Problem:
Score: 200 points

Problem Statement

We have N cards. A number a_i is written on the i-th card.

Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first.

The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.

Constraints

N is an integer between 1 and 100 (inclusive).

a_i \ (1 \leq i \leq N) is an integer between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 a_3 ... a_N

Output

Print Alice's score minus Bob's score when both players take the optimal strategy to maximize their scores.

Sample Input 1

2
3 1

Sample Output 1

2

First, Alice will take the card with 3. Then, Bob will take the card with 1.
The difference of their scores will be 3 - 1 = 2.

Sample Input 2

3
2 7 4

Sample Output 2

5

First, Alice will take the card with 7. Then, Bob will take the card with 4. Lastly, Alice will take the card with 2. The difference of their scores will be 7 - 4 + 2 = 5. The difference of their scores will be 3 - 1 = 2.

Sample Input 3

4
20 18 2 18

Sample Output 3

18

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using System.Text;
using System.Globalization;
using System.Numerics;
using System.Text.RegularExpressions;
//byte 255 int<2147483647 ≒ 2 * 10^9 long < 9223372036854775807 ≒ 9 * 10^18
//decimal 29桁

class Myon
{
    //chokudaiさんのコードをコピペ
    static Scanner cin;
    public Myon() { }
    public static int Main()
    {
        //Console.SetOut(new Printer(Console.OpenStandardOutput()));
        cin = new Scanner();
        new Myon().calc();

        return 0;
    }

    public void calc()
    {
        int n = cin.nextInt();
        int[] a = cin.nextIntArr(n);
        Array.Sort(a);
        int alice = 0, bob = 0;
        for (int i = 0; i < n; i++)
        {
            if(i%2 == 0) { alice += a[n - 1 - i]; }
            else { bob += a[n - 1 - i]; }
        }

        showL(alice - bob);
    }
    
    public void calc_ABC076D()
    {
        int n = cin.nextInt();
        int[] t = new int[n + 1]; for (int i = 0; i < n; i++) t[i] = cin.nextInt();
        int[] v = new int[n + 1]; for (int i = 0; i < n; i++) v[i] = cin.nextInt();
        t[n] = 0; v[n] = 0;

        //k番目の拘束条件(t = Σti(i=0 ~ k-1) ~ Σti + tk) Σti(i=0 ~ k-1) を stk とかく
        //          (言い換えればk番目の拘束が始まる時間)
        //  は、言い換えればt = x において (v(x)を実際の速度として) 
        //  x <= stk-1 ⇒ v(x) <= vk + (stk-1 + x) 
        //  x >= stk ⇒ v(x) ∈ [vk - (stk - x), vk + (stk + x) ]

        int[] sumt = new int[n+2]; for (int i = 1; i < n + 2; i++) sumt[i] = sumt[i - 1] + t[i-1];
        int[] maxv = new int[sumt[n+1]+1];

        for(int tt = 1; tt < sumt[n+1]; tt++) 
        {
            maxv[tt] = maxv[tt - 1] + 1;
            for(int i = 0; i < n+1; i++)
            {
                if (tt <= sumt[i]) maxv[tt] = Math.Min(maxv[tt], v[i] + (sumt[i] - tt));
                else if( tt <= sumt[i] + t[i]) maxv[tt] = Math.Min(maxv[tt], v[i]);
            }
        }

        double ans = 0;
        int imadokonokousoku = 0;
        int tempt = 1;
        for (int tt = 1; tt < sumt[n]; tt++)
        {
            if(tempt == t[imadokonokousoku]) { tempt = 0; imadokonokousoku++; }
            ans += ((double)maxv[tt] + maxv[tt - 1]) / 2;
            if (maxv[tt] < v[imadokonokousoku] &&  maxv[tt] == maxv[tt + 1])
            {
                ans += 0.25;//t= a.5 に最大値がくる場合の補正
            }
            tempt++;
        }
        ans += 0.5;//最後の一秒 v = 1 to 0なので
        showL(ans);

    }
    public void calcC()
    {
        int n = cin.nextInt();

        //System.Diagnostics.Stopwatch sw = new System.Diagnostics.Stopwatch();
        //sw.Start();

        int m = cin.nextInt();
        List<int> a = new List<int>(m);
        List<int> b = new List<int>(m);
        for (int i = 0; i < m; i++)
        {
            a.Add(cin.nextInt());
            b.Add(cin.nextInt());
        }

        int ans = 0;
        List<int> bridge = new List<int>();
        for(int i = 0; i < m; i++)
        {
            tempConnected = new bool[n];
            AddConnected(1, a, b, i);
            bool allconnected = true; for(int j = 0;j<n;j++) if (!tempConnected[j]) { allconnected = false;break; }
            if (!allconnected)
            {
                bridge.Add(i + 1);
                ans++;
            }
            //else
            //{
            //    showL($"{i + 1} is not bridge");
            //}
        }
        showL(ans);

        //sw.Stop();
        //showL($"処理時間: {sw.ElapsedMilliseconds} [msec]");


    }

    
    bool[] tempConnected;
    public void AddConnected(int index, List<int>a,List<int>b, int ignoreingBridge = -1)
    {
        //index番目のノードにつながるノードをconnectedにして、さらにaddconnectedを呼び出す
        for(int i = 0; i < a.Count; i++)
        {
            if (i == ignoreingBridge) continue;
            if (a[i] -1 == index && !tempConnected[b[i]-1]) { tempConnected[b[i]-1] = true; AddConnected(b[i]-1, a, b,ignoreingBridge); }
            else if (b[i]-1 == index && !tempConnected[a[i]-1]) { tempConnected[a[i]-1] = true; AddConnected(a[i]-1, a, b,ignoreingBridge); }
        }
   
