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 main(){
    int x[3];
    int count=0;
    scanf("%d %d %d",&x[0],&x[1],&x[2]);
    for(int i=0;i<2;i++){
        if(x[i]>x[i+1]){
            int t=x[i]+x[i+1];
            x[i]=t-x[i];
            x[i+1]=t-x[i+1];
        }
    }
    for(int i=0;i<2;i++){
        if(x[i]>x[i+1]){
            int t=x[i]+x[i+1];
            x[i]=t-x[i];
            x[i+1]=t-x[i+1];
        }
    }
    int xa=x[0]%2,xb=x[1]%2,xc=x[2]%2;
    if((xa!=xb)&&((xb==xc)||(xa==xc))){
        count+=2;
    }else if((xa==xb)&&(xb==xc)){

    }else{
        count++;
    }
    count +=(x[2]-x[1])/2;
    count +=(x[2]-x[0])/2;
    printf("%d",count);
    return 0;
}