解题思路

定义数据结构来存储摄入物质的信息，包括物质类型（巧克力或咖啡）、摄入时间、摄入数量。对于每个查询时间，遍历已记录的摄入信息，根据摄入时间和查询时间的关系，使用相应公式（巧克力：$H(t) = 8*n - t/12$；咖啡：$H(t) = 2*n - t*t/79$）计算该物质对亢奋程度的贡献。将所有物质的贡献累加，若结果为负则取 0，最终结果与 1 比较，取较大值作为亢奋程度。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <iomanip>
#include <string>
#include <cmath>
#include <sstream>

using namespace std;

// 定义结构体存储摄入物质信息
struct Substance {
    string type;
    int time;
    double quantity;
};

int main() {
    vector<Substance> substances;
    vector<int> queryTimes;
    string line;

    while (getline(cin, line)) {
        if (line.substr(0, 5) == "Query") {
            int queryTime;
            istringstream iss(line);
            string queryStr;
            iss >> queryStr >> queryTime;
            queryTimes.push_back(queryTime);
        } else {
            Substance substance;
            istringstream iss(line);
            iss >> substance.type >> substance.time >> substance.quantity;
            substances.push_back(substance);
        }
    }

    // 对查询时间排序
    sort(queryTimes.begin(), queryTimes.end());

    for (int queryTime : queryTimes) {
        double hyperactivityLevel = 0;
        for (const Substance& substance : substances) {
            double contribution = 0;
            int elapsedTime = queryTime - substance.time;
            if (elapsedTime >= 0) {
                if (substance.type == "Chocolate") {
                    contribution = 8 * substance.quantity - elapsedTime / 12.0;
                } else if (substance.type == "Coffee") {
                    contribution = 2 * substance.quantity - pow(elapsedTime, 2) / 79.0;
                }
                // 确保 contribution 不为负
                contribution = max(contribution, 0.0);
            }
            hyperactivityLevel += contribution;
        }
        // 确保 hyperactivityLevel 不小于 1
        hyperactivityLevel = max(hyperactivityLevel, 1.0);

        cout << queryTime << " " << fixed << setprecision(1) << hyperactivityLevel << endl;
    }

    return 0;
}