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.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		int N = sc.nextInt();

		int[] a = new int[N];
		long lcm_a = 1;
		for(int i=0;i<N;i++) {
			a[i] = sc.nextInt();
			lcm_a = lcm(lcm_a,a[i]);
		}

		long m = lcm_a - 1;

		long ans=0;
		long ans2=0;
		for(int i=0;i<N;i++) {
			ans += a[i];
		}

		System.out.println(ans - N);
	}

	//最小公倍数
	public static long lcm (long lcm_a, long b) {
			long temp;
			long c = lcm_a;
			c *= b;
			while((temp = lcm_a%b)!=0) {
				lcm_a = b;
				b = temp;
			}
			return c/b;
	}

}