HDU 5882 Balanced Game

Posted 2020-09-03 Angel_Kitty

Balanced Game

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 508    Accepted Submission(s): 428

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

2016 ACM/ICPC Asia Regional Qingdao Online

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5882

分析：

大体题意：

石头剪刀布这个游戏是公平的，每个手势攻防都是一样的！，现在给你n 个手势判断能否公平！

思路：

其实读完题目 猜到了，，没敢立马写= =

思路很简单，要想每一个手势攻防都一样，那么必须n-1个手势有一半攻击自己，有一半防守自己，那么显然n-1是偶数，

当n 是奇数是成立的，否则不成立！

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int n;
 6     int m;
 7     int i,j;
 8     while(scanf("%d",&n)!=EOF)
 9     {
10         while(n--)
11         {
12             scanf("%d",&m);
13             if(m%2==1)
14             printf("Balanced\n");
15             else printf("Bad\n");
16         }
17     }
18     return 0;
19 }