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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef vector<long long> vll;
typedef vector<int> vi;
typedef pair<int, int> pi;

#define yes "Yes"
#define yse "Yes"
#define no "No"

#define pb push_back
#define rep(i, l, n) for (int i = (l); i < n; i++)
#define rep_cs(i, l, n) for (int i = (l); i <= n; i++)
#define rep_d(j, n) for (int j = n; j >= 0; j--)
#define rep_d_op(j, n, a) for (int j = n; j >= a; j--)

template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}

const int MOD = 1000000007;
const vector<int> dx = {1, 0};
const vector<int> dy = {0, 1};
const ll INF = 1LL << 60;

// int max = *std::max_element(v.begin(), v.end());
// int sum = accumulate(v, v + v.size(), 0);

int GCD(int a, int b) { return b ? GCD(b, a % b) : a; }

using Graph = vector<vector<int>>;
Graph G;
vector<bool> seen;
void dfs(const Graph &G, int v)
{
    seen[v] = true;

    for (auto next_v : G[v])
    {
        if (seen[next_v])
            continue;
        dfs(G, next_v);
    }
}

ll dp[100100][3];
ll v[110];
ll w[110];

int main()
{
    vll v(3);
    rep(i, 0, 3) cin >> v[i];

    int odd, even = 0;

    rep(i, 0, 3)
    {
        if (v[i] % 2 == 0)
        {
            even++;
        }
        else
        {
            odd++;
        }
    }
    ll ans = 0;

    if (even == 2)
    {
        ans++;
        rep(i, 0, 3)
        {
            if (v[i] % 2 == 0)
            {
                v[i]++;
            }
        }
    }
    else if (odd == 2)
    {
        ans++;
        rep(i, 0, 3)
        {
            if (v[i] % 2 != 0)
            {
                v[i]++;
            }
        }
    }

    int max = *std::max_element(v.begin(), v.end());

    rep(i, 0, 3)
    {
        ans += (max - v[i]) / 2;
    }
    cout << ans << endl;
}