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 A, B, C;
	int max;
	int min;
	int mid;
	int count = 0;
	scanf("%d %d %d", &A, &B, &C);
	//最大値、中央値を求める
	if (A > B) {
		max = A;
		mid = B;
	}
	else {
		max = B;
		mid = A;
	}
	if (max < C) {
		mid = max;
		max = C;
	}
	else {
		if (mid < C) {
			mid = C;
		}
	}
	//最小値を求める
	if (A < B) {
		min = A;
	}
	else {
		min = B;
	}
	if (min > C) {
		min = C;
	}
	//小さい数字二つの差/2 (2づつ足す)
	count = (mid - min)/2;
	//はみ出す場合は+1(minとmax)
	if ((mid - min) % 2 == 1) {
		count++;
		max++;
	}
	count = count + (max - mid);
	printf("%d\n", count);
	return 0;
}