/* * 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
     
      #include
      
       using namespace std;#define LL long long#define NMAX 100010int arr[NMAX];LL binary(int n, int nmax, LL sum) {    LL mid = -1, left = nmax, right = sum;    while (left <= right) {        mid = left + (right - left) / 2;        if (sum <= (mid * n)) {            right = mid - 1;        } else {            left = mid + 1;        }    }    return left;}int main() {    freopen("data.in", "r", stdin);    int n, nmax;    LL sum;    while(~scanf("%d",&n)) {        nmax = -1;        sum = 0;        for(int i = 0;i < n;i++) {            scanf("%d",arr + i);            if(nmax < arr[i]) {                nmax = arr[i];            }            sum += arr[i];        }        printf("%I64d\n",binary(n - 1,nmax,sum));    }    return 0;}