Codeforces 348A：Mafia（二分+思维）

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

42 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).

题目大意：n个人打牌，每局需要一个人当裁判，然后给出了n个人每人想当玩家的次数，问如果这些人的愿望都满足时，最少需要玩多少局。

解题思路：判断n*mid-sum>=mid是否符合条件，这个式子怎么来的：mid-a1+mid-a2+mid-a3+mid-a4+...+mid-an，其中 mid-ai是每人当裁判的场数，变形可得上式为n*mid-sum，如果所有人当裁判的场数多余当前场数，那么压缩区间。。。

代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
#define LL __int64int main()
{
LL n;
scanf("%I64d",&n);
LL l=-1,r,sum=0;
LL tmp;
for(LL i=0;i<n;i++)
{
scanf("%I64d",&tmp);
sum=sum+tmp;
l=max(l,tmp);
}
r=sum;
LL mid,ans;
while(l<=r)
{
mid=(l+r)/2;
if(n*mid-sum>=mid)
{
ans=mid;
r=mid-1;
}
else
{
l=mid+1;
}
}
printf("%I64d\n",ans);
return 0;
}