Balanced Game

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 309    Accepted Submission(s): 275

Problem Description
Rock-paper-scissors is a zero-sum hand game usually played between two people, in which each player simultaneously forms one of three shapes with an outstretched hand. These shapes are "rock", "paper", and "scissors". The game has only three possible outcomes other than a tie: a player who decides to play rock will beat another player who has chosen scissors ("rock crushes scissors") but will lose to one who has played paper ("paper covers rock"); a play of paper will lose to a play of scissors ("scissors cut paper"). If both players choose the same shape, the game is tied and is usually immediately replayed to break the tie.

Recently, there is a upgraded edition of this game: rock-paper-scissors-Spock-lizard, in which there are totally five shapes. The rule is simple: scissors cuts paper; paper covers rock; rock crushes lizard; lizard poisons Spock; Spock smashes scissors; scissors decapitates lizard; lizard eats paper; paper disproves Spock; Spock vaporizes rock; and as it always has, rock crushes scissors.

Both rock-paper-scissors and rock-paper-scissors-Spock-lizard are balanced games. Because there does not exist a strategy which is better than another. In other words, if one chooses shapes randomly, the possibility he or she wins is exactly 50% no matter how the other one plays (if there is a tie, repeat this game until someone wins). Given an integer N, representing the count of shapes in a game. You need to find out if there exist a rule to make this game balanced.

 
Input
The first line of input contains an integer t, the number of test cases. t test cases follow.
For each test case, there is only one line with an integer N (2≤N≤1000), as described above.

Here is the sample explanation.

In the first case, donate two shapes as A and B. There are only two kind of rules: A defeats B, or B defeats A. Obviously, in both situation, one shapes is better than another. Consequently, this game is not balanced.

In the second case, donate two shapes as A, B and C. If A defeats B, B defeats C, and C defeats A, this game is balanced. This is also the same as rock-paper-scissors.

In the third case, it is easy to set a rule according to that of rock-paper-scissors-Spock-lizard.

 
Output
For each test cases, output "Balanced" if there exist a rule to make the game balanced, otherwise output "Bad".
 
Sample Input
3
2
3
5
 
Sample Output
Bad
Balanced
Balanced
 
Source
 
 
 
解析:维持平衡,即每种策略的胜率均为50%。而每种手势对其他手势不是必胜就是必败,要维持平衡,显然手势的种数必须是奇数;否则不能平衡。当手势种数为奇数时,很容易构造出平衡的情况。因此,当手势种数为奇数可以平衡,为偶数时不能平衡。
 
 
 
#include <cstdio>

int main()
{
int t, n;
scanf("%d", &t);
while(t--){
scanf("%d", &n);
printf(n&1 ? "Balanced\n" : "Bad\n");
}
return 0;
}

  

HDU 5882 Balanced Game的更多相关文章

  1. hdu 5882 Balanced Game 2016-09-21 21:22 80人阅读 评论(0) 收藏

    Balanced Game Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) To ...

  2. HDU 5882 Balanced Game (水题)

    题意:问 nnn 个手势的石头剪刀布游戏是否能保证出每种手势胜率都一样. 析:当每种手势的攻防个数完全相等才能保证平衡,所以容易得出 nnn 是奇数时游戏平衡,否则不平衡. 也就是说打败 i 的和 i ...

  3. HDU 3709 Balanced Number (数位DP)

    Balanced Number Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) ...

  4. hdu 3709 Balanced Number(平衡数)--数位dp

    Balanced Number Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) ...

  5. HDU 6299.Balanced Sequence-贪心、前缀和排序 (2018 Multi-University Training Contest 1 1002)

    HDU6299.Balanced Sequence 这个题就是将括号处理一下,先把串里能匹配上的先计数去掉,然后统计左半边括号的前缀和以及右半边括号的前缀和,然后结构体排序,然后遍历一遍,贪心策略走一 ...

  6. hdu 3709 Balanced Number(数位dp)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3709 题意:给定区间[a,b],求区间内平衡数的个数.所谓平衡数即有一位做平衡点,左右两边数字的力矩相 ...

  7. hdu 6299 Balanced Sequence (贪心)

    Balanced Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others ...

  8. HDU - 3709 - Balanced Number(数位DP)

    链接: https://vjudge.net/problem/HDU-3709 题意: A balanced number is a non-negative integer that can be ...

  9. HDU 3709 Balanced Number

    发现只要Σa[i]*i%Σa[i]==0就可以. #include<iostream> #include<cstdio> #include<cstring> #in ...

随机推荐

  1. POJ 1734

    #include<iostream> #include<stdio.h> #define MAXN 105 #define inf 123456789 using namesp ...

  2. C Primer Plus之高级数据表示

     抽象数据类型(ADT)    类型是由什么组成?一个类型(type)指定两类信息:一个属性集和一个操作集. 所以您想定义一个新的数据类型.首先,您需要提供存储数据的方式,可能是通过设计一个结构.第二 ...

  3. cojs QAQ的矩阵 题解报告

    题目描述非常的清晰 首先我们考虑(A*B)^m的求法,这个部分可以参考BZOJ 杰杰的女性朋友 我们不难发现(A*B)^m=A*(B*A)^(m-1)*B A*B是n*n的矩阵,而B*A是k*k的矩阵 ...

  4. 为什么重写equals方法还要重写hashcode方法?

    我们都知道Java语言是完全面向对象的,在java中,所有的对象都是继承于Object类.Ojbect类中有两个方法equals.hashCode,这两个方法都是用来比较两个对象是否相等的. 在未重写 ...

  5. java如何得到GET和POST请求URL和参数列表

    转载:http://blog.csdn.net/yaerfeng/article/details/18942739 在servlet中GET请求可以通过HttpServletRequest的getRe ...

  6. Xamarin.Android 入门之:Listview和adapter

    一.引言 不管开发什么软件,列表的使用是必不可少的,而本章我们将学习如何使用Xamarin去实现它,以及如何使用自定义适配器.关于xamarin中listview的基础和适配器可以查看官网https: ...

  7. 每用户订阅上的所有者 SID 不存在 (异常来自 HRESULT:0x80040207)

    出现这个问题是因为pQueryFilter.WhereClause = "RoomNumber=" +cmbFromPoint.SelectedItem;中的cmbFromPoin ...

  8. .md文件 Markdown 语法说明

    Markdown 语法说明 (简体中文版) / (点击查看快速入门) 概述 宗旨 兼容 HTML 特殊字符自动转换 区块元素 段落和换行 标题 区块引用 列表 代码区块 分隔线 区段元素 链接 强调 ...

  9. 56. Merge Intervals

    题目: Given a collection of intervals, merge all overlapping intervals. For example,Given [1,3],[2,6], ...

  10. ServletContentLIstener接口演示ServletContext的启动和初始化

    ServletContextListener接口中包含两个方法,一个是contextInitialized()方法, 用来监听ServletContext的启动和初始化:一个是contextDestr ...