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>

int a[3];

void detect(int *max, int *min)
{
    int i;
    *max = 0;
    *min = 0;
    for (i = 1; i < 3; i++)
        if (a[*max] < a[i]) *max = i;
    for (i = 1; i < 3; i++)
        if (a[*min] > a[i]) *min = i;
}

int other(int max, int min)
{
    int res;
    for (res = 0; res < 3; res++)
        if (res != max && res != min) break;
    return res;
}

int main()
{
    int max;
    int min;
    int ans;

    scanf("%d %d %d", a, a+1, a+2);

    ans = 0;
    while (1) {
        detect(&max, &min);
        if (a[max] - a[min] <= 1) break;
        a[min] += 2;
        ans++;
    } 

    detect(&max, &min);
    if (a[max] == a[min]) printf("%d\n", ans);
    else if(a[min] == a[other(max, min)])
        printf("%d\n", ans+1);
    else printf("%d\n", ans+2);

    return 0;
}