Find the customer id of the customers who have made a payment on 2005-05-25 but the amount does not exceed 3.

您所描述的是一个数据库查询问题，通常可以通过SQL查询来解决。在SQL中，您可以使用SELECT语句结合WHERE子句来筛选出满足特定条件的记录。假设我们有一个名为`payments`的表格，其中包含`customer_id`、`payment_date`和`amount`三个字段，我们可以使用如下查询来找出在指定日期支付且金额不超过3的顾客ID：

SELECT customer_id
FROM payments
WHERE payment_date = '2005-05-25'
AND amount <= 3;

这个查询做了以下几件事情：
1. `SELECT customer_id`指定了我们想从表中选择哪些字段，这里我们只想得到`customer_id`。
2. `FROM payments`指定了我们操作的表名，即`payments`。
3. `WHERE payment_date = '2005-05-25'`指定了筛选条件，即支付日期必须是`2005-05-25`。
4. `AND amount <= 3`添加了另一个筛选条件，即支付金额必须小于或等于3。

运行这个查询后，您将得到所有在2005年5月25日支付且支付金额不超过3的顾客的ID。

相关问题

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。

You are given an array a1,a2,…,an, where all elements are different. You have to perform exactly k operations with it. During each operation, you do exactly one of the following two actions (you choose which to do yourself): find two minimum elements in the array, and delete them; find the maximum element in the array, and delete it. You have to calculate the maximum possible sum of elements in the resulting array. Input The first line contains one integer t (1≤t≤104) — the number of test cases. Each test case consists of two lines: the first line contains two integers n and k (3≤n≤2⋅105; 1≤k≤99999; 2k<n) — the number of elements and operations, respectively. the second line contains n integers a1,a2,…,an (1≤ai≤109; all ai are different) — the elements of the array. Additional constraint on the input: the sum of n does not exceed 2⋅105. Output For each test case, print one integer — the maximum possible sum of elements in the resulting array.

这是一道贪心算法的题目。我们可以将数组 a 排序，然后每次操作时，选择最小的两个元素进行删除，或者选择最大的元素进行删除。根据贪心的思想，我们应该优先选择最小的元素进行删除，因为这样可以保留更多的大元素，从而获得更大的总和。因此，我们可以考虑使用一个小根堆来维护当前数组中最小的两个元素，每次操作时，从堆中取出最小的两个元素进行删除，并将删除后的元素插入堆中。当堆中元素个数不足两个时，我们就从数组中选择最大的元素进行删除。

具体实现时，我们可以使用一个变量 sum 记录当前数组中所有元素的总和，使用一个变量 min_sum 记录当前堆中最小的两个元素的和，使用一个变量 max_num 记录当前数组中最大的元素。每次操作时，如果堆中元素不足两个，就删除最大的元素，否则就从堆中取出最小的两个元素进行删除。删除元素时，需要将 sum 减去被删除元素的值，并将删除后的元素插入堆中。最终，返回 sum 减去 min_sum 即可。

时间复杂度为 O(k log n)。