HDU5882

Posted 2020-08-09 Penn000

Balanced Game

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 115    Accepted Submission(s): 99

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

 

签到水题

思路：把一个形状抽象成一个点，要与其他所有点有向相连，出度为胜，入度为负，只有当入度与出度相等，即胜负概率相同，才为平衡。故有奇数个形状平衡，偶数不平衡。

 1 //2016.9.17
 2 #include <iostream>
 3 #include <cstdio>
 4 
 5 using namespace std;
 6 
 7 int main()
 8 {
 9     int T, n;
10     scanf("%d", &T);
11     while(T--)
12     {
13         scanf("%d", &n);
14         if(n&1)printf("Balanced\n");
15         else printf("Bad\n");
16     }
17 
18     return 0;
19 }