    }
    public void backetSort(ref int[] a)
    {
        //max a << a.len に対して有効なソート
        //よってa = int[] とする
        long n = a.Length;
        var count = new int[n];
        var loc = new long[n];//loc[i] は i が終わる 位置 max index where c[index] = loc[i]
        for (long i = 0; i < n; i++) count[a[i]]++;
        loc[0] = count[0] - 1;
        for (long i = 0; i < n-1; i++) loc[i+1] = loc[i] + count[i+1]; 
        long index = 0;

        for(int i = 0; i < n; i++)
        {
            while (index <= loc[i]) a[index++] = i; //incex <= loc[i] == index < loc[i] + 1
        }
    }

    public static void showL() { Console.WriteLine(""); }
    public static void showL<T>(T s) { Console.WriteLine(s); }
    public static void show<T>(T s) { Console.Write(s); }

}

class Printer : StreamWriter
{
    public override IFormatProvider FormatProvider { get { return CultureInfo.InvariantCulture; } }
    public Printer(Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
    public Printer(Stream stream, Encoding encoding) : base(stream, encoding) { }
}

class Scanner
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner()
    {
        s = new string[0];
        i = 0;
    }

    public string next()
    {
        while (i >= s.Length)
        {
            string st = Console.ReadLine();
            while (st == "") st = Console.ReadLine();
            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            i = 0;
        }
        return s[i++];
    }

    public int nextInt()
    {
        return int.Parse(next());
    }

    public int[] nextIntArr(long n)
    {
        int[] output = new int[n];
        for (int i = 0; i < n; i++) output[i] = nextInt();
        return output;
    }

    public long nextLong()
    {
        return long.Parse(next());
    }

    public double nextDouble()
    {
        return double.Parse(next());
    }

}

static class SMath
{
    
    /// <summary>
    /// 素因数分解して、n番目に見つかった素因数をoutput[n-1]に返します。
    /// 12 なら 2,2,3 です。
    /// </summary>
    /// <param name="n"></param>
    /// <returns></returns>
    public static List<long> Factorization(long n)
    {
        if (n <= 0) return null;
        List<long> output = new List<long>();
        int i = 2;
        long max = (long)Math.Sqrt(n); ;
        while (n > 1)
        {
            if (i > max) { output.Add(n); break; }
            while (n % i == 0)
            {
                output.Add(i);
                n /= i;
            }
            i++;
        }
        return output;
    }

    /// <summary>
    /// 素因数分解して、n番目に見つかった素因数をoutput[n-1]に返します。
    /// 84 = (2^2 * 3 * 7) なら 2, 1, 0, 1 　remainder = 1です。(打ち切られない）
    /// 13 = 0,0,...0 remainder = 2019
    /// ただし、試行を打ち切った時に限り残りをremainderに返す。(remainderは素数または1となる)
    /// </summary>
    /// <param name="n"></param>
    /// <param name="remainder"></param>
    /// <returns></returns>
    public static List<long> Factorization(long n, out long remainder)
    {
        //output[i] が i-1番目の素数の数を示す
        //ただし、試行を打ち切った時に限り残りをremainderに返す。(remainderは素数または1となる)
        remainder = 1;
        if (n <= 0) return null;
        List<long> output = new List<long>();
        long tempn = 2;
        long max = (long)Math.Sqrt(n);
        bool[] isNOTPrime = new bool[max+1];
        while (n > 1)
        {

            if (tempn > max) { remainder = n; break; }
            if (!isNOTPrime[tempn])
            {
                output.Add(0);
                while (n % tempn == 0)
                {
                    output[output.Count - 1]++;
                    n /= tempn;
                }
                for (long j = tempn * 2; j < isNOTPrime.Length; j += tempn) isNOTPrime[j] = true;
            }
            tempn++;
        }
        return output;
    }

    /// <summary>
    /// i番目の素数をoutput[i-1]に返します。
    /// ただしnより大きい素数については判定しません。
    /// </summary>
    /// <param name="n"></param>
    /// <returns></returns>
    public static List<long> GetPrimeNumbers(long n)
    {
        if (n <= 0) return null;
        List<long> output = new List<long>();
        long tempn = 2;
        bool[] isNOTPrime = new bool[n + 1];
        while (n > 1)
        {
            if (tempn > n) { break; }
            if (!isNOTPrime[tempn])
            {
                output.Add(tempn);
                for (long j = tempn * 2; j < isNOTPrime.Length; j += tempn) isNOTPrime[j] = true;
            }
            tempn++;
        }
        return output;
    }


    private static long[][] combMemo;
    private static int maxn_combination = -1;
    public static long Combination(int n, int k)
    {
        if (n < 0 || k > n) return 0;
        if (n == k || k == 0) return 1;
        if (k > n/2) return Combination(n, n - k);
        if (maxn_combination < n)
        {
            maxn_combination = n;
            combMemo = new long[n+1][];
        }
        if (combMemo[n] == null) combMemo[n] = new long[n/2 + 1];
        if (combMemo[n][k] != 0) return combMemo[n][k];
        else
        {
            combMemo[n][k] = Combination(n - 1, k) + Combination(n - 1, k - 1);
            return combMemo[n][k];
        }
        
    }
    public static long Gcd(long a, long b)
    {
        if (a <= 0) return -1;
        //if (a > b) return gcd(b, a);
        if (b == 0) return a;
        return Gcd(b, a % b);
    }
    public static long Lcm(long a,long b)
    {
        return a / Gcd(a, b) * b;
    }


    public class pairComparer : IComparer<int[]>
    {
        int index = 0;
        public pairComparer(int n)
        {
            index = n;
        }
        public pairComparer()
        {
            index = 0;
        }
        public int Compare(int[] x, int[] y)
        {
            return y[index] - y[index];
        }
    }

}