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<stdlib.h>
#include<string.h>
#include<math.h>
#include<limits.h>

int max(int a, int b){
  return a >= b ? a : b;
}
int min(int a, int b){
  return b >= a ? a : b;
}
int sei(int a){
  return a>0 ? a : 0;
}
int factorial(int n) {
    if (n > 0) return n*factorial(n - 1);
		else return 1;
}
int compare_up_int(const void *a, const void *b){
   return *(int*)a - *(int*)b;
}
int compare_down_int(const void *a, const void *b){
   return *(int*)b - *(int*)a;
}
int euclid(int a, int b){
  if(a < b){int tmp = a;a = b;b = tmp;}
  int r = a % b;
  if(r < 0) r += b;
  while(r != 0){
    a = b;b = r;r = a % b;
    if(r < 0) r += b;
  }
  return b;
}

int main(){
  int num[3],count=0;
  scanf("%d %d %d",&num[0],&num[1],&num[2]);
  qsort(num,3,sizeof(int),compare_up_int);
  if((num[2]-(num[0]+num[2]-num[1]))%2 == 0){
    int a = num[2]-num[1];
    printf("%d\n",a+(num[2]-num[0]-a)/2);
  }
  else{
    num[0]+=1;num[1]+=1;
    int a = num[2]-num[1];
    printf("%d\n",a+(num[2]-num[0]-a)/2 + 1);    
  }
  return 0;
}