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 (Clang 3.8.0) (judge verdict: Accepted)

#include<stdio.h>
int main()
{
    int a[100],i,n,j,sum=0;
    while(scanf("%d",&n)==1)
    {
        sum=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(i=0;i<n;i++)
        {
            for(j=i+1;j<n;j++)
            {
                if(a[i]<a[j])
                {
                    int tem=a[i];
                    a[i]=a[j];
                    a[j]=tem;
                }
            }
        }
        for(i=1;i<n;i++)
        {
            sum=sum+a[i];
        }
        sum=sum+a[0]/2;
        printf("%d",sum);
    }
    return 0;
}