节来输入指令測试: 首先前提是Hadoop安装并启动,mahout已经安装了. <strong>< ...
- 9.5 Binder系统_驱动情景分析_transaction_stack机制
参考文章:http://www.cnblogs.com/samchen2009/p/3316001.html test_server服务进程可能有多个线程,而在发送数据的时候handle只表示了那个进 ...