将秒转换为天、分钟和秒

Question

我已经设置了将秒转换为天、分钟和秒格式的“挑战”。

例如：31600000 = 365 天 46 分钟 40 秒。

using namespace std;
const int hours_in_day = 24;
const int mins_in_hour = 60;
const int secs_to_min = 60;

long input_seconds;
cin >> input_seconds;

long seconds = input_seconds % secs_to_min;
long minutes = input_seconds / secs_to_min % mins_in_hour;
long days = input_seconds / secs_to_min / mins_in_hour / hours_in_day;

cout << input_seconds << " seconds = "
     << days << " days, "
     << minutes << " minutes, "
     << seconds << " seconds ";

return 0;

它有效并给出了正确的答案，但完成后我查看了其他人是如何解决这个问题的，他们的答案有所不同。我想知道我是否错过了什么。

Answer 1

在我看来，这是将秒转换为 DD/hh/mm/ss 的最简单方法：

#include <time.h>
#include <iostream>
using namespace std;    

time_t seconds(1641); // you have to convert your input_seconds into time_t
tm *p = gmtime(&seconds); // convert to broken down time

cout << "days = " << p->tm_yday << endl;
cout << "hours = " << p->tm_hour << endl;
cout << "minutes = " << p->tm_min  << endl;
cout << "seconds = " << p->tm_sec << endl;

Answer 2

我认为这是斯蒂芬·普拉塔（Stephen Prata）书中的挑战。我是这样做的：

#include <iostream>

using namespace std;

int main()
{
    long input_seconds = 31600000;

    const int cseconds_in_day = 86400;
    const int cseconds_in_hour = 3600;
    const int cseconds_in_minute = 60;
    const int cseconds = 1;

    long long days = input_seconds / cseconds_in_day;
    long hours = (input_seconds % cseconds_in_day) / cseconds_in_hour;
    long minutes = ((input_seconds % cseconds_in_day) % cseconds_in_hour) / cseconds_in_minute;
    long seconds = (((input_seconds % cseconds_in_day) % cseconds_in_hour) % cseconds_in_minute) / cseconds;
    cout << input_seconds << " seconds is " << days << " days, " << hours << " hours, " << minutes << " minutes, and " << seconds << " seconds.";

    cin.get();
    return 0;
}

Answer 3

关于编程的事情之一是，做某事从来没有只有一种方法。事实上，如果我下定决心，我也许能够想出十几种完全不同的方法来实现这一目标。如果您的代码满足要求，您就不会遗漏任何内容。

为了您的娱乐，这里有一种在 Windows 下格式化小时：分钟：秒的方法（elapsed 是一个双精度值，表示自...某事以来经过的秒数）

sprintf_s<bufSize>(buf, "%01.0f:%02.0f:%02.2f", floor(elapsed/3600.0), floor(fmod(elapsed,3600.0)/60.0), fmod(elapsed,60.0));

Answer 4

例如：31600000 = 365 天 46 分 40 秒。

真的吗？

$ bc
365*24*60*60 + 46*60 + 40
31538800

365*24*60*60 + 1066*60 + 40
31600000

您的意思是“将输入转换为天、小时、分钟和秒，然后丢弃小时”还是“将输入转换为天、一天内的总分钟（即可以超过 60）和秒”？

在第二种情况下，我认为你应该将几分钟的指令替换为

long minutes = input_seconds / secs_to_min % (mins_in_hour * hours_in_day);

Answer 5

将秒转换为天、小时、分钟和秒；它运行完美，无需消耗太多资源：

#include <iostream>
using namespace std;

int main()
{

    const int Seconds_Per_Day = 86400;
    const short Seconds_Per_Hour = 3600;
    const short Seconds_Per_Minute = 60;
    int Input_Seconds;

    cout << "Please, enter seconds: \n"; 
    cin >> Input_Seconds;

    short Number_Of_Days = floor(Input_Seconds / Seconds_Per_Day);
    int Reminder_1 = Input_Seconds % Seconds_Per_Day;
    short Number_Of_Hours = floor(Reminder_1 / Seconds_Per_Hour);
    short Reminder_2 = Reminder_1 % Seconds_Per_Hour;
    short Number_Of_Minutes = floor(Reminder_2 / Seconds_Per_Minutes);
    short Number_Of_Seconds = Reminder_2 % Seconds_Per_Minutes;

    cout << "The number of days is " << Number_Of_Days << ", of hours is " << 
    Number_Of_Hours << ", of minutes is " << Number_Of_Minutes << ", and of 
    seconds is " << Number_Of_Seconds << endl;

    return 0;
}