Problem:
Score : 200 points

Problem Statement

In some other world, today is the day before Christmas Eve.

Mr. Takaha is buying N items at a department store. The regular price of the i-th item (1 \leq i \leq N) is p_i yen (the currency of Japan).

He has a discount coupon, and can buy one item with the highest price for half the regular price. The remaining N-1 items cost their regular prices. What is the total amount he will pay?

Constraints

2 \leq N \leq 10

100 \leq p_i \leq 10000

p_i is an even number.

Input

Input is given from Standard Input in the following format:

N
p_1
p_2
:
p_N

Output

Print the total amount Mr. Takaha will pay.

Sample Input 1

3
4980
7980
6980

Sample Output 1

15950

The 7980-yen item gets the discount and the total is 4980 + 7980 / 2 + 6980 = 15950 yen.

Note that outputs such as 15950.0 will be judged as Wrong Answer.

Sample Input 2

4
4320
4320
4320
4320

Sample Output 2

15120

Only one of the four items gets the discount and the total is 4320 / 2 + 4320 + 4320 + 4320 = 15120 yen.

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
int main(void){
    int N, i, ans = 0, max = 0;
    int p[10000];
    
    scanf("%d", &N);
    
    for(i = 0; i < N; i++){
        scanf("%d", &p[i]);
    }
    
    for(i = 0; i < N; i++){
        if(max < p[i]){
            max = p[i];
            p[i] = max;
        }
    }
    
    for(i = 0; i < N; i++){
        if(max == p[i]){
            p[i] = p[i] / 2;
            break;
        }
    }
    
    for(i = 0; i < N; i++){
        ans = ans + p[i];
    }
    printf("%d\n", ans);
    
}