【CodeForces】348A - Mafia（二分 & 思维）

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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 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 airounds.

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

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).

这道题能想出是二分就好做了。

l 等于给出的数中最大的（至少进行这么多次），r 等于总和。

然后二分，需要的这个 x 回合，如果 x*n - sum >= x 的话，那么就满足。为什么呢？我们把式子分解一下：

x - num [ 1 ] + x - num [ 2 ] + x - num [ n ] （这个式子表示每一个人可以当裁判的回合数的总和，然后合并一下就是上面的式子了，让它大于总回合数 x 就行了）

代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
	__int64 sum;
	__int64 n;
	__int64 l,r,mid;
	while (~scanf ("%I64d",&n))
	{
		sum = 0;
		l = 0;
		for (int i = 1 ; i <= n ; i++)
		{
			__int64 t;
			scanf ("%I64d",&t);
			l = max (l , t);
			sum += t;
		}
		r = sum;
		while (r >= l)
		{
			mid = (l + r) >> 1;
			if (mid * n - sum >= mid)
				r = mid - 1;
			else
				l = mid + 1;
		}
		printf ("%I64d\n",l);
	}
	return 0;
}