cf Codeforces Round #544 C. Balanced Team

原題：

C. Balanced Team
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are a coach at your local university. There are n
students under your supervision, the programming skill of the i-th student is ai.

You have to create a team for a new programming competition. As you know, the more students some team has the more probable its victory is! So you have to create a team with the maximum number of students. But you also know that a team should be balanced. It means that the programming skill of each pair of students in a created team should differ by no more than 5.

Your task is to report the maximum possible number of students in a balanced team.
Input

The first line of the input contains one integer n
(1≤n≤2⋅10^5) — the number of students.

The second line of the input contains n
integers a1,a2,…,an (1≤ai≤10^9), where ai is a programming skill of the i-th student.
Output

Print one integer — the maximum possible number of students in a balanced team.

Examples
Input

6
1 10 17 12 15 2

Output

3

Input

10
1337 1337 1337 1337 1337 1337 1337 1337 1337 1337

Output

10

Input

6
1 1000 10000 10 100 1000000000

Output

1

Note

In the first example you can create a team with skills [12,17,15].

In the second example you can take all students in a team because their programming skills are equal.

In the third example you can create a team consisting of a single student (and you cannot create a team consisting of at least two students).

中文：

給你一堆數，讓你挑出其中的一部分數，這些數之間的差的絕對值不超過5，最多能有多少個數。

代碼：

#include<bits/stdc++.h>
using namespace std;
 
typedef long long ll;
 
const int maxn=200005;
int a[maxn];
 
int main()
{
    ios::sync_with_stdio(false);
    int n;
    while(cin>>n)
    {
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        sort(a+1,a+1+n);
        int s=1;
        int ans=0;
        for(int i=1;i<=n&&s<=n;)
        {
            if(a[i]-a[s]<=5)
            {
                ans=max(ans,i-s+1);
                i++;
            }
            else
            {
                s++;
            }
        }
 
        cout<<ans<<endl;
    }
    return 0;
}

思路：

排序後是直接用雙指針（尺取法）即可。