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≤1000 ) — the number of test cases. The first line of each test case contains two positive integers n (1≤n≤2⋅105 ) and k (1≤k≤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≤109 ) — the difficulty of each problem. Note that the sum of n over all test cases doesn't exceed 2⋅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.用cpp怎么写

你好！对于这个问题，你可以使用以下的C++代码来解决：

#include <iostream>
#include <vector>
#include <algorithm>

int minimumProblemsToRemove(int n, int k, std::vector<int>& difficulties) {
    int minProblemsToRemove = 0;
    std::sort(difficulties.begin(), difficulties.end());

    for (int i = 1; i < n; i++) {
        int diff = difficulties[i] - difficulties[i-1];
        int problemsToRemove = (diff - 1) / k;
        minProblemsToRemove += problemsToRemove;
    }

    return minProblemsToRemove;
}

int main() {
    int t;
    std::cin >> t;

    while (t--) {
        int n, k;
        std::cin >> n >> k;

        std::vector<int> difficulties(n);
        for (int i = 0; i < n; i++) {
            std::cin >> difficulties[i];
        }

        int minProblems = minimumProblemsToRemove(n, k, difficulties);
        std::cout << minProblems << std::endl;
    }

    return 0;
}

这段代码首先读取测试用例的数量 t，然后循环 t 次，每次读取问题的数量 n 和最大允许的两个连续问题之间的难度差值 k。接下来，读取每个问题的难度，并将它们存储在 difficulties 向量中。然后，调用 minimumProblemsToRemove 函数来计算最小需要移除的问题数量，并输出结果。

在 minimumProblemsToRemove 函数中，首先对 difficulties 向量进行排序。然后，使用一个循环遍历每个问题，计算当前问题与前一个问题之间的难度差值 diff。根据差值 diff 和最大允许的难度差值 k，计算需要移除的问题数量，并将其累加到 minProblemsToRemove 变量中。最后，返回 minProblemsToRemove。

希望对你有所帮助！如果有任何疑问，请随时问我。