codeforces Mafia

/*
 * Mafia.cpp
 *
 *  Created on: 2013-10-12
 *      Author: wangzhu
 */

/**
 * 每个人都想玩若干场，求至少需要玩几场才可以满足大家的需求。
 * 结果必然在某个人想玩的次数nmax（此人想玩的是最多的）与所有人想玩的次数和sum之间，
 * 故二分，left = nmax,right = sum,
 * 只需要需要玩的次数 * （总人数-1） >= 大家想玩的次数和即可
 *
 */
#include<cstdio>
#include<iostream>
using namespace std;
#define LL long long
#define NMAX 100010
int arr[NMAX];
LL binary(int n, int nmax, LL sum) {
    LL mid = -, left = nmax, right = sum;
    while (left <= right) {
        mid = left + (right - left) / ;
        if (sum <= (mid * n)) {
            right = mid - ;
        } else {
            left = mid + ;
        }
    }
    return left;
}
int main() {
    freopen("data.in", "r", stdin);
    int n, nmax;
    LL sum;
    while(~scanf("%d",&n)) {
        nmax = -;
        sum = ;
        for(int i = ;i < n;i++) {
            scanf("%d",arr + i);
            if(nmax < arr[i]) {
                nmax = arr[i];
            }
            sum += arr[i];
        }
        printf("%I64d\n",binary(n - ,nmax,sum));

    }
    return ;
}