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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
 
#define rep(i, n) for (int i = (int)(0); i < (int)(n); ++i)
#define reps(i, n) for (int i = (int)(1); i <= (int)(n); ++i)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; i--)
#define rreps(i, n) for (int i = ((int)(n)); i > 0; i--)
#define irep(i, m, n) for (int i = (int)(m); i < (int)(n); ++i)
#define ireps(i, m, n) for (int i = (int)(m); i <= (int)(n); ++i)
#define FOR(e, c) for (auto &e : c)
#define SORT(v, n) sort(v, v + n);
#define vsort(v) sort(v.begin(), v.end());
#define rvisort(v) sort(v.begin(), v.end(), greater<int>());
#define all(v) v.begin(), v.end()
#define mp(n, m) make_pair(n, m);
#define cout(d) cout<<d<<endl;
#define coutd(d) cout<<std::setprecision(10)<<d<<endl;
#define cinline(n) getline(cin,n);
 
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using pii = pair<int, int>;
using vpii = vector<pii>;
using vs = vector<string>;
using vd = vector<double>;
using ul = unsigned long;
 
template<class T, class C> void chmax(T& a, C b){ a>b?:a=b; }
template<class T, class C> void chmin(T& a, C b){ a<b?:a=b; }
 
const int mod=1e9+7;
 
struct mint {
  ll x;
  mint(ll x=0):x(x%mod){}
  mint& operator+=(const mint a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator-=(const mint a) {
    if ((x += mod-a.x) >= mod) x -= mod;
    return *this;
  }
  mint& operator*=(const mint a) {
    (x *= a.x) %= mod;
    return *this;
  }
  mint operator+(const mint a) const {
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const {
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const {
    mint res(*this);
    return res*=a;
  }
};
 
mint c[4005][4005];
void init() {
  c[0][0] = 1;
  for (int i = 0; i <= 4000; i++) {
    for (int j = 0; j <= i; j++) {
      c[i+1][j] += c[i][j];
      c[i+1][j+1] += c[i][j];
    }
  }
}
mint comb(int n, int k) {
  return c[n][k];
}
 
bool IsPrime(int num)
{
    if (num < 2) return false;
    else if (num == 2) return true;
    else if (num % 2 == 0) return false;
 
    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            return false;
        }
    }
 
    return true;
}
 
map<ll, ll> primeFact(ll n) {
  map<ll, ll> res;
  for (ll i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      res[i]++;
      n /= i;
    }
  }
  if (n != 1) res[n]++;
  return res;
}
 
void printv(vi& v){
  rep(i,v.size()) cout << v[i] << " ";
  cout << endl;
}
void printv(vll& v){
  rep(i,v.size()) cout << v[i] << " ";
  cout << endl;
}
int vtotal(vi& v){
  int total=0;
  rep(i,v.size()) total+=v[i];
  return total;
}
 
int main()
{
  //init(); // mint初期化
  cin.tie( 0 );
  ios::sync_with_stdio( false );
 
  int n;
  cin>>n;
  vi p(n);
  rep(i,n)cin>>p[i];
  rvisort(p);
  int ans=vtotal(p)-p[0]/2;
  cout(ans);
 
  return 0;
}