Codeforces Round #202 (Div. 1) A. Mafia二分查找，玩游戏

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

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 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.

Sample Input

Input

3
3 2 2

Output

4

Input

42 2 2 2

Output

3

Hint

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个人在玩游戏，但在玩游戏中必须要有人监督，给出n个数，表示每个人要玩的次数，问至少要玩多少次游戏，才能满足所有人要求

#include<cstdio>
int n;
__int64 a, sum, mx, low, top, mid;
int main(){
while(~scanf("%d", &n)){
mx=sum=0;
for(int i=0; i<n; i++){
scanf("%I64d", &a);
sum+=a;
if(a>mx)	mx=a;//找出二分时最小值
}
low=mx;
top=sum;
while(low<top){
mid = (low+top)/2;
if(mid*n-sum>=mid)//判断是否每人都有一次监督的机会
top=mid;
else
low=mid+1;
}
printf("%I64d\n", low);
}
return 0;
}