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

#define rep(i,begin,end) for(int i=begin; i<end; i++)
#define revrep(i,begin,end) for(int i=begin; i>end; i--)
#define lld long long int

int main(){
    int a[3];
    scanf("%d %d %d", &a[0], &a[1], &a[2]);
    revrep(i, 2, 0){
        rep(j, 0, i){
            if(a[j] > a[j+1]){
                int b = a[j];
                a[j] = a[j+1];
                a[j+1] = b;
            }
        }
    }
    int ans = 0;
    if((a[2] - a[0])%2 == 1 && (a[2] - a[1])%2 == 1){
        a[0]++; a[1]++; ans++;
    }
    if((a[2] - a[0])%2 == 1 && (a[2] - a[1])%2 == 0){
        a[2]++; a[1]++; ans++;
    }
    if((a[2] - a[0])%2 == 0 && (a[2] - a[1])%2 == 1){
        a[0]++; a[2]++; ans++;
    }
    ans += (a[2] - a[0]) / 2;
    ans += (a[2] - a[1]) / 2;
    printf("%d", ans);

    return 0;
}