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

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 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 integersa₁, 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_i 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
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).

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;
}