【CodeForces-348A】解题报告(直接或二分)

原始题目

A. Mafia

  • time limit per test2 seconds
  • memory limit per test256 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 - 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 $n (3 ≤ n ≤ 10^5)$. The second line contains $n$ space-separated integers $a_1, a_2  \cdots a_n (1 ≤ a_i ≤ 10^9)$ — 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

$n$ 个人进行游戏,每局游戏需要1人监管,剩下 $n-1$ 人参与游戏,现在给出 $n$ 人每人想至少参与游戏的此数,求满足每个人要求的最少游戏盘数。

解题思路

水题,可以直接解或者二分

解题代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
long long a[200010];
int main(){
int n;
long long sum=0,MAX=0;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%I64d",a+i);
sum+=a[i]; MAX=max(a[i],MAX);
}
long long ans;
if(sum%(n-1)==0) ans=sum/(n-1);
else ans=sum/(n-1)+1;
ans=max(MAX,ans);
printf("%I64d\n",ans);
}

收获与反思

  • 二分做法,下界为数列最大值,上界为数列总和,做下面划分
    1
    2
    if(mid*(n-1)>=sum) ri=mid;
    else le=mid+1;