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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.math.*;

public class Main{
	
	public static void main(String args[]){
		int n;
		int vet[] = new int[5010];
		Scanner sc = new Scanner(System.in);
		n = sc.nextInt();
		
		for (int i = 0; i < n; i++)
		{
			vet[i] = sc.nextInt();
		}
		
		
		BigInteger b = new BigInteger(String.valueOf(vet[0]));
		BigInteger aux = new BigInteger("0");
		BigInteger aux2 = new BigInteger("0");
		for (int i = 1; i < n; i++)
		{
			aux = b.multiply(BigInteger.valueOf(vet[i]));
			aux2 = b.gcd(BigInteger.valueOf(vet[i]));
			b = aux.divide(aux2);
		}
		//~ System.out.println(b);
		
		b = b.subtract(BigInteger.ONE);
		long ans = 0;
		for (int i = 0; i < n; i++)
		{
			//~ System.out.println(b.mod(BigInteger.valueOf(vet[i])));
			ans += b.mod(BigInteger.valueOf(vet[i])).longValue();
		}
		System.out.println(ans);
	}
	
}