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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <string.h>
int main (){
    int N,temp;
    int anc =0;
    int p[10];
    scanf("%d",&N);
    for(int i =0; i<N; i++) scanf("%d",&p[i]);

    for(int i= 0; i<N; i++){
        for(int j= N-1; j>i; j--){
            if(p[j] < p[j-1]){
                temp =p[j];
                p[j] =p[j-1];
                p[j-1]= temp;
            }
        }
    } 
  
    anc = p[N-1] / 2;
    for(int i=0; i<N-1; i++)    anc += p[i];
    printf("%d", anc);


    
    

    
   
}
 //chcp 65001 cd contest