palindrom_date.txt

/*
 8. Write a program to generate all palindrome dates by taking the beginning and the ending dates as an input from the user. The format of the date is given as MMDDYYYY. 

That is, given a beginning datae and a ending date, you need to output all the palindrome dates between the beginning and ending date.
*/

#include <iostream>
using namespace std;

int date[4]={0};
/*The argument y is the year in the main function.*/
bool isPalindrome(int y){
    /*we first need to get the 4 digit of the year, assume year=dgt4_dgt3_dgt2_dgt1.*/
    int dgt1 = y%10;
    int dgt2 = int(y/10)%10;
    int dgt3 = int(y/100)%10;
    int dgt4 = int(y/1000)%10;
    date[0] = dgt1;
    date[1] = dgt2;
    date[2] = dgt3;
    date[3] = dgt4;
    
    /*nMonthLeap and nMonthNonLeap are the number of days in each month.*/
    int nMonthLeap[] = {31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    int nMonthNonLeap[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    
    /*Now determine date like MMDD=dgt1_dgt2_dgt3_dgt4 is valid.*/
    int month = 10*dgt1 + dgt2;
    int day = 10*dgt3 + dgt4;
    
    /*if num is leap year*/
    if (y%400==0 || (y%4==0 && y%100!=0)){
        if (month<=12 && month>=1 && day<=nMonthLeap[month-1] && day>0)
            return true;
        else return false;
    }
    else{
        if (month<=12 && month>=1 && day<=nMonthNonLeap[month-1] && day>0)
            return true;
        else return false;
    }
}

int main(){
    int year1, year2;
    cout<<"input the first year:"<<endl;
    cin>>year1;
    cout<<"input the second year:"<<endl;
    cin>>year2;
    /*check if year is appropriate.*/
    if (year1<0 || year2<0)
        cout<<"wrong year"<<endl;
    /*loop through each year. each year corrensponds to a date.*/
    for (int i=year1; i<=year2; ++i){
        /*We check if the date corresponding to the year i is a palindrom.*/
        if (isPalindrome(i)){
            /*if the date is a palindrome, we output it.*/
            for (int j=0; j<4; ++j)
                cout<<date[j];
            cout<<i;
            cout<<endl;
        }
    }
    return 0;
}
REMARK:
1. First, the wrong idea is to loop through date from year1 to year2, eg: from 01012001 to 01012004. The correct idea is to loop through year. Each year corresponds to a unique palindrome number, we just need to verify if the palindrome is a valid date. Example: we have year 2001, then we reverse it to get 1002, then we determine if 10022001 is valid.