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>
#include<stdlib.h>
#include<string.h>

typedef long long ll;

int max(int a, int b){
  if(a > b) return a;
  else return b;
}
int min(int a, int b){
  if(a < b) return a;
  else return b;
}

int sort(const void *a ,const void *b){
  return *(int*)a - *(int*)b;
}

int main(){

  int n,sum = 0;
  scanf("%d",&n);

  int p[n];
  for(int i = 0; i < n; ++i){
    scanf("%d",&p[i]);
  }

  qsort(p,n,sizeof(int),sort);

  for(int i = 0; i < n-1; ++i){
    sum += p[i];
  }
  sum += p[n-1] / 2;

  printf("%d",sum);


  return 0;
}