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;
/** 復習用 */
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		final int N = sc.nextInt();
		// 解くべき問題が間違っていた。
		// f(m) = 0となるようなmは、 mがすべてのaの倍数になるようなm
		// つまり m0 = a1 * a2 *...* aN と定めれば良い
		// このとき、m%aiの各項が最大になるようなmは、m0-1
		// なぜなら m mod ai のすべての項が ai-1になるから
		// よってf(m)の最大値は f(m0-1) = (a1 - 1) + (a2 - 1) +..+ (aN - 1)
		int fm_max = 0;
		for (int i = 0; i < N; i++) {
			int ai = sc.nextInt();
			fm_max += ai - 1;
		}
		System.out.println(fm_max);
	}
}