Question

这是一项练习;我禁止使用C ++的内置日期功能或任何现成的库。

我有这段代码：

#include <iostream>
#include <vector>

using namespace std;


bool IsLeap(unsigned int  year)
{
    if (year % 4 != 0) return false;
    else
        if (year % 100 != 0) return true;
        else
            if (year % 400 != 0) return false;
            else
                return true;
}

class Date
{
public:
    unsigned short int Day;
    unsigned short int Month;
    unsigned short int Year;
    unsigned short int DayOfWeek;
};

int MonthLimit(int Month, int year)
{
    switch (Month)
    {
    case 4:
    case 6:
    case 9:
    case 11:
    {
        return 30;
        break;
    }
    case 1:
    case 3:
    case 5:
    case 7:
    case 8:
    case 12:
    {
        return 31;
        break;
    }
    case 2:
        if (IsLeap(year))
        {
            return 29;
            break;
        }
        else
        {
            return 28;
            break;
        }

    }
}



const Date FirstDayEver = { 1, 1, 1900, 1 };  //January 1, 1900 was on a Monday.

int main()
{
    unsigned int years;
    cin >> years;

    vector<int> counters(7);   //whenever a 13th falls on a day X, we increment the Xth counter. It's that simple


    for (Date i = FirstDayEver; !((i.Day == 31) && (i.Month == 12) && (i.Year == 1900 + years - 1)); i.Day++)
    {
        i.DayOfWeek = (i.DayOfWeek + 1) % 7;
        if (i.Month == MonthLimit(i.Month, i.Year))
        {
            i.Month++;
            i.Day = 1;
            cout << "Proceeded to " << i.Day << "." << i.Month << "." << i.Year << "\n";
        }
        if ((i.Day == 31) && (i.Month == 12))
        {
            i.Year++;
            i.Day = 1;
            i.Month = 1;
            cout << "Proceeded to " << i.Day << "." << i.Month << "." << i.Year << "\n";
        }

        if (i.Day == 13)
        {
            counters[i.DayOfWeek]++;
            cout << i.Day << "." << i.Month << "." << i.Year << " was a " << i.DayOfWeek << "\n";
        }
    }

    cout << counters[6] << " " << counters[7] << " " << counters[1] << " " << counters[2] << " " << counters[3] << " " << counters[4] << " " << counters[5] << "\n";

    exit(0);
}

调试信息（cout s）用于查看循环中是否发生任何事情。到目前为止，唯一发生的事情是很多线路说＆＃34; 13.1.1900是＆＃34;数量不一。我认为增加日期的理想逻辑在代码中是相当清楚的，但我试图增加错误。哪里是我的错？

Answer 1

首先，

if (i.Month == MonthLimit(i.Month, i.Year))

将月份与月份中的天数进行比较由于第一个月是1，而那个月有31天，因此您永远不会增加月份由于你永远不会增加月份，你永远不会增加年份（如果你已经命名了函数＆＃34; DaysInMonth＆＃34;或者其他更清楚它返回的东西，你可能会发现这更容易。）

你想要

if (i.Day == MonthLimit(i.Month, i.Year))

其次，MonthLimit缺少10月的案例。

Answer 2

您的增量程序太复杂了我实际上还在努力弄清楚它们是如何组合在一起的（边缘情况等），但似乎你试图独立地每天/每月/每年进行。

你可以在这样一个简单的函数中完成所有的递增：

void increment_Date(Date &d)
{
    d.DayOfWeek = (d.DayOfWeek + 1) % 7; //increase weekday
    if (++d.Day > MonthLimit(d.Month, d.Year)//increase day, if larger than month size...
    {
        d.Day = 1;                           //reset day to 1
        if (++d.Month > 12)                  //increase month, if larger than year size
        {
          d.Month = 1;                       //reset month to 1
          d.Year++;                          //increase year
        }
    }
}

逻辑更简单，因为我以一种与你似乎完全不同的方式接近它在上面的函数中，我首先增加最小单位（天），检查溢出，然后在需要时移动到月（等等）。

这种方式的工作原理与汽车里程拨号的数字仅在拨号到达10并且返回0时才会打勾。或者，添加1到9和＆＃ 39;溢出＆＃39;进入数字的10位数。

否则你的代码很好。 MonthLimit的Switch语句是一个不错的选择。我在函数中添加一个switch语句，以返回一周中的字符串：

std::String printable_DOW(int DOW)
{
    switch (DOW)
    case 0:
        return "Sunday";
    case 1:
        return "Monday";
    case 2:
        return "Tuesday";
    case 3:
        return "Wednesday";
    case 4:
        return "Thursday";
    case 5:
        return "Friday";
    case 6:
        return "Saturday";
}

