World-wide symmetric daily computing (C ++)

Today is the world's fully symmetric Day (2011 1102), so I want to write an algorithm to calculate all the fully symmetric days in a period of time. Let's see how many days are the world's fully symmetric day.

 

Description:
Enter the start and end years to obtain all the fully symmetric dates.

Input:

Enter the start year and end year endyear (startyear <endyear );

Output:

The exact symmetric day required by the output.

 

Solution:

1) Calculate the date of the corresponding year based on the month and day in reverse order (for example, in the case of January 1-> 0101-> 1010, the year is 1010)

2) Determine whether the calculated year is between the input year

3) exclude the illegal date of January 1, February 29 when a non-leap year is exceeded.

 

Code:

#include <vector>
#include <string>
#include <sstream>
#include <iomanip>
#include <algorithm>
using namespace std;

const int MonthDays[] ={
31,29,31,30,31,30,31,31,30,31,30,31
};

class SymmetricalDay
{
public:

bool isLeap(int year)
    {
return (( year % 4 == 0 ) && ( year % 100 != 0 ) || ( year % 400 == 0 ));
    }

    vector<string> getDays(int startYear, int endYear)
    {
        vector<string> results;

for (int curMonth = 1; curMonth <= 12; ++curMonth)
        {
for (int curDay = 1; curDay <= MonthDays[curMonth-1]; ++curDay)
            {        
                ostringstream tempValue;

                tempValue << setw(2) << setfill('0') << curMonth;
                tempValue << setw(2) << setfill('0') << curDay;

string strData(tempValue.str());
string strReverse(strData.rbegin(), strData.rend());

                istringstream yearValue(strReverse);            

int curYear = 0;
                yearValue >> curYear;

if (curYear >= startYear && curYear <= endYear)
                {                    
if (!isLeap(curYear) && curMonth==2 && curDay==29)
                    {
continue;
                    }

string tempResult = yearValue.str() + "" + tempValue.str();
                    results.push_back(tempResult);
                }
            }
        }

        sort(results.begin(), results.end());

return results;
    }
};