Codeforces Round #202 (Div. 1) A. Mafia 【二分】

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 - 1 people 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 integersa1, 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 thecin, cout streams
or the %I64d specifier.

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).

如果可以进行游戏，那么必定每次游戏都会有一个人当裁判。所以判断是否可以进行游戏的条件就是检查对于n次游戏，是否都有n个人充当裁判。进行二分就十分简单了。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

__int64 n, a[100100];

bool judge(__int64 x) {
__int64 cnt = 0;
for (__int64 i = 0; i < n; i++) {
//脑残，没注意这个条件，导致一直WA
if (x < a[i])   return false;
cnt += x - a[i];
}
return x <= cnt;
}
int main() {
while (cin>>n) {
for (__int64 i = 0; i < n; i++) cin>>a[i];
__int64 lb = 0, ub = 1e12;
__int64 ans = 0;
while (ub >= lb) {
__int64 mid = (lb + ub)>>1;
if(judge(mid)) {
ans = mid;
ub = mid - 1;
}
else lb = mid + 1;
}
cout<<ans<<endl;
}
return 0;
}