CodeForces 348A Mafia （思維）

A. Mafia

time limit per test:2 seconds

memory limit per test:256 megabytes

input:standard input

output:standard output

One day n friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and othern - 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: thei-th person wants to playa_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 containsn space-separated integersa₁, a₂, ..., a_n(1 ≤ a_i ≤ 10⁹) — thei-th number in the list is the number of rounds thei-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 thei-th person play at leasta_i rounds.

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

Sample test(s)

Input

3
3 2 2

Output

4

Input

4
2 2 2 2

Output

3

Note

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).

題意：

N個人玩遊戲，每論遊戲需要1個主持人，N-1個玩家，列出每個人想當玩家的局數，求滿足條件的最少局數。

思路：

1.局數最少可能爲這N個人中想當玩家局數的最大值

2.若每個人當主持人的局數之和大於等於這個最大值，則這個最大值滿足條件

3.若2不滿足，設此時需增加x局，剩下沒人當主持人的局數爲y，則

y + x = N * x;

=> x = y / (N - 1);

y 可求出，則 x 的向上取整即爲答案。

/*************************************************************************
	> File Name: GGG.cpp
	> Author: BSlin
	> Mail:  
	> Created Time: 2013年10月03日 星期四 19時52分01秒
 ************************************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <iterator>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#define MP make_pair
#define INF (1<<30)
#define PI acos(-1.0)
#define esp 1e-8
const int dx[4]={0,0,0,0};
using namespace std;
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#if defined (_WIN32) || defined (__WIN32) || defined (WIN32) || defined (__WIN32__)
#define LL __int64
#define LLS "%" "I" "6" "4" "d"
#else
#define LL long long
#define LLS "%" "l" "l" "d"
#endif


#define M 100010

LL a[M];

int main(int argc, char** argv) {
    int n;
    LL max,sum,ans;
    while(scanf("%d",&n) != EOF) {
        max = 0;
        for(int i=0; i<n; i++) {
            scanf(LLS,&a[i]);
            if(max < a[i]) max = a[i];
        }
        sum = 0;
        for(int i=0; i<n; i++) {
            sum += (max - a[i]);
        }
        if(sum >= max) {
            ans = max;
        } else {
            ans = max - sum;
            if(ans % (n - 1) == 0) ans = max + ans / (n - 1);
            else ans = max + ans / (n - 1) + 1;
            //ans = max + (max - sum + n - 2) / (n - 1);  //可取代前3句
        }
        printf(LLS"\n",ans);
    }
    return 0;
}