解决思路

1. 输入处理：读取输入的测试用例数量，然后逐个处理每个测试用例。
2. 提取数据字段：对于每个测试用例的字符串，我们需要找到两个数据字段。数据字段的格式是固定的，可以通过查找=和单位（A、V、W）来定位。
3. 解析数据字段：对于每个数据字段，提取其概念（P、U、I）、数值和前缀，并将数值转换为标准单位（无前缀）。
4. 计算缺失的量：根据已知的两个量，计算第三个量。
5. 输出结果：按照要求的格式输出结果，注意数值的格式化和单位的正确性。

解决代码

#include <iostream>
#include <string>
#include <iomanip>
#include <sstream>
#include <vector>
#include <cctype>
#include <cstdlib> // for atof

using namespace std;

struct DataField {
    char concept;
    double value;
    char prefix;
    char unit;
};

DataField parseDataField(const string& field) {
    DataField data;
    size_t eq_pos = field.find('=');
    data.concept = field[eq_pos - 1];
    
    size_t unit_pos = field.find_first_of("AVW", eq_pos);
    data.unit = field[unit_pos];
    
    string num_str = field.substr(eq_pos + 1, unit_pos - eq_pos - 1);
    data.prefix = '\0';
    if (!num_str.empty() && (num_str[num_str.size()-1] == 'm' || num_str[num_str.size()-1] == 'k' || num_str[num_str.size()-1] == 'M')) {
        data.prefix = num_str[num_str.size()-1];
        num_str = num_str.substr(0, num_str.size() - 1);
    }
    // Replace stod with atof (C-style conversion)
    data.value = atof(num_str.c_str());
    
    return data;
}

double getValueWithPrefix(double value, char prefix) {
    switch (prefix) {
        case 'm': return value * 0.001;
        case 'k': return value * 1000;
        case 'M': return value * 1000000;
        default: return value;
    }
}

void solve() {
    int T;
    cin >> T;
    cin.ignore(); // Ignore the newline after T
    
    for (int k = 1; k <= T; ++k) {
        string line;
        getline(cin, line);
        
        vector<DataField> fields;
        size_t pos = 0;
        while (pos < line.size()) {
            size_t eq_pos = line.find('=', pos);
            if (eq_pos == string::npos) break;
            
            // Find the start of the field (start of concept)
            size_t start = eq_pos - 1;
            while (start > 0 && !isspace(line[start - 1])) {
                --start;
            }
            
            // Find the end of the field (end of unit)
            size_t unit_pos = line.find_first_of("AVW", eq_pos);
            if (unit_pos == string::npos) break;
            size_t end = unit_pos + 1;
            
            string field_str = line.substr(start, end - start);
            DataField field = parseDataField(field_str);
            fields.push_back(field);
            
            pos = end;
        }
        
        double P = -1, U = -1, I = -1;
        for (vector<DataField>::const_iterator it = fields.begin(); it != fields.end(); ++it) {
            const DataField& field = *it;
            double value = getValueWithPrefix(field.value, field.prefix);
            if (field.concept == 'P') {
                P = value;
            } else if (field.concept == 'U') {
                U = value;
            } else if (field.concept == 'I') {
                I = value;
            }
        }
        
        cout << "Problem #" << k << endl;
        if (P == -1) {
            P = U * I;
            cout << fixed << setprecision(2) << "P=" << P << "W" << endl;
        } else if (U == -1) {
            U = P / I;
            cout << fixed << setprecision(2) << "U=" << U << "V" << endl;
        } else if (I == -1) {
            I = P / U;
            cout << fixed << setprecision(2) << "I=" << I << "A" << endl;
        }
        cout << endl;
    }
}

int main() {
    solve();
    return 0;
}