销售系统问题

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**Question**

A distributor distributes energy drinks in 34 oz cans to customers who each have a unique ID. It distributes the drinks according to the demand for them. It developed a system that handles all the transaction queries of the store. The system will handle two types of queries (i.e., query 1 and query 2). 

Write an algorithm to help the distributor find the answer for all the type 2 queries.

**Input**
- The first line of the input consists of an integer `numOfCustomer`, representing the number of customers (C).
- The second line consists of two space-separated integers - `numOfQueries` and `queryDesc` representing the number of queries (Q) and the description of queries (queryDesc = 3 always), respectively.
- The next Q lines consist of three space-separated integers, wherein the first integer represents the type of the query and can only be 1 or 2. If the first integer is 1, then it is a type 1 query and the second and third integers represent the customer ID and the quantity of drinks purchased, respectively. If the first integer is 2, then it is a type 2 query, and the second and third integers represent the starting and ending customer ID range (both inclusive).

**Output**
Print space-separated integers representing the answer to all the type 2 queries. If no type 2 query is present, then do not print anything in the output.

**Constraints**
- \( 0 \leq numOfCustomer \leq 10^5 \)
- \( 0 \leq numOfQueries \leq 10^5 \)
- queryDesc = 3

**Example**
Input: 
4 
5 3 
1 3 12 
2 0 2 
1 1 4 
1 3 2 
2 1 4

问题

一个分销商以34盎司的罐装向每位拥有唯一ID的客户分发能量饮料。它根据需求分发饮料。它开发了一个处理商店所有交易查询的系统。该系统将处理两种类型的查询（即查询1和查询2）。第一个查询确定客户的ID是否已经被记录，如果是，则将新销售的量添加到之前的数量并更新记录。如果客户的ID之前没有被记录过，则会创建一个新条目。查询2检索给定客户ID范围内购买的罐数。在一天结束时，分销商必须提交所有类型2查询的报告。

编写一个算法来帮助分销商找到所有类型2查询的答案。

输入

• 输入的第一行包含一个整数numOfCustomer，代表客户数量（C）。

• 第二行包含两个空格分隔的整数 - numOfQueries和queryDesc，分别代表查询数量（Q）和查询描述（queryDesc始终为3）。

• 接下来的Q行包含三个空格分隔的整数，其中第一个整数代表查询类型，只能是1或2。如果第一个整数是1，则它是类型1查询，第二和第三个整数分别代表客户ID和购买的饮料数量。如果第一个整数是2，则它是类型2查询，第二和第三个整数代表起始和结束客户ID范围（两者都包括在内）。

输出

打印代表所有类型2查询答案的空格分隔整数。如果没有类型2查询，则不在输出中打印任何内容。

约束条件

• ( 0 \leq numOfCustomer \leq 10^5 )

• ( 0 \leq numOfQueries \leq 10^5 )

• queryDesc = 3

示例

输入:
4
5 3
1 3 12
2 0 2
1 1 4
1 3 2
2 1 4

---

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#include <iostream>
#include <map>
#include <vector>
using namespace std;

// 处理查询的函数
vector<int> process_queries(int numOfCustomer,  vector<vector<int>>& queries) {
    map<int, int> customer_data;  // 存储客户ID和购买数量
    vector<int> results;  // 存储类型2查询的结果

    // 使用传统的for循环，基于索引访问queries
    int numOfQueries = queries.size();
    for (int i = 0; i < numOfQueries; ++i) {
        int query_type = queries[i][0];

        if (query_type == 1) {
            // 处理类型1查询 (记录销售)
            int customer_id = queries[i][1];
            int quantity = queries[i][2];

            // 如果客户已存在，则累加，否则创建新记录
            if (customer_data.find(customer_id) != customer_data.end()) {
                customer_data[customer_id] += quantity;
            } else {
                customer_data[customer_id] = quantity;
            }
        } else if (query_type == 2) {
            // 处理类型2查询 (范围查询)
            int start_id = queries[i][1];
            int end_id = queries[i][2];
            int total_quantity = 0;

            // 累加在给定范围内客户的购买数量
            for (int cid = start_id; cid <= end_id; ++cid) {
                if (customer_data.find(cid) != customer_data.end()) {
                    total_quantity += customer_data[cid];
                }
            }
            results.push_back(total_quantity);  // 将结果加入结果集
        }
    }

    return results;
}

int main() {
    // 示例输入
    int numOfCustomer = 4;

    vector<vector<int>> queries = {
        {1, 3, 12},
        {2, 0, 2},
        {1, 1, 4},
        {1, 3, 2},
        {2, 1, 4}
    };

    // 调用函数处理查询
    vector<int> results = process_queries(numOfCustomer, queries);

    // 使用传统的for循环输出结果
    for (int i = 0; i < results.size(); ++i) {
        cout << results[i] << " ";
    }
    cout << endl;

    return 0;
}