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* max(int a[3]){
    if(a[0] >= a[1] && a[0] >= a[2]) return &a[0];
    else if(a[1] >= a[0] && a[1] >= a[2]) return &a[1];
    else return &a[2];
}

int find_even(int ar[3]){
    int num = 0;
    for(int i = 0; i < 3; i++){
        if(ar[i] % 2 == 0) num++;
    }
    return num;
}

int main(void){
    int a[3];
    scanf("%d %d %d", &a[0], &a[1], &a[2]);
    
    int *target;
    target = max(a);
    int num_even = find_even(a);
    int ans;
    if(num_even == 3 || num_even == 0){
        ans = ((*target) * 3 - a[0] - a[1] - a[2]) / 2;
    }else{
        if((*target) % 2 != num_even % 2){
            ans = (*target - a[0]) / 2 + (*target - a[1]) / 2 + (*target - a[2]) / 2 + 1;
        }else{
            for(int i = 0; i < 3; i++){
                if(a[i] != (*target) && a[i] % 2 == 0){
                    a[i]++;
                    break;
                }
            }
            (*target)++;
            ans = 1 + ((*target) * 3 - a[0] - a[1] - a[2]) / 2;
        }
    }
    printf("%d\n", ans);
    return 0;
}