思维逆向

题目：
We’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?

Input
The 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.

Output
Output a single integer — the minimum number of points you have to remove.

Example
Input
3 1
2 1 4
Output
1
Input
3 0
7 7 7
Output
0
Input
6 3
1 3 4 6 9 10
Output
3
Note
In 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. The remaining points will have coordinates 3, 4 and 6, so the diameter will be equal to 6 - 3 = 3.
思维：逆向。这题如果正向求解的话不仅要将输入的数排序，还要将输入的相同的数的次数排序，再进行比较。但是如果逆向求解，求可以保留的最大数目，一个二维循环即可
ac代码：

#include<iostream>
#include<algorithm>
#include<stdio.h> 
using namespace std;
int main()
{
    int m,n,co=0;
    int a[110];
    while(~scanf("%d%d",&m,&n))
    {
        for(int i=0;i<m;i++)
    {
        cin>>a[i];
    } 
    sort(a,a+m);

    for(int i=0;i<m;i++)
    {
        for(int j=i;j<m;j++)
           {
            if(a[j]-a[i]<=n)  co=max(co,j-i+1);
           }
    }
    cout<<m-co<<endl;
    }

    return 0;
}

其他可参考的逆向：http://blog.youkuaiyun.com/u013007900/article/details/49493967