CF 348A(Mafia-判定性问题)

最新推荐文章于 2020-07-12 11:17:30 发布

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本文探讨了在Mafia游戏中如何通过优化算法来确定最少的游戏轮次，确保每位参与者都能达到其期望的参与次数。针对具体问题，文章提供了一种有效的解决方案，并通过示例验证了方法的有效性。

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 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).

这题如果硬解没思路，但是判断一个答案就是很简单。

然后你懂得。。

因为用了int惨遭Hack，各种NC。。。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n;
ll a[MAXN];
int main()
{
	cin>>n;
	For(i,n) cin>>a[i];
	ll k=0;
	For(i,n) k=max(k,a[i]);
	ll t=0;
	For(i,n) t+=k-a[i];
	if (t>=k) {cout<<k<<endl;return 0;	}
	else 
	{
		double p=((double)(k-t))/((double)(n-1));
		k+=ceil(p);
		cout<<k<<endl;
	}
	
	return 0;
}