time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are the author of a Codeforces round and have prepared n� problems you are going to set, problem i� having difficulty ai��. You will do the following process:
- remove some (possibly zero) problems from the list;
- rearrange the remaining problems in any order you wish.
A round is considered balanced if and only if the absolute difference between the difficulty of any two consecutive problems is at most k� (less or equal than k�).
What is the minimum number of problems you have to remove so that an arrangement of problems is balanced?
Input
The first line contains a single integer t� (1≤t≤10001≤�≤1000) --- the number of test cases.
The first line of each test case contains two positive integers n� (1≤n≤2⋅1051≤�≤2⋅105) and k� (1≤k≤1091≤�≤109) --- the number of problems, and the maximum allowed absolute difference between consecutive problems.
The second line of each test case contains n� space-separated integers ai�� (1≤ai≤1091≤��≤109) --- the difficulty of each problem.
Note that the sum of n� over all test cases doesn't exceed 2⋅1052⋅105.
Output
For each test case, output a single integer --- the minimum number of problems you have to remove so that an arrangement of problems is balanced.
Example
input
Copy
7
5 1
1 2 4 5 6
1 2
10
8 3
17 3 1 20 12 5 17 12
4 2
2 4 6 8
5 3
2 3 19 10 8
3 4
1 10 5
8 1
8 3 1 4 5 10 7 3
output
Copy
2
0
5
0
3
1
4Note
For the first test case, we can remove the first 22 problems and construct a set using problems with the difficulties [4,5,6][4,5,6], with difficulties between adjacent problems equal to |5−4|=1≤1|5−4|=1≤1 and |6−5|=1≤1|6−5|=1≤1.
For the second test case, we can take the single problem and compose a round using the problem with difficulty 1010.
解题说明：此题是一道数学题，分析后能发现直接对数列排序，然后再遍历一遍找出符合两个数之间差 <= k 的元素数量是多少， 再用总数量减去符合两个数之间差 <= k 的元素最长连续数量。
cpp
#include<bits/stdc++.h>
using namespace std;
int t, n, k, ans, temp, a[200000];
int main()
{
cin >> t;
while (t--)
{
cin >> n >> k;
for (int i = 0; i < n; i++)
{
cin >> a[i];
}
sort(a, a + n);
temp = 1;
ans = 1;
for (int i = 1; i < n; i++)
{
if (a[i] - a[i - 1] <= k)
{
temp++;
}
else
{
temp = 1;
}
ans = max(ans, temp);
}
cout << n - ans << '\n';
}
return 0;
}