CodeForces - 348A

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 - 1people 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 integers a1, a2, ..., an (1 ≤ ai ≤ 109) — thei-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 airounds.

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 %I64dspecifier.

Sample test(s)

input

33 2 2

output

4

input

42 2 2 2

output

3

如果把焦点集中在n-1个人上，是不好想到方法的，至少我没有想到，不过，如果把焦点集中在那一个当裁判（好像是）的人上，那么就很容易想到。

比如第一组数据，首先次数肯定是不能小于最大的数，假设是3，那么第一个人要最多当0次裁判，第二个人要最多当1次裁判，第三个人要最多当1次裁判，且加起来的次数要至少达到3才行。就像是有三个桶，每个桶的容量分别是0，1，1，要装3升水，只要容量加起来大于等于3，肯定是能办到的。不过显然不大于3，所以还要继续尝试.........

当为4的时候，痛的容量分别是1，2，2，加起来大于4，所以可以.....

当然，枚举的过程也可以省略掉，即我们可以通过数学方法计算出来假设一开始桶的容量和是s_t,目标的容量是s_{e，我们每次都要让所有桶的容量全部加1，目标容量加1，直到能够装这么多水，求出最小的次数，即st+n*k>=se+1*k，其中k是次数，求出最小的k，移项后得st+(n-1)*k>=se，所以k很容易直接计算出来。}

_{最后，这里可以不用开数组，因为我们只需用到和}

#include <cstdio>int n;long long h1,h2,h3;int main(){    while(scanf("%d",&n)!=EOF){        h2=0;h3=0;        for(int i=0;i<n;i++){            scanf("%I64d",&h1);            h2-=h1;            if(h3<h1)   h3=h1;        }        h2+=(n*h3);        if(h2>=h3)  {printf("%I64d\n",h3);continue;}        printf("%I64d\n",h3+(h3-h2)/(n-1)+int(((h3-h2)%(n-1))>0));    }    return 0;}