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>
#include<string.h>


int main(){
  int a,b,c,k,max,mx,my;
  int ans = 0;
  scanf("%d %d %d",&a,&b,&c);
  
  if(a%2 == b%2 && a%2 != c%2){
    a = a+1;
    b = b+1;
    ans += 1;
  }
  
  if(a%2 == c%2 && a%2 != b%2){
    a = a+1;
    c = c+1;
    ans += 1;
  }
  
  if(b%2 == c%2 && c%2 != a%2){
    c = c+1;
    b = b+1;
    ans += 1;
  }
  
  if(a>b && a>c){
    max = a;
    mx = b;
    my = c;
    k=(a-b)+(a-c);
  }
  if(b>a && b>c){
    max = b;
    mx = a;
    my =c;
    k=(b-a)+(b-c);
  }
  if(c>a && c>b){
    max = c;
    mx = a;
    my = b;
    k=(c-a)+(c-b);
  }
  
    while(max!=mx && max!=my){
      mx = mx + 1;
      my = my + 1;
      ans = ans + 1;
    }
    while(max!=mx){
      mx = mx + 2;
      ans = ans + 1;
    }
    while(max!=my){
      my = my + 2;
      ans = ans + 1;
    } 
    
    printf("%d",ans);
    return 0;
}