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: Accepted)

#include <stdio.h>
#include <stdlib.h>

/* ソート関数 */
int int_sort( const void * a , const void * b ) {
  /* 引数はvoid*型と規定されているのでint型にcastする */
  if( *( int * )a < *( int * )b ) {
    return -1;
  }
  else
  if( *( int * )a == *( int * )b ) {
    return 0;
  }
  return 1;
}




int main(void){
    int a[3];
    scanf("%d %d %d",&a[0],&a[1],&a[2]);
      qsort(( void * )a , 3 , sizeof( a[0] ) , int_sort );
    int ans=0;
    while(a[2]-a[0]>1){a[0]+=2;ans++;}
    while(a[2]-a[1]>1){a[1]+=2;ans++;}
    if(a[0]==a[1]&&a[1]!=a[2])ans++;
    else if(a[0]!=a[1])ans+=2;
    printf("%d",ans);
    return 0;
}