Interfaces as Plugins

Lab C++ Advanced 30 points

Write an interface class Transform { virtual string apply(const string&) const = 0; } with implementations Upper, Reverse and Repeat(k). Read a word and a pipeline of transforms, apply them in order, and print the result after each step.

Input

Line 1: a word. Line 2: n. Then n transforms (upper, reverse, repeat K).

Output

n lines.

Example

Input

abc
3
upper
reverse
repeat 2

Output

ABC
CBA
CBACBA

Sample input available Sample output available

Editor

C++

Hints

Store the pipeline as a vector of base-class pointers.

Solution explanation

New transforms can be added without touching the pipeline code - the open/closed principle.

Show solution

#include <iostream>
#include <memory>
#include <string>
#include <vector>
using namespace std;

class Transform {
public:
    virtual ~Transform() = default;
    virtual string apply(const string &s) const = 0;
};

class Upper : public Transform {
public:
    string apply(const string &s) const override {
        string r = s;
        for (char &c : r) c = toupper(c);
        return r;
    }
};

class Reverse : public Transform {
public:
    string apply(const string &s) const override { return string(s.rbegin(), s.rend()); }
};

class Repeat : public Transform {
public:
    explicit Repeat(int k) : k_(k) {}
    string apply(const string &s) const override {
        string r;
        for (int i = 0; i < k_; i++) r += s;
        return r;
    }
private:
    int k_;
};

int main() {
    string word;
    int n;
    cin >> word >> n;
    vector<unique_ptr<Transform>> pipeline;
    for (int i = 0; i < n; i++) {
        string kind;
        cin >> kind;
        if (kind == "upper") pipeline.push_back(make_unique<Upper>());
        else if (kind == "reverse") pipeline.push_back(make_unique<Reverse>());
        else {
            int k;
            cin >> k;
            pipeline.push_back(make_unique<Repeat>(k));
        }
    }
    for (const auto &t : pipeline) {
        word = t->apply(word);
        cout << word << "\n";
    }
    return 0;
}

Output

Answer