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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>

int main(){
    int N = 0;
    int j = 1;
    int ja = 0;
    int md = 0;
    int summd = 0;
    int f = 0;

    if(scanf("%d", &N) < 1){
        exit(1);
    }

    int a[N];

    for(int i= 0; i < N;i++){
        if(scanf("%d", &a[i]) < 1){
            exit(1);
        }
    }

    while(j != 0){
        summd = 0;
        ja = j * a[0];
        for(int i=1; i < N;i++){
            md = ja % a[i];
            summd += md;
            if(md != 0){
                break;
            }
        }
        if(summd != 0){
            j++;
        }else{
            j = 0;
        }
    }

    ja -= 1;

    for(int i = 0;i< N;i++){
        f += ja % a[i];
    }
    printf("%d\n", f);
}