Question

该计划的目标是计算开始时间和结束时间之间的时差。

开始和结束时间将作为4位数字整数输入，并且需要计算时间差。输入的时间表示为hh：mm，没有“：”。

function ValidName(){
    var regex = /^[a-zA-Z]$/;
    var name = $("#txtNome").val();

    if(name == ""){
        alert("Please, type your name.");
        name.focus();
        return false;
    }
    else if(name =! name.regex){
        alert("Please, type your name.");
        name.focus();
        return false;
    }
}

这是我的代码：

Example: 
first time: 0800
second time: 1755
Time elapsed: 9 hr and 55 min

问题是它确实以小时为单位输出时间。任何有关如何改进该计划的建议将不胜感激。

Answer 1

最好在几分钟内转换所有内容，计算差异，然后转换为小时和分钟：

#include <iostream>

int main() 
{
    int first_time;
    int second_time;

    std::cout << "Enter first time" << std::endl;
    std::cin >> first_time;

    std::cout << "Enter second time" << std::endl;
    std::cin >> second_time;

    int first_time_min = first_time / 100 * 60 + first_time % 100;
    int second_time_min = second_time / 100 * 60 + second_time % 100;
    int diff_time_min = second_time_min - first_time_min;

    std::cout << "Time elapsed: " << diff_time_min / 60 << " hours " << diff_time_min % 60 << " minutes" << std::endl;
}

Answer 2

可能的解决方案：

#include <iostream>

int main() {
    int first_time;
    int second_time;
    int dif_time_in_minutes;
    int hours, minutes;

    std::cout << "Enter first time:";
    std::cin >> first_time;

    std::cout << "Enter second time:";
    std::cin >> second_time;

    dif_time_in_minutes = (second_time / 100)*60 + (second_time % 100) - 
        (first_time / 100) * 60 + (first_time % 100);

    hours = dif_time_in_minutes / 60;
    minutes = dif_time_in_minutes % 60;

    std::cout << "Time elapsed: " << hours << " hours " << minutes << " minutes" << std::endl;
}

C ++两次之间的时差

2 个答案: