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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
const int N = 202, INF = 1000000009;

int dp[N][N][N];
bool cc[N][N][N];

int rec(int a, int b, int c)
{
    if (a >= N || b >= N || c >= N)
        return INF;
    if (cc[a][b][c])
        return dp[a][b][c];
    cc[a][b][c] = true;
    if (a == b && b == c)
    {
        return dp[a][b][c] = 0;
    }
    dp[a][b][c] = INF;
    dp[a][b][c] = min(dp[a][b][c], rec(a + 2, b, c) + 1);
    dp[a][b][c] = min(dp[a][b][c], rec(a, b + 2, c) + 1);
    dp[a][b][c] = min(dp[a][b][c], rec(a, b, c + 2) + 1);
    dp[a][b][c] = min(dp[a][b][c], rec(a + 1, b + 1, c) + 1);
    dp[a][b][c] = min(dp[a][b][c], rec(a + 1, b, c + 1) + 1);
    dp[a][b][c] = min(dp[a][b][c], rec(a, b + 1, c + 1) + 1);
    return dp[a][b][c];
}

int main()
{
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    printf("%d\n", rec(a, b, c));
    return 0;
}