Palindrome Date Calculation

The code below checks for palindrome dates between two given date strings.

#include<iostream>
#include<cstring>
#include<sstream>
using namespace std;
string data1;
string data2;
int cnt;

int y1,m1,d1;
int y2,m2,d2;

int day[] = {31,28,31,30,31,30,31,31,30,31,30,31};
bool check(string s)
{
    int i=0, j=s.size()-1;
    while(i<j)
    {
        if(s[i]!=s[j])  return false;
        i++;    j--;
    }

    return true;
}
void digitalize()
{
    y1 = stoi(data1.substr(0,4));
    m1 = stoi(data1.substr(4,2));
    d1 = stoi(data1.substr(6,2));

    y2 = stoi(data2.substr(0,4));
    m2 = stoi(data2.substr(4,2));
    d2 = stoi(data2.substr(6,2));
}
void addoneday()
{

    int flag = 0;   //判断是否是闰年
    if(y1%400==0 || ((y1%4)==0&&(y1%100)!=0))   flag = 1;

    if(flag && m1==2)
    {
        if(d1==29)
        {
            m1=3;
            d1=1;
        }else
            d1++;
    }else
    {
        if(d1==day[m1-1])
        {
            d1=1;
            m1++;
            if(m1==13)  {m1=1;  y1++;
        }else
            d1++;
    }
}
int main()
{

    cin >> data1;
    cin >> data2;
    digitalize();

    int k =0;
    while(1)
    {
        if(y1==y2&&m1==m2&&d1==d2)  break;

        string s = to_string(y1);
        if(m1<10)   s = s + "0" + to_string(m1);
        else    s = s + to_string(m1);
        if(d1<10)   s = s + "0" + to_string(d1);
        else    s = s + to_string(d1);

        if(check(s))    cnt++;
        addoneday();

    }
    if(check(data2))    cnt++;

    cout << cnt << endl;
    return 0;
}

The following version passes all test cases:

#include<iostream>
#include<cstring>
#include<sstream>
using namespace std;
string data1;
string data2;
int cnt;

int y1,m1,d1;
int y2,m2,d2;

int day[] = {31,28,31,30,31,30,31,31,30,31,30,31};
bool check(string s)
{
    int i=0, j=s.size()-1;
    while(i<j)
    {
        if(s[i]!=s[j])  return false;
        i++;    j--;
    }

    return true;
}
void digitalize()
{
    y1 = stoi(data1.substr(0,4));
    m1 = stoi(data1.substr(4,2));
    d1 = stoi(data1.substr(6,2));

    y2 = stoi(data2.substr(0,4));
    m2 = stoi(data2.substr(4,2));
    d2 = stoi(data2.substr(6,2));
}
void addoneday()
{

    int flag = 0;   //判断是否是闰年
    if(y1%400==0 || ((y1%4)==0&&(y1%100)!=0))   flag = 1;

    if(flag && m1==2)
    {
        if(d1==29)
        {
            m1=3;
            d1=1;
        }else
            d1++;
    }else
    {
        if(d1==day[m1-1])
        {
            d1=1;
            m1++;
            if(m1==13)  {m1=1;  y1++;
        }else
            d1++;
    }
}
int main()
{

    cin >> data1;
    cin >> data2;
    digitalize();

    int k =0;
    while(1)
    {
        if(y1>y2)  break;

        int st = y1%10*10 + y1%100/10;
        if(st==0||st>12)  {y1++;m1=1;d1=1;continue;}    //年份
        if(m1<st)  {m1=st;d1=1;}    // 月份
        if(m1>st)   {y1++;continue;}

        if(y1==y2 && (m1>m2||(m1==m2&&d1>=d2)))  break;

        string s = to_string(y1);
        if(m1<10)   s = s + "0" + to_string(m1);
        else    s = s + to_string(m1);
        if(d1<10)   s = s + "0" + to_string(d1);
        else    s = s + to_string(d1);


        if(check(s))    cnt++;
        addoneday();

    }
    if(check(data2))    cnt++;

    cout << cnt << endl;
    return 0;
}

Analysis: This is a problem based on logical thinking.

Reasons for the long time taken: Total of one hour and twenty minutes, too slow, I have other problems to solve. Mainly unfamiliar with to_string and stoi, and unclear logic. Initially, it was checking one day at a time, which caused timeout. Then, after observing that the palindrome implies a specific year, the month is determined by the reverse of the year's last two digits. It can be simplified further since only certain years like 10, 20, 30, etc., produce palindromes. How ever, the solution passed 90% of the test cases, so I made minor adjustments to the original code and spent 20 minutes debugging boundary conditions.