CF#348 A. Mafia（二分，思维）

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.

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).

题意：n个人玩游戏，每次选出一位主持人（不参与游戏），第i个人至少要玩a[ i ]次游戏，问总共至少需要玩多少次。

思路一：用二分枚举最小值。

#include<bits/stdc++.h>
using namespace std;
const int N = 100000 + 10;
int n;
long long a[N];
int judge(long long mid)
{
    long long ans=0;
    for(int i=1; i<=n; ++i)
    {
        if(a[i]>mid)return 0;
        ans+=mid-a[i];
    }
    if(ans>=mid) return 1;
    return 0;
}
int main()
{
    while(scanf("%d",&n)==1)
    {
        long long sum=0;
        for(int i=1; i<=n; ++i)
        {
            scanf("%I64d",&a[i]);
            sum+=a[i];
        }
        long long ans=0,left=0,right=sum;
        while(left<=right)
        {
            long long mid=(left+right)>>1;
            if(judge(mid))  ans=mid,right=mid-1;
            else    left=mid+1;
        }
        printf("%I64d\n",ans);
    }
}

思路二：总共n人，每次n - 1人参与游戏，那么理想情况就为 所有人想参加的次数sum /( n - 1 ) + ( sum % ( n - 1)  > 0) ，可实际情况可能会有极端值，所以我们只需将理想情况和极端情况比较，输出较大者即可。

#include<bits/stdc++.h>
using namespace std;
const int N = 100000 + 10;
int n;
long long a[N];

int main()
{
    while(scanf("%d",&n)==1)
    {
        long long sum=0,maxi=-0x3f3f3f;
        for(int i=0;i<n;i++)
        {
            scanf("%I64d",&a[i]);
            sum+=a[i];
            if(maxi<a[i])
                maxi=a[i];
        }
        long long ans=sum/(n-1)+(sum%(n-1)>0);
        printf("%I64d\n",max(ans,maxi));
    }
}

PS： 这是队长古宇大大的思路，佩服佩服~

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