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>
#define max(a, b) ((a) > (b) ? (a) : (b))
#define min(a, b) ((a) > (b) ? (b) : (a))

int
main(int argc, char *argv[])
{
  int que[3][100000], d[100000];
  int first, tail;
  int a, b, c;
  scanf("%d %d %d", &a, &b, &c);
  que[0][0] = a; que[1][0] = b; que[2][0] = c;
  d[0] = 0;
  first = 0; tail = 1;
  while(1){
    for(int i = 0; i < 3; i++){
      if(que[i][first] >= max(que[(i + 1) % 3][first], que[(i + 2) % 3][first]))
	continue;
      que[i][tail] = que[i][first] + 2;
      for(int j = 0; j < 3; j++){
	if(j != i) que[j][tail] = que[j][first];
      }
      d[tail] = d[first] + 1;
      if(que[0][tail] == que[1][tail] && que[1][tail] == que[2][tail]){
	goto sdf;
      }
      if(tail == 99999) tail = 0;
      else tail++;
    }
    for(int i = 0; i < 3; i++){
      
      que[i][tail] = que[i][first] - 1;
      for(int j = 0; j < 3; j++){
	if(j != i) que[j][tail] = que[j][first];
      }
      d[tail] = d[first] + 1;
      if(que[0][tail] == que[1][tail] && que[1][tail] == que[2][tail]){
	goto sdf;
      }
      if(tail == 99999) tail = 0;
      else tail++;
    }
    if(first == 99999) first = 0;
    else first++;
  }
 sdf:
  for(int i = 0; i <= tail; i++){
    printf("(%d, %d, %d)\n", que[0][i], que[1][i], que[2][i]);
  }
  printf("%d\n", d[tail]);

  return 0;
}