如何找到最大值和最小值

Question

我想在一個函數中找到最大值和最小值。 我試過了，但是我只得到了最小值。 這是因為代碼不是從第一行讀取的，並且第一行包含最高值。 這是我的代碼：

void price (ifstream & infile)
{

   infile.clear();
   infile.seekg(0);

   int year, mall, cafe, expenses;
   infile >> year >> mall >> cafe >> expenses;


   int high_mall {} , high_year {}, high_expenses {};
   int low_mall {} , low_year {}, low_expenses {};

   low_mall = mall;
   low_year = year;
   low_expenses = expenses;

    for (int year {}, mall {}, cafe {}, expenses {};
       nameFile >> year >> mall >> cafe >> expenses; )
    {

        if (expenses > high_expenses)
        {
            high_expenses = expenses;
            high_year = year;
            high_mall = mall;
        }
         if (expenses < low_expenses)
        {
            low_expenses = expenses;
            low_year = year;
            low_mall = mall;
        }
    }

        cout << "Highest expenses are " << high_expenses << "at" << high_year << endl;
        cout << "Total mall that year are " << high_mall << endl ;
        
        cout << "Lowest expenses are " << low_expenses << "at" << low_year << endl;
        cout << "Total mall that year are " << low_mall << endl ;

    }

這是我的代碼。 我試圖刪除第一個內聯並獲得最高值但最小值變為 0。有誰知道如何解決它？ 我在考慮std::numeric_limits 。 但是我可以不使用它來解決這個問題嗎？

Answer 1

由於low_expenses的初始值為零，您將永遠找不到任何費用較低的年份，除非費用為負數。 您應該將low_expenses初始化為最大可能的整數，如下所示：

low_expenses{ std::numeric_limits<int>::max() };

如果費用可能是負數，將high_expenses初始化為盡可能低的值也有幫助：

high_expenses{ std::numeric_limits<int>::min() };

您需要包含<limits>標頭才能使其工作。

此外，正如 WhozCraig 指出的那樣，這條線

infile >> year >> mall >> cafe >> expenses;

將丟棄一行輸入。 沒有必要。 您可以將low_...和high_...初始化為第一行的值，但是您可能還需要檢查第一個 I/O 操作是否成功。 （如果沒有行）

Answer 2

在進入循環讀取其余內容之前，在初始讀取后初始化最高和最低內容。

例子：

void price(ifstream &infile)
{
    infile.clear();
    infile.seekg(0);

    int year, mall, cafe, expenses;
    if (infile >> year >> mall >> cafe >> expenses)
    {
        int high_mall{mall}, high_year{year}, high_expenses{expenses};
        int low_mall{mall}, low_year{year}, low_expenses{expenses};

        while (infile >> year >> mall >> cafe >> expenses)
        {
            if (expenses > high_expenses)
            {
                high_expenses = expenses;
                high_year = year;
                high_mall = mall;
            }
            else if (expenses < low_expenses)
            {
                low_expenses = expenses;
                low_year = year;
                low_mall = mall;
            }
        }

        cout << "Highest expenses are " << high_expenses << "at" << high_year << endl;
        cout << "Total mall that year are " << high_mall << endl;

        cout << "Lowest expenses are " << low_expenses << "at" << low_year << endl;
        cout << "Total mall that year are " << low_mall << endl;
    }
}

Answer 3

一個好的做法是使用標准庫已經提供的 Thins。 有std::minmax_element可以完全滿足您的需求。

也可以將流轉換為某些特定數據的迭代器： std::istream_iterator 。

這兩件事可以組合成以下代碼：

struct Data
{
    int year, mall, cafe, expenses;
};

std::istream& operator>>(std::istream& in, Data& data)
{
    return in >> data.year >> data.mall >> data.cafe >> data.expenses;
}

std::ostream& operator<<(std::ostream& out, const Data& data)
{
    return out << data.year << ", " << data.mall << ", " << data.cafe << ", " << data.expenses;
}


int main()
{
    auto result = std::minmax_element(std::istream_iterator<Data>{std::cin}, {}, 
        [](const auto& a, const auto& b) { return a.expenses < b.expenses; });

    std::cout << *result.first << '\n';
    std::cout << *result.second << '\n';

    return 0;
}

https://godbolt.org/z/EMvjx5Yrq

Answer 4

我試試這個，它起作用了

void price (ifstream & infile)
{

   infile.clear();
   infile.seekg(0);

  int year, mall, cafe, expenses;

  int high_mall {} , high_year {}, high_expenses {};


  int low_mall;
  int low_year;
  int low_expenses;

  for (int year {}, mall {}, cafe {}, expenses {};
   nameFile >> year >> mall >> cafe >> expenses; )
{

    if (expenses > high_expenses)
    {
        high_expenses = expenses;
        high_year = year;
        high_mall = mall;
    }
     if (expenses < low_expenses)
    {
        low_expenses = expenses;
        low_year = year;
        low_mall = mall;
    }
}

    cout << "Highest expenses are " << high_expenses << "at" << high_year << endl;
    cout << "Total mall that year are " << high_mall << endl ;
    
    cout << "Lowest expenses are " << low_expenses << "at" << low_year << endl;
    cout << "Total mall that year are " << low_mall << endl ;

}