CodeForces - 348A Mafia
题目描述：
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 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) — 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 ai 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
4
Input
4
2 2 2 2
Output
3
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).
题目大意：一共有n个人，给出每个人参与游戏的最小次数，每场游戏都需要一个人当裁判，问最少需要进行多少场游戏。
思路：二分模拟咯。最少进行多少场。。。最多进行多少场。模拟mid时，每个人满足了游戏最少次数就把他流放去放裁判。。。最后如果被流放总和太少，小于mid，low=mid，不然high=mid。
上代码！！！
#include <iostream>
#include <cctype>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
using namespace std;
int n,a[100010];
int main(){
while(~scanf("%d",&n)){
long long r = 0 , l = 0, t = 0 , mid , ans;
for(int i = 0 ; i < n ; ++i){
scanf("%d",&a[i]);
r += a[i];
}
sort(a,a+n);
l = a[n-1];
while( l < r){
mid = (r + l) / 2.0;
t = 0;
for(int i = 0 ; i < n ; ++i){
long long temp = mid - a[i];
t += temp;
}
if(t < mid) l = mid + 1;
else{
ans = mid;
r = mid;
}
}
printf("%I64d\n",ans);
}
return 0;
}