C++ 如何限制减量？

最初的游戏难度是

game_difficulty=5   //Initial

每3次如果你做对了，你的难度就会上升到无穷大，但每3次你做错了，你的难度就会下降，但不会低于5。因此，在该代码中，例如：

if(user_words==words)   win_count+=1;
else()  incorrect_count+=1;


if(win_count%3==0) /*increase diff*/;
if(incorrect_count%3==0) /*decrease difficulty*/;

---

我该怎么做呢

您可以将此作为一个条件添加：

if (user words==words) {
    win_count += 1;
    if (win_count %3 == 0) {
        ++diff;
    }
} else {
    incorrect_count += 1;
    if (incorrect_count % 3 == 0 && diff > 5) {
        --diff
    }
}

例如：

if(win_count%3==0) difficulty++;
if(incorrect_count%3==0 && difficulty > 5) difficulty--;

简单回答：

if(incorrect_count%3==0) difficulty = max(difficulty-1, 5);

但就我个人而言，我会把它放在一个小类中，然后你可以包含所有的逻辑，并在你继续的过程中进行扩展，比如：

class Difficulty
{
public:
    Difficulty() {};

    void AddWin()
    {
        m_IncorrectCount = 0; // reset because we got one right?

        if (++m_WinCount % 3)
        {
            m_WinCount = 0;
            ++m_CurrentDifficulty;
        }
    }

    void AddIncorrect()
    {
        m_WinCount = 0; // reset because we got one wrong?

        if (++m_IncorrectCount >= 3 && m_CurrentDifficulty > 5)
        {
            m_IncorrectCount = 0;
            --m_CurrentDifficulty;
        }
    }

    int GetDifficulty()
    {
        return m_CurrentDifficulty;
    }

private:
    int m_CurrentDifficulty = 5;
    int m_WinCount = 0;
    int m_IncorrectCount = 0;
};

---

这可以成为定制数据类型的激励示例

创建一个类，将难度

int

包装为私有成员变量，并在公共成员函数中确保满足所谓的约定。您将得到一个始终保证满足您的规格的值。以下是一个例子：

class Difficulty
{
public:

    // initial values for a new Difficulty object:
    Difficulty() :
        right_answer_count(0),
        wrong_answer_count(0),
        value(5)
    {}

    // called when a right answer should be taken into account:
    void GotItRight()
    {
        ++right_answer_count;
        if (right_answer_count == 3)
        {
            right_answer_count = 0;
            ++value;
        }
    }

    // called when a wrong answer should be taken into account:
    void GotItWrong()
    {
        ++wrong_answer_count;
        if (wrong_answer_count == 3)
        {
            wrong_answer_count = 0;
            --value;
            if (value < 5)
            {
                value = 5;
            }
        }
    }

    // returns the value itself
    int Value() const
    {
        return value;
    }

private:
    int right_answer_count;
    int wrong_answer_count;
    int value;
};

一旦您掌握了这些类型是如何创建和使用的，您就可以开始研究运算符重载，以便该类型可以像真正的

int

一样使用，即使用

+

，

-

等等

我该怎么做呢

<>你把这个问题标记为C++。iMHO C++的方法是创建一个封装所有问题的类。 也许是这样的：

class GameDifficulty
{
public:
    GameDifficulty () : 
       game_difficulty (5), win_count(0), incorrect_count(0)
    {}

    ~GameDifficulty () {}

    void update(const T& words)
    {
       if(user words==words)   win_count+=1;
       else              incorrect_count+=1;

       // modify game_difficulty as you desire
       if(win_count%3 == 0)      
          game_difficulty += 1 ; // increase diff no upper limit

       if((incorrect_count%3 == 0) && (game_difficulty > 5)) 
          game_difficulty -= 1; //decrease diff;
    }

   inline int gameDifficulty() { return (game_difficulty); }
   // and any other access per needs of your game

private:
   int game_difficulty;
   int win_count;
   int incorrect_count;
}

//注-未编译或测试

用途如下：

// instantiate
GameDiffculty  gameDifficulty;

// ...

// use update()
gameDifficulty.update(word);

// ...

// use access
gameDifficulty.gameDifficulty();

优点：封装

此代码位于一个地方，不会污染代码中的其他地方

---

你可以在这里更改这些政策，对代码的其余部分没有影响。

使用

if

子句，以确保在递减之前难度不会太低。@MattiVirkkunen这是一个很好的建议，我考虑过它，但我的整个程序有很多条件，我正试图尽可能减少条件，这样逻辑就不会太复杂太令人困惑了。想知道是否还有其他方法！谢谢你的回答！：）在本例中，使用while（n-->最小值），因为语法有效，但更混乱。如果您确信游戏难度>5-=1 : 5;看起来你比我快了！：）那…那真是无中生有。真的没有必要上课。谢谢你的回答！我是编程新手，刚开始学习基本的东西，所以我对类的用法不太了解。但这个概念肯定是可行的。

class GameDifficulty
{
public:
    GameDifficulty () : 
       game_difficulty (5), win_count(0), incorrect_count(0)
    {}

    ~GameDifficulty () {}

    void update(const T& words)
    {
       if(user words==words)   win_count+=1;
       else              incorrect_count+=1;

       // modify game_difficulty as you desire
       if(win_count%3 == 0)      
          game_difficulty += 1 ; // increase diff no upper limit

       if((incorrect_count%3 == 0) && (game_difficulty > 5)) 
          game_difficulty -= 1; //decrease diff;
    }

   inline int gameDifficulty() { return (game_difficulty); }
   // and any other access per needs of your game

private:
   int game_difficulty;
   int win_count;
   int incorrect_count;
}

// instantiate
GameDiffculty  gameDifficulty;

// ...

// use update()
gameDifficulty.update(word);

// ...

// use access
gameDifficulty.gameDifficulty();