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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    // 直接用数组映射难度数字到比例，索引即难度数字
    double ratio[11] = {0.0};
    ratio[2] = 0.4; ratio[3] = 0.5; ratio[4] = 0.6;
    ratio[6] = 0.7; ratio[7] = 0.8; ratio[9] = 0.9; ratio[10] = 1.0;

    int T;
    cin >> T;
    while (T--) {
        int m;
        cin >> m;
        double total = 0.0;
        for (int i = 0; i < m; i++) {
            char op;
            cin >> op;
            if (op == 'A') {
                int x;
                cin >> x;
                total += (x <= 10) ? x : (x % 10) * 10;
            } 
            else if (op == 'B') {
                int n, c;
                cin >> n >> c;
                // B 操作整数部分先算好，最后 floor
                double coins = floor(ratio[n] * 50.0 * c * 0.1);
                total += coins;
            } 
            else { // C 操作
                int n, c;
                cin >> n >> c;
                total += ratio[n] * 20.0 * c * 0.01;
            }
        }
        cout << fixed << setprecision(2) << total << '\n';
    }
    return 0;
}

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    // 难度比例表
    unordered_map<int, double> ratio = {
        {2, 0.4}, {3, 0.5}, {4, 0.6},
        {6, 0.7}, {7, 0.8}, {9, 0.9}, {10, 1.0}
    };

    int T;
    cin >> T;
    while (T--) {
        int m;
        cin >> m;
        double total = 0.0;
        for (int i = 0; i < m; i++) {
            char op;
            cin >> op;
            if (op == 'A') {
                int x;
                cin >> x;
                if (x >= 1 && x <= 10) total += x;
                else total += (x % 10) * 10;
            } 
            else if (op == 'B') {
                int n, c;
                cin >> n >> c;
                total += ratio[n] * 50 * c * 0.1;
            } 
            else if (op == 'C') {
                int n, c;
                cin >> n >> c;
                total += ratio[n] * 20 * c * 0.01;
            }
        }
        cout << fixed << setprecision(2) << total << "\n";
    }
    return 0;
}