Codeforces 940-A

A. Points on the linetime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputWe've got no test cases. A big olympiad is coming up. But the problemsetters' number one priority should be adding another problem to the round.The diameter of a multiset of points on the line is the largest distance between two points from this set. For example, the diameter of the multiset {1, 3, 2, 1} is 2.Diameter of multiset consisting of one point is 0.You are given n points on the line. What is the minimum number of points you have to remove, so that the diameter of the multiset of the remaining points will not exceed d?InputThe first line contains two integers n and d (1 ≤ n ≤ 100, 0 ≤ d ≤ 100) — the amount of points and the maximum allowed diameter respectively.The second line contains n space separated integers (1 ≤ xi ≤ 100) — the coordinates of the points.OutputOutput a single integer — the minimum number of points you have to remove.ExamplesinputCopy

3 1
2 1 4

output

1

inputCopy

3 0
7 7 7

output

0

inputCopy

6 31 3 4 6 9 10

output

3

NoteIn the first test case the optimal strategy is to remove the point with coordinate 4. The remaining points will have coordinates 1 and 2, so the diameter will be equal to 2 - 1 = 1.In the second test case the diameter is equal to 0, so its is unnecessary to remove any points.In the third test case the optimal strategy is to remove points with coordinates 1, 9 and 10.
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The remaining points will have coordinates 3, 4and 6, so the diameter will be equal to 6 - 3 = 3.题意：一个数轴上有n个点（点的位置大于等于1），然后点与点之间存在一定的距离（点可以重复，即距离可为0），给定一个数k，问你最少删除掉几个点之后，剩余所有点之间的最大距离小于等于k。题解：直接把它放在一个数轴上来处理。将其分为长度为k的小区间，然后找存在最多的点的区间，最后用n减去该区间内的点的个数就能得到答案。        如图，依次遍历每个区间，最终找到点数最多的区间。



#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int book[105];
int a[105];
int main()
{
int n,d;
cin>>n>>d;
memset(book, 0, sizeof(book));
for(int i = 1; i<=n; i++)
{
cin>>a[i];
book[a[i]]++;
}

sort(a+1, a+n+1);
int m = a
;
int cnt = 0;
int num = 0;
int mm = 0;

for(int i = 1; i<=m; i++)
{
mm = 0;
for(int j = i; j<=i+d ; j++)
{

if(book[j]!=0)
{
mm = mm+book[j];

}
}
if(mm>num)
num = mm;
}
cnt = n - num;
cout<<cnt<<endl;
return 0;
}