Change the ball

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

Total Submission(s): 564    Accepted Submission(s): 198

Problem Description

Garfield has three piles of balls, each pile has unique color of following: yellow, blue, and red. Now we also know Garfield has Y yellow balls, B blue balls, and R red balls. But Garfield just wants to change all the balls to one color. When he puts two balls of different color togather, the balls then change their colors automatically into the rest color. For instance, when Garfield puts a red one and a yellow one togather, the two balls immediately owns blue color, the same to other situations. But the rule doesn’t work when the two balls have the same color.

  Garfield is not able to estimate the minimal steps to achieve the aim. Can you tell him?

 

 

Input

For each line, there are three intergers Y, B, R(1<=Y,B,R<=1000),indicate the number refered above.

 

 

Output

For each case, tell Garfield the minimal steps to complete the assignment. If not, output the symbol “):”.

 

 

Sample Input

1 2 3

1 2 2

 

 

Sample Output

):

2

 

 

Source

 

 

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分析：三个数中只要有两个数之差为3的倍数就可以完成转化； 转化次数为num=max(a, b); 结果为min(num) ; 

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #define max(a,b) a>b?a:b  5 using namespace std; 6 int main() 7 { 8  9     int a, b, c;10     int i, total;11     12     while(~scanf("%d %d %d", &a, &b, &c))13     {14         total = 0;15         int num[3] = {
    0, 0, 0};16         if(abs(a-b)%3==0)17             num[total] += max(a,b);18         total++;19         if(abs(b-c)%3==0)20             num[total] += max(b,c);21         total++;22         if(abs(a-c)%3==0)23             num[total] += max(a,c);24         total++;25         if(num[0]==0 && num[1]==0 && num[2]==0)26         　　printf("):\n");27         else28         {29             sort(num, num+3);30             for(i=0; i<3; i++)31             {32                 if(num[i] != 0)33                 {34                     printf("%d\n",num[i]);35                     break; 36                 }37             }38         }39     }40     return 0;41 }