A block of 64 contiguous LAPs is reserved for inquiry operations; one LAP common to all devices is reserved for general inquiry, the remaining 63 LAPs are reserved for dedicated inquiry of specific classes of devices (see Assigned Numbers). The same LAP values are used regardless of the contents of UAP and NAP. Consequently, none of these LAPs can be part of a user BD_ADDR. The reserved LAP addresses are 0x9E8B00 to 0x9E8B3F. The general inquiry LAP is 0x9E8B33. All addresses have the LSB at the rightmost position, hexadecimal notation. The default check initialization (DCI) is used as the UAP whenever one of the reserved LAP addresses is used. The DCI is defined to be 0x00 (hexadecimal).

有一组64个连续的LAP保留用于查询操作；其中一个LAP是所有设备共用的，用于通用查询，剩下的63个LAP保留用于针对特定设备类别的专用查询（请参阅分配的编号）。无论UAP和NAP的内容如何，都使用相同的LAP值。因此，这些LAP都不能成为用户BD_ADDR的一部分。保留的LAP地址为0x9E8B00至0x9E8B3F。通用查询LAP为0x9E8B33。所有地址的最低有效位位于最右边的位置，使用十六进制表示。当使用保留的LAP地址之一时，使用默认检查初始化（DCI）作为UAP。DCI被定义为0x00（十六进制）。

相关问题

Jonathan is fighting against DIO's Vampire minions. There are n of them with strengths a1,a2,…,an. Denote (l,r) as the group consisting of the vampires with indices from l to r. Jonathan realizes that the strength of any such group is in its weakest link, that is, the bitwise AND. More formally, the strength level of the group (l,r) is defined as f(l,r)=al&al+1&al+2&…&ar. Here, & denotes the bitwise AND operation. Because Jonathan would like to defeat the vampire minions fast, he will divide the vampires into contiguous groups, such that each vampire is in exactly one group, and the sum of strengths of the groups is minimized. Among all ways to divide the vampires, he would like to find the way with the maximum number of groups. Given the strengths of each of the n vampires, find the maximum number of groups among all possible ways to divide the vampires with the smallest sum of strengths. Input The first line contains a single integer t (1≤t≤104) — the number of test cases. The description of test cases follows. The first line of each test case contains a single integer n (1≤n≤2⋅105) — the number of vampires. The second line of each test case contains n integers a1,a2,…,an (0≤ai≤109) — the individual strength of each vampire. The sum of n over all test cases does not exceed 2⋅105. Output For each test case, output a single integer — the maximum number of groups among all possible ways to divide the vampires with the smallest sum of strengths.c++实现

您的问题是如何使用C++实现解决这个问题。

以下是使用C++编写的解决方案：

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

using namespace std;

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

    while (t--) {
        int n;
        cin >> n;

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

        // Sort the strengths in non-decreasing order
        sort(strengths.begin(), strengths.end());

        int minStrength = strengths[0];
        int groupCount = 1;

        // Count the number of groups with the smallest strength
        for (int i = 1; i < n; i++) {
            if (strengths[i] == minStrength) {
                groupCount++;
            } else {
                break;
            }
        }

        cout << groupCount << endl;
    }

    return 0;
}

这段代码首先读取测试用例的数量t。然后，对于每个测试用例，它读取吸血鬼的数量n和每个吸血鬼的强度。然后，它对吸血鬼的强度进行排序，并找到最小的强度minStrength。接下来，它遍历强度数组并计算具有最小强度的组数groupCount。最后，将groupCount输出到标准输出。

请注意，这只是一种可能的解决方案，并且可能不是最优解。还有其他方法可以解决这个问题。

You are given a string s consisting of the characters 0, 1 and/or ?. Let's call it a pattern. Let's say that the binary string (a string where each character is either 0 or 1) matches the pattern if you can replace each character ? with 0 or 1 (for each character, the choice is independent) so that the strings become equal. For example, 0010 matches ?01?, but 010 doesn't match 1??, ??, or ????. Let's define the cost of the binary string as the minimum number of operations of the form "reverse an arbitrary contiguous substring of the string" required to sort the string in non-descending order. You have to find a binary string with the minimum possible cost among those that match the given pattern. If there are multiple answers, print any of them. Input The first line contains a single integer t (1≤t≤3⋅104 ) — the number of test cases. The first and only line of each test case contains the string s (1≤|s|≤3⋅105 ) consisting of characters 0, 1, and/or ?. The sum of the string lengths over all test cases does not exceed 3⋅105 . Output For each test case, print a binary string with the minimum possible cost among those that match the given pattern. If there are multiple answers, print any of them. Example

这是一道编程题，需要你编写程序来解决。以下是C++的代码实现，可以供参考：

#include <bits/stdc++.h>
using namespace std;

int main() {
    int t;
    cin >> t;
    while (t--) {
        string s;
        cin >> s;
        int n = s.size();
        vector<int> q;
        for (int i = 0; i < n; i++) {
            if (s[i] == '?') {
                q.push_back(i);
            }
        }
        if (q.empty()) {
            int cnt = 0;
            for (int i = 1; i < n; i++) {
                if (s[i] != s[i-1]) {
                    cnt++;
                }
            }
            if (cnt == 0) {
                cout << s << endl;
            } else if (cnt == 1) {
                cout << s << endl;
            } else {
                cout << "NO" << endl;
            }
            continue;
        }
        bool ok = false;
        string ans(n, '0');
        for (int i = 0; i < (1<<q.size()); i++) {
            for (int j = 0; j < q.size(); j++) {
                ans[q[j]] = (i>>j)&1 ? '1' : '0';
            }
            int cnt = 0;
            for (int j = 1; j < n; j++) {
                if (ans[j] != ans[j-1]) {
                    cnt++;
                }
            }
            if (cnt == q.size() || cnt == q.size()+1) {
                ok = true;
                break;
            }
        }
        if (ok) {
            cout << ans << endl;
        } else {
            cout << "NO" << endl;
        }
    }
    return 0;
}

该代码先处理出原始字符串中所有的问号的位置，然后枚举所有可能的填充方案，对于每个方案，计算出填充后的字符串的最小排序代价，判断是否符合要求，如果符合，则输出该方案，否则输出NO。