One day n friends gathered together to play "Mafia". During each round of the game some player must be the supervisor and othern - 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: thei-th person wants to playai 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?
The first line contains integer n
(3 ≤ n ≤ 105). The second line containsn space-separated integersa1, a2, ..., an(1 ≤ ai ≤ 109)
— thei-th number in the list is the number of rounds thei-th person wants to play.
In a single line print a single integer — the minimum number of game rounds the friends need to let thei-th person play at leastai 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.
3 3 2 2
4
4 2 2 2 2
3
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).
题意：
N个人玩游戏，每论游戏需要1个主持人，N-1个玩家，列出每个人想当玩家的局数，求满足条件的最少局数。
思路：
1.局数最少可能为这N个人中想当玩家局数的最大值
2.若每个人当主持人的局数之和大于等于这个最大值，则这个最大值满足条件
3.若2不满足，设此时需增加x局，剩下没人当主持人的局数为y，则
y + x = N * x;
=> x = y / (N - 1);
y 可求出，则 x 的向上取整即为答案。
/*************************************************************************
> File Name: GGG.cpp
> Author: BSlin
> Mail:
> Created Time: 2013年10月03日 星期四 19时52分01秒
************************************************************************/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <iterator>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <queue>
#define MP make_pair
#define INF (1<<30)
#define PI acos(-1.0)
#define esp 1e-8
const int dx[4]={0,0,0,0};
using namespace std;
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)
#if defined (_WIN32) || defined (__WIN32) || defined (WIN32) || defined (__WIN32__)
#define LL __int64
#define LLS "%" "I" "6" "4" "d"
#else
#define LL long long
#define LLS "%" "l" "l" "d"
#endif
#define M 100010
LL a[M];
int main(int argc, char** argv) {
int n;
LL max,sum,ans;
while(scanf("%d",&n) != EOF) {
max = 0;
for(int i=0; i<n; i++) {
scanf(LLS,&a[i]);
if(max < a[i]) max = a[i];
}
sum = 0;
for(int i=0; i<n; i++) {
sum += (max - a[i]);
}
if(sum >= max) {
ans = max;
} else {
ans = max - sum;
if(ans % (n - 1) == 0) ans = max + ans / (n - 1);
else ans = max + ans / (n - 1) + 1;
//ans = max + (max - sum + n - 2) / (n - 1); //可取代前3句
}
printf(LLS"\n",ans);
}
return 0;
}