HDU 5882 Balanced Game

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

題目連接： http://acm.hdu.edu.cn/showproblem.php?pid=5882

分析：

大致題意：

石頭剪刀布這個遊戲是公平的，每一個手勢攻防都是同樣的！，如今給你n 個手勢判斷可否公平！

思路：

其實讀完題目 猜到了，，沒敢立馬寫= =

思路很簡單，要想每個手勢攻防都同樣，那麼必須n-1個手勢有一半攻擊本身，有一半防守本身，那麼顯然n-1是偶數，

當n 是奇數是成立的，不然不成立！

下面給出AC代碼：

 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 }