Question

这是一个课程的作业，我几乎到了我想要的地方，只是想要一些关于最佳方法的建议。

用户将以24小时格式输入时间，以小时和分钟为单位输入延迟时间。我必须在延迟结束时以AM / PM格式计算和显示时间。我必须使用函数，不能使用全局变量。我还需要在为小时部分输入999时退出条件。

基本上，我正在尝试决定继续计算延迟的最佳方法。我尝试了几种不同的方法，但是，对它们中的任何一种都没有真正的满意。现在，程序只是循环询问小时，分钟和延迟，直到我能够决定最佳的继续方式。

#include <iostream>

using namespace std;

//Function Prototypes
void convert(int&, int&, int&, char&);
void output(int&, int&, char&);

int main()
{

    int hours, minutes, delay;
    char ampm; 

    cout<<"Enter the value for Hours (999 to quit):  "<<endl;
    cin >> hours;



    while (hours != 999)
    {
        cout << "Enter minutes: ";
        cin >> minutes;

        cout << "Enter delay: ";
        cin >> delay;

        cout<<"Enter the value for Hours (999 to quit):  "<<endl;
    cin >> hours;


    }
return 0;   
}


void convert(int& hours, int& minutes,  int& delay, char& ampm)
{

    if(hours > 12)
    {
        hours  =  hours - 12;
        ampm = 'p';
    }
    else if(hours  == 12) ampm = 'p'; 
    else ampm = 'a'; 
}


void output(int& hours, int& minutes, char& ampm)
{
    if(ampm == 'p')
    {
        (minutes  < 10); cout << hours << ":0" << minutes << " P.M."; 
         cout << hours << ":" << minutes << " P.M.";
    }
    else
    {
        (minutes < 10); cout << hours << ":0" << minutes << " A.M."; 
         cout << hours << ":" << minutes << " A.M.";
        }

}

Answer 1

我希望它有所帮助。我试图让它保持干净简洁。

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

//Function Prototypes
void convert(int&, int&, int&, string&);
void output(int&, int&, string&);

int main()
{

    int hours, minutes, delay;
    string ampm;

    cout << "Enter hours(0-24): ";
    cin >> hours;

    do{
        cout << "Enter minutes: ";
        cin >> minutes;

        cout << "Enter delay(in minutes): ";
        cin >> delay;

        convert(hours, minutes, delay, ampm);
        output(hours, minutes, ampm);

        cout << endl << "Enter the value for Hours (999 to quit):  ";
        cin >> hours;
    } while (hours != 999);

    return 0;
}


void convert(int& hours, int& minutes, int& delay, string& ampm)
{
    int temp = 0;

    minutes += delay;
    if (minutes > 60) {
        temp = minutes / 60;
        hours += temp;
        minutes = minutes - temp * 60;
    }

    if (hours > 12){
        hours -= 12;
        ampm = "PM";
    }
    else {
        ampm = "AM";
    }
}


void output(int& hours, int& minutes, string& ampm)
{
    if (minutes > 0 && minutes < 10) {
        cout << hours << ":0" << minutes << " " << ampm << endl;;
    }
    else if (minutes == 0) {
        cout << hours << ":0" << minutes << " " << ampm << endl;;
    }
    else {
        cout << hours << ":" << minutes << " " << ampm << endl;
    }
}

C ++时间转换程序

1 个答案: