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 d[3];
    int di[3]={};
    int max=-100;
    for(int i=0;i<3;i++){
        scanf("%d",&d[i]);
        if(max<d[i]){
            max=d[i];
        }
    }

    int tmp;
    for(int i=0;i<2;i++){
        for(int j=i+1;j<3;j++){
            if(d[i]>d[j]){
                tmp=d[i];
                d[i]=d[j];
                d[j]=tmp;
            }
        }
    }

    int ans=0;
    ans=d[2]-d[1];
    d[0]+=d[2]-d[1];
    d[1]=d[2];
    ans+=(d[2]-d[0])/2;
    d[0]+=2*((d[2]-d[0])/2);
    //printf("%d %d %d\n",d[0],d[1],d[2]);
    printf("%d\n",(d[2]-d[0])%2==0?ans:ans+2);
    return 0;
}