Problem:
Score : 300 points

Problem Statement

You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order:

Choose two among A, B and C, then increase both by 1.

Choose one among A, B and C, then increase it by 2.

It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.

Constraints

0 \leq A,B,C \leq 50

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the minimum number of operations required to make A, B and C all equal.

Sample Input 1

2 5 4

Sample Output 1

2

We can make A, B and C all equal by the following operations:

Increase A and C by 1. Now, A, B, C are 3, 5, 5, respectively.

Increase A by 2. Now, A, B, C are 5, 5, 5, respectively.

Sample Input 2

2 6 3

Sample Output 2

5

Sample Input 3

31 41 5

Sample Output 3

23

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int main(){
        int num[3],i,tmp,add=0;
        scanf("%d %d %d",&num[0],&num[1],&num[2]);
        for(i=0;i<2;i++){
                if(num[i] < num[i+1]){
                        tmp = num[i];
                        num[i] = num[i+1];
                        num[i+1] = tmp;
                }
        }
        while(1){
                if(num[0]==num[1] && num[1]==num[2])    break;
                if((num[0]-num[1])%2 == 0 && num[0]!=num[1]){
                        add+=(num[0]-num[1])/2;
                        num[1]=num[0];
                }else{
                        num[1]++;
                        num[2]++;
                        add++;
                        continue;
                }
                if((num[0]-num[2])%2 == 0){
                        add+=(num[0]-num[2])/2;
                        num[2]=num[0];
                        break;                                                                 
                }else{
                        num[0]++;
                        num[1]++;
                        add++;
                }
        }
        printf("%d\n",add);
        return 0;
}