我需要计算给定日期的天数。一年有 366 天。然而,每个月都有不同的值,我必须分配这些值。有没有比我正在做的更快的方法来做到这一点?
#include<iostream>
using namespace std;
int main()
{
int day, month, year, dayNumber;
cout<< "Please enter the month, by numerical value:";
cin>> month;
cout<<"Please enter the day, by numerical value:";
cin>> day;
cout<<"Please enter the year, by numerical value:";
cin>> year;
if (month == 1)
{
dayNumber= day;
cout<< "Month;" << '\t'<< month << '\n'
<< "Day:"<<'\t'<< day<< '\n'
<< "Year:"<<'\t'<< year<<'\n'
<< "Day Number:"<< '\t'<< dayNumber<< endl;
}
else if(month==2)
{
dayNumber= day+31;
}
}
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只需使用mktime:
tm date = {};
date.tm_year = year - 1900;
date.tm_mon = month - 1;
date.tm_mday = day;
mktime( &date );
dayNumber = date.tm_yday;
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否则,您将需要一个二维表:
int daysToMonth[2][12] =
{
{ 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334 },
{ 0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335 },
};
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和一个功能:
bool isLeapYear( int year )
{
return year % 4 == 0 && (year % 100 != 0 || year % 400 == 0);
}
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那么年日是:
daysToMonth[isLeapYear( year ) ? 1 : 0][month] + day;
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