C

 

Description

One day n friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and other n - 1people 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 a_i 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, ..., a_n(1 ≤ a_i ≤ 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 a_irounds.

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.

Sample Input

Input

33 2 2

Output

4

Input

42 2 2 2

Output

3

Hint

You don't need to know the rules of "Mafia" to solve this problem. If you're curious, it's a game Russia got from the Soviet times: http://en.wikipedia.org/wiki/Mafia_(party_game).

题意：

每局游戏1个主管，n-1个陪玩的，看至少多少局，能够满足每个人是玩伴的次数

分析：

先找出最大数，假设就来这些局的，看满足所有人是玩伴的要求是否还差主管，现在主管的和是最大值减去每个人的需求相加的总和，如果最大值大就直接输出，否则一个一个的加，没加一局游戏想当于把其他人的需求往下移，上面的主管数就加了n-1，不断加1，直到大于等于最大值，然后加上加一的次数，输出

代码：

#include<bits/stdc++.h>using namespace std;int main(){    long long int n,i,t,s,d[100005],r;    while(cin>>n)    {        for(i=0,s=0,t=0;i<n;i++){            cin>>d[i];            if(t<d[i])                t=d[i];        }        for(i=0;i<n;i++){            d[i]=t-d[i];            s+=d[i];        }        if(s>=t)            cout<<t<<endl;        else        {            r=0;            while(s<t)            {                r++;                s+=n-1;            }            cout<<t+r<<endl;        }    }}

感受：

这个题一开始没有思路。。。。根本就是一片懵逼。。然后看了网上的分析，才有了思路。。还不错，至少理解了