Codeforces Round #202 (Div. 1) A. Mafia 推公式 + 二分答案-优快云博客

本文介绍了一个关于多人游戏“Mafia”的问题解决方案，通过数学建模和二分查找的方法，确定了最少的游戏轮数，使得每位参与者都能达到其期望的参与次数。

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http://codeforces.com/problemset/problem/348/A

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 integers a₁, 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 the cin, cout streams or the %I64d specifier.

Examples

Input

3
3 2 2

Output

Input

4
2 2 2 2

Output

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).

这题用了组合数学的表示方法，这个方法经常用而且很好用，其实就是设未知数。

设x[i]表示第i个人，做了多少次裁判，那么ans = sigma(x[i])

对于第i个人，它参赛了ans - x[i]次，那么需要

ans - x[1] >= a[1]

ans - x[2] >= a[2]

ans - x[3] >= a[3]

.....

ans - x[n] >= a[n]

求和，得到(n - 1) * ans >= sum(a[i])

二分ans即可，

注意ans要大于等于max(a[i])

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 1e5 + 20;
LL a[maxn], sum;
LL mx;
int n;
bool check(LL ans) {
    return ans >= (LL)ceil(1.0 * sum / (n - 1)) && ans >= mx;
}
void work() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
        sum += a[i];
        mx = max(mx, a[i]);
    }
    LL be = 1, en = 1e18L;
    while (be <= en) {
        LL mid = (be + en) >> 1;
        if (check(mid)) {
            en = mid - 1;
        } else be = mid + 1;
    }
    cout << be << endl;
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    IOS;
    work();
    return 0;
}