和print_date功能：

void print_date(Date d)
{
    std::cout<< printable_DOW(d.DayOfWeek) 
             << ", " << d.Day << "." << d.Month << "." d.Year;
}

并且很容易将它们放在一起并使用它们：

int main
{

    Date my_date;
    my_date.Day = 1;
    my_date.Month = 1;
    my_date.Year = 2000;
    my_date.DayOfWeek = 0;  //assuming 0 == Sunday, 1 = Monday...
    for (int daycount = 0; daycount < 5114 /*days since 1/1/2000 to today*/; daycount++)
    {
        increment_date(my_date);
        //debug:
        print_date(my_date);
    }
    //print "today's" date:
    print_date(my_date);
}

作为一个班级： 它主要在这里和（大多数）工作。随意编辑人物！

class date
{
private:
    unsigned short int day;
    unsigned short int month;
    unsigned short int year;
    unsigned short int day_of_week;
    int month_limit();
    bool is_leap_year();
public:
    date();
    date(int in_day, int in_month, int in_year, int in_day_of_week);
    date& date::operator=(date rhs)
    date& operator++();    //pre-increment (++date)
    date operator++(int);  //post-increment (date++)

    //these are "special" they don't work exactly as you might think and are slightly broken right now,
    //but I'm putting them in regardless
    //they also don't work on DOW right now.
    date& operator+=(const date& rhs)               
    date& operator-=(const date& rhs)
    inline date operator+(date lhs, const date& rhs)
    inline date operator-(date lhs, const date& rhs)
}
//phwew

date::date(int in_day, int in_month, int in_year, int in_day_of_week)
{
    this->day = in_day;
    this->month = in_month;
    this->year = in_year;
    this->day_of_week = in_doay_of_week;
}

date::month_limit()
{
    switch (this->month)
    {
    case 4:
    case 6:
    case 9:
    case 11:
    {
        return 30;
        break;
    }
    case 1:
    case 3:
    case 5:
    case 7:
    case 8:
    case 12:
    {
        return 31;
        break;
    }
    case 2:
        if (is_leap_year(this->year))
        {
            return 29;
            break;
        }
        else
        {
            return 28;
            break;
        }

    }
}

bool is_leap_year()
{
    if (this->year % 4 != 0) return false;
    else
        if (this->year % 100 != 0) return true;
        else
            if (this->year % 400 != 0) return false;
            else
                return true;
}

date& date::operator=(date rhs)
{
  swap(rhs);
  return *this;
}

date& date::operator++()
{
    this->day_of_week = (this->day_of_week + 1) % 7; //increase weekday
    this->day++;
    if (++(this->day) > month_limit())           //increase day, if larger than month size...
        {
            this->day = 1;                           //reset day to 1
            if (++this->month > 12)                  //increase month, if larger than year size
            {
              this->month = 1;                       //reset month to 1
              this->year++;                          //increase year
            }
        }
    return *this;
}

date date::operator++(int)
{
    date tmp(*this);
    operator++();
    return tmp;
}

//adds years on, then months, then days
date& date::operator+=(const date& rhs)
{
    this->year  += rhs.year;
    this->month += rhs.month;
    this->day   += rhs.day;
    if (this->month > 12)  //get to the right month
    {
        this->year  = this->month / 12;
        this->month = this->month % 12;
    }
    if (this->day > month_limit())
    {
        this->month = this->day / month_limit();
        this->day   = this->day % month_limit();
        if (this->month > 12)  //recalculate  **yes, I know this is currently wrong if more than one month is added on in days**
        {
            this->year  = this->month / 12;
            this->month = this->month % 12;
        }
    }
    return *this;
}

inline date date::operator+(date lhs, const date& rhs)
{
  lhs += rhs;
  return lhs;
}

//subtracts years, then months, then days
date& date::operator-=(const date& rhs)
{
    if ((rhs.year < this->year) || ((rhs.year == this->year) && (rhs.month < this->month))
       || (((rhs.year == this->year) && (rhs.month == this->month)) && (rhs.day < this->day)
    {
        swap(rhs);
    }
    this->year -= rhs.year;
    this->month -= rhs.month;
    this->day -= rhs.day;

    return *this;
}

inline date date::operator+(date lhs, const date& rhs)
{
  lhs += rhs;
  return lhs;
}

循环中增加日期

2 个答案: