Codeforces 348A. Mafia

One day n friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other n - 1 people take part in the game. For each person we know in how many rounds he wants to be a player, not the supervisor: the i-th person wants to play ai rounds. What is the minimum number of rounds of the "Mafia" game they need to play to let each person play at least as many rounds as they want?

Input

The first line contains integer n (3 ≤ n ≤ 105). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109) — the i-th number in the list is the number of rounds the i-th person wants to play.

Output

In a single line print a single integer — the minimum number of game rounds the friends need to let the i-th person play at least ai rounds.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

二分搜索

关于判定函数就是将数组排序，设当前共有n盘然后对于i来说，他可以最多当n-a[i]盘裁判，所以求出总数是否比n大就可以了

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;

LL a[100005],sum[100005];
LL ans,sums = 0;
LL maxa,n;

bool oks(LL now){
  LL index = upper_bound(a+1,a+1+n,now) - (a+1);
  // cout << now << index << endl;
  if (now * index - sum[index] >= now){
    return true;
  }
  return false;
}
LL search(LL l,LL r){
  // cout << l << r << endl;
  if (l == r){
    return l;
  }
  if (l+1 == r){
    if (oks(l)) return l;
    return r;
  }
  LL mid = l + (r - l) / 2;
  if (oks(mid)){
    return search(l,mid);
  }
  else {
    return search(mid+1,r);
  }
}

int main(){
  // freopen("test.in","r",stdin);
  cin >> n;
  maxa = 0;
  sum[0] = 0;
  for (int i=1;i<=n;i++){
    cin >> a[i];
    sums += a[i];
    sum[i] = sum[i-1] + a[i];
    maxa = max(maxa,a[i]);
  }
  sort(a+1,a+1+n);
  ans = search(maxa,sums);
  cout << ans;
  // cout << oks(3) << endl;
}