You are a coach at your local university. There are n n students under your supervision, the programming skill of the i i -th student is a i 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 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 n ( 1≤n≤2⋅ 10 5 1≤n≤2⋅105 ) — the number of students. The second line of the input contains n n integers a 1 , a 2 ,…, a n a1,a2,…,an ( 1≤ a i ≤ 10 9 1≤ai≤109 ), where a i ai is a programming skill of the i i -th student. Output Print one integer — the maximum possible number of students in a balanced team. Examples Input Copy 6 1 10 17 12 15 2 Output Copy 3 Input Copy 10 1337 1337 1337 1337 1337 1337 1337 1337 1337 1337 Output Copy 10 Input Copy 6 1 1000 10000 10 100 1000000000 Output Copy 1 #####题解:题意给n个数,求子集的最小值和最大值之差不超过5,求最大的子集的元素数. #####我的思路:先对数组排序,这里如果直接暴力的话,O(n^2)的复杂度,肯定会tle, #####优化一下,因为是排过序的,所以子集的左端和右端都是极值,如果右值与左值之差大于5,左值的位置右移, ####再次优化,如果前一值与后一值之差大于5,I 移动的后一值的位置.

#include <bits/stdc++.h>
const int N=2e5+5;
using namespace std;
int a[N],b[N];
int ans=0,cnt=0,flag=0,maxn=0;
void check(int m){
    if(m>maxn) maxn=m;
}
int cal(int i,int j){
    return a[j]-a[i];
}
int main(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    sort(a+1,a+1+n);
    int i=1,j=i+1;
    while(i<=n-1&&j<=n){
        if(cal(j-1,j)>5){
            check(j-1-i);
            i=j;
            j++;
        }
        else if(cal(i,j)>5){
            check(j-1-i);
            i++;
            if(i==j) j++;
            //cout<<"jjg"<<endl;
        }
        else{
            j++;
            //cout<<"jj"<<endl;
        }
        //if(!flag) break;
    }
    check(j-1-i);
    printf("%d\n",maxn+1);
    return 0;
}