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 <stdlib.h>

int main()
{
    int N, p[100], tz, j, k, max, s=0, m;
    scanf("%d", &N);
    for(int i=1;i<=N;i++)
        scanf("%d", &p[i]);
   for(int f=1;f<N;f++)
       { for(j=1;j<N;j++)
        if(p[j]>p[j+1])
       {

        tz=p[j];
        p[j]=p[j+1];
        p[j+1]=tz;
        }}

    for(m=1;m<N;m++)
    {
     s=s+p[m];
    }
     s=s+(p[N]/2);
     printf("%d", s);
    return 0;
}