Code forces 348A-----思维题

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本文介绍了一种算法，用于计算让每个参与者达到其期望参与次数所需的最少游戏轮数。通过输入每个人的期望参与次数，该算法能够高效地计算出最少轮数。

Mafia

Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 348A

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 %I64dspecifier.

Sample Input

Input

3
3 2 2

Output

Input

4
2 2 2 2

Output

就是有n个人玩游戏，每轮需要一个裁判之类的东西，其他人就是玩家，给出每个人想要当玩家的局数，问最少要玩多少轮，才能全部满足

如样例1，三个人A,B,C分别相当3,2,2次玩家，第一场A裁判，B,C都玩一场，第二局B裁判，第三局C裁判，最后一句，B，C任意一个当裁判都可以

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define maxn 500005
__int64 s[maxn];
int main(){
	int n;
	__int64 ans = 0, cnt = 0, t;
	scanf("%d", &n);
	for(int i = 0; i < n; i++){
		scanf("%I64d", &s[i]);
		cnt += s[i];
		ans = ans > s[i] ? ans : s[i];
	}
	t = ceil(cnt / (n-1.0));//ceil返回不小于表达式的最小整数，n-1个人平均每人玩几场，自己弄几个数试试就能找出规律
	t = t > ans ? t : ans;
	printf("%I64d\n", t);
	return 0;
}