Cf386B

/*
B - Fly, freebies, fly!
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
Submit
 
Status
 
Practice
 
CodeForces 386B
Description
Everyone loves a freebie. Especially students.

It is well-known that if in the night before exam a student opens window, opens the student's record-book and shouts loudly three times "Fly, freebie, fly!" — then flown freebie helps him to pass the upcoming exam.

In the night before the exam on mathematical analysis n students living in dormitory shouted treasured words. The i-th student made a sacrament at the time ti, where ti is the number of seconds elapsed since the beginning of the night.

It is known that the freebie is a capricious and willful lady. That night the freebie was near dormitory only for T seconds. Therefore, if for two students their sacrament times differ for more than T, then the freebie didn't visit at least one of them.

Since all students are optimists, they really want to know what is the maximal number of students visited by the freebie can be.

Input
The first line of the input contains integer n (1 ≤ n ≤ 100), where n — the number of students shouted "Fly, freebie, fly!" The second line contains n positive integers ti (1 ≤ ti ≤ 1000).

The last line contains integer T (1 ≤ T ≤ 1000) — the time interval during which the freebie was near the dormitory.

Output
Print a single integer — the largest number of people who will pass exam tomorrow because of the freebie visit.

Sample Input
Input
6
4 1 7 8 3 8
1
Output
3

By Grant Yuan
2014.7.11
*/
#include<iostream>
#include<stdio.h>

#include<stdlib.h>
#include<string.h>
using namespace std;
int n,T,t[102],sum[102];
int cmp(const void *a,const void *b)
{
    return(*(int*)a-*(int*)b);
}
int main()
{
    cin>>n;
    memset(sum,0,102);
    for(int i=0;i<n;i++)
      cin>>t[i];
      cin>>T;
    qsort(t,n,sizeof(t[0]),cmp);
    for(int i=0;i<n;i++)
       for(int j=i;j<n;j++)
         if(t[i]+T>=t[j])
           sum[i]++;
    int max=sum[0];
    for(int i=0;i<n;i++)
       if(sum[i]>max)
         max=sum[i];
    cout<<max<<endl;
    return 0;
}