Codeforces 349C - Mafia

最新推荐文章于 2019-07-22 16:13:45 发布

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本文探讨了一种特殊的游戏策略问题——一群朋友玩黑手党游戏时，如何安排最少轮次让每个人都能达到他们想要的游戏轮数。通过分析每个玩家希望参与的轮数，提供了一个算法解决方案。

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 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 %I64dspecifier.

Examples

input
3
3 2 2

output
4

input
4
2 2 2 2

output
3

题目：

第一步：

那么，

那么我们就可得以下代码：

#include<cstdio>
long long max(long long a,long long b){
	if(a>b) return a;
	else return b;
}
int a[100003];
int main()
{
	int n;scanf("%d",&n);
	
	int max_ai=0;
	long long sum=0;
	for(int i=1;i<=n;i++) {
		scanf("%d",&a[i]);
		sum+=a[i];
		if(max_ai<a[i]) max_ai=a[i];
	}
	
	if( sum/(double)(n-1) == sum/(n-1) ) sum=sum/(n-1);
	else sum=sum/(n-1)+1;
	
	long long ans=max((long long)max_ai,sum);
	printf("%I64d\n",ans);
}