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: Wrong Answer)

#include<stdio.h>

void sort(int* A,int* B,int* C){
	int backup;
	if(*B<=*C){
		int backup=*B;
		*B=*C;
		*C=backup;
	}
	if(*A<=*B){
		int backup=*A;
		*A=*B;
		*B=backup;
	}
	if(*B<=*C){
		int backup=*B;
		*B=*C;
		*C=backup;
	}
}

int main(void){
	int A;
	int B;
	int C;
	scanf("%d",&A);
	scanf("%d",&B);
	scanf("%d",&C);
	sort(&A,&B,&C);
	int kaisu=A-B;
	B=A;
	C+=kaisu;
	if(B-C>=2){
		B+=2;
		kaisu++;
	}
	if(B-C==0)
		printf("%d",kaisu);
	else
		printf("%d",kaisu+2);
}