Problem:
Score : 300 points

Problem Statement

You are given N positive integers a_1, a_2, ..., a_N.

For a non-negative integer m, let f(m) = (m\ mod\ a_1) + (m\ mod\ a_2) + ... + (m\ mod\ a_N).

Here, X\ mod\ Y denotes the remainder of the division of X by Y.

Find the maximum value of f.

Constraints

All values in input are integers.

2 \leq N \leq 3000

2 \leq a_i \leq 10^5

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the maximum value of f.

Sample Input 1

3
3 4 6

Sample Output 1

10

f(11) = (11\ mod\ 3) + (11\ mod\ 4) + (11\ mod\ 6) = 10 is the maximum value of f.

Sample Input 2

5
7 46 11 20 11

Sample Output 2

90

Sample Input 3

7
994 518 941 851 647 2 581

Sample Output 3

4527

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Security.Cryptography;

namespace cswork
{
    class Program
    {
        class Scanner
        {
            string[] s;
            int i;

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

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

            public string next()
            {
                if (i < s.Length) return s[i++];
                string st = Console.ReadLine();
                while (st == "") st = Console.ReadLine();

                s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
                i = 0;
                return next();
            }

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

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

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

        public Program() { }

        static void Main(string[] args)
        {
            new Program().Omame();
        }

        Scanner cin;
        const int MAXIN = 1024;
        bool[] used = new bool[MAXIN];
        int[] perm = new int[MAXIN];

        List<int[]> permList = new List<int[]>();

        public void Permutation(int pos, int n, int length)
        {
            if (pos == n)
            {
                permList.Add((int[])perm.Clone());
            }

            for (int i = 0; i < n; i++)
            {
                if (!used[i])
                {
                    perm[pos] = i;
                    used[i] = true;
                    Permutation(pos + 1, n, length);
                    used[i] = false;
                }
            }
        }

        public static List<long> GeneratePrime(long max)
        {
            long prime;
            double sqrtMax = Math.Sqrt(max);
            var primeList = new List<long>();

            List<long> searchList = new List<long>();
            for (long i = 2; i <= max - 1; i++)
            {
                searchList.Add(i);
            }

            do
            {
                prime = searchList.First();
                primeList.Add(prime);
                searchList.RemoveAll(n => n % prime == 0);
            } while (prime < sqrtMax);

            primeList.AddRange(searchList);

            return primeList;
        }

        // 最小公倍数
        public static long Lcm(long a, long b)
        {
            return a * b / Gcd(a, b);
        }

        // ユークリッドの互除法 
        public static long Gcd(long a, long b)
        {
            if (a < b)
                return Gcd(b, a);
            while (b != 0)
            {
                var remainder = a % b;
                a = b;
                b = remainder;
            }
            return a;
        }

        public static readonly string Yes = "Yes";
        public static readonly string YES = "YES";
        public static readonly string No = "No";
        public static readonly string NO = "NO";
        public static readonly string OK = "OK";
        public static readonly string NG = "NG";

        void Omame()
        {
            cin = new Scanner();
            
            long N = cin.nextLong();
            List<long> aList = new List<long>();
            long lcm = 1;
            for (int i = 0; i < N; i++)
            {
                long a = cin.nextLong();
                aList.Add(a);
                lcm = Lcm(lcm, a);
            }

            long ans = 0;
            for (int i = 0; i < N; i++)
            {
                ans += (lcm - 1) % aList[i];
            }
            Console.WriteLine(ans);
        }
    }
}