2016 ACM/ICPC 青岛区域赛网络赛 1005 Balanced Game （找规律）

Balanced Game

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2126    Accepted Submission(s): 1448

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

    
    
    
    题意： 给出n个手势，判断是否能否达到平衡，比如n=3（剪子、包袱、锤），可以达到平衡，再比如题目中给出的n=5, 也能够达到平衡，，， 
    
    
    
    思路： 一开始读错题，看成了N个人，首先想到找规律，发现N=3、4、5时可以，n=6的时候可以看做两个3,之后的都可 以，于是只判断n是否大于2，，结果秒错。。 最后想到，平衡状态应该是每个手势克与被克的次数是一样的，即总次数是偶数，如果n个手势都要出现，那么每 个手势与其他n-1个手势都应该有克与被克的关系，就像题目中的n=5一样，每个手势与其他4个都有克与被克的关系 ，就可以设置一种规则，让每个手势赢2次，输2次，无论怎么出，只要5个手势都出现，一定是平衡，， 如果n=6,那么每个手势与其他5个手势都有关系，无论怎么设置规则，至少有一个手势不会平衡，， 想到这里就会发现，如果n是偶数，n-1为奇数，不能达到平衡，，反之可以。。 以下AC代码： 
    
    
    
    
#include<stdio.h>
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        if(n%2)
            printf("Balanced\n");
        else
            printf("Bad\n");
    }
    return 0;
}