CF1042A Benches

最新推荐文章于 2020-04-25 16:34:36 发布

ShadyPi

最新推荐文章于 2020-04-25 16:34:36 发布

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本文解析了CodeForces竞赛中A类题目“长凳”的解题思路与代码实现。介绍了如何通过排序来找到使额外进入公园的人数达到最大与最小分布的方法。

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原题链接：http://codeforces.com/contest/1042/problem/A

Benches

There are $n$ benches in the Berland Central park. It is known that $a_i$ people are currently sitting on the $i$ -th bench. Another $m$ people are coming to the park and each of them is going to have a seat on some bench out of $n$ available.

Let $k$ be the maximum number of people sitting on one bench after additional $m$ people came to the park. Calculate the minimum possible $k$ and the maximum possible $k$ .

Nobody leaves the taken seat during the whole process.

Input

The first line contains a single integer $n (1 \leq n \leq 100)$ — the number of benches in the park.

The second line contains a single integer $m (1 \leq m \leq 10000)$ — the number of people additionally coming to the park.

Each of the next $n$ lines contains a single integer $a_i(1≤a_i≤100)$ — the initial number of people on the $i$ -th bench.

Output

Print the minimum possible $k$ and the maximum possible $k$ , where $k$ is the maximum number of people sitting on one bench after additional $m$ people came to the park.

Examples

input

4
6
1
1
1
1

output

3 7

input

1
10
5

output

15 15

input

3
6
1
6
5

output

6 12

input

3
7
1
6
5

output

7 13

Note

In the first example, each of four benches is occupied by a single person. The minimum $k$ is $3$ . For example, it is possible to achieve if two newcomers occupy the first bench, one occupies the second bench, one occupies the third bench, and two remaining — the fourth bench. The maximum $k$ is $7$ . That requires all six new people to occupy the same bench.

The second example has its minimum $k$ equal to $15$ and maximum $k$ equal to $15$ , as there is just a single bench in the park and all $10$ people will occupy it.

题解

我都不知道我为什么要排序 ~~（可能是身处排序强校的缘故吧）~~。

代码

#include<bits/stdc++.h>
using namespace std;
int a[105],maxn,n,m;
void in(){scanf("%d%d",&n,&m);for(int i=1;i<=n;++i)scanf("%d",&a[i]);}
void ac()
{
	sort(a+1,a+1+n);maxn=a[n]+m;
	for(int i=1;i<=n;++i)m-=a[n]-a[i];
	if(m<0)printf("%d",a[n]);
	else printf("%d",a[n]+m/n+(bool)(m%n));
	printf(" %d",maxn);
}
int main(){in();ac();}