#include<iostream>
using namespace std;
int nian(int m)
{
    return m / 10000000 % 10 * 1000 + m / 1000000 % 10 * 100 + m / 100000 % 10 * 10 + m / 10000 % 10;
}
int yue(int m)
{
    return m / 1000 % 10 * 10 + m / 100 % 10;
}
int ri(int m)
{
    return m / 10 % 10 * 10 + m % 10;
}
int dao(int a)
{
    int rt=0,cnt=1000;
    while (a != 0)
    {
        rt += a % 10 * cnt;
        a /= 10;
        cnt /= 10;

    }
    return rt;
}
int s[13] = { 0,31,29,31,30,31,30,31,31,30,31,30,31 };
int main()
{
    int a, b,cnt=0;
    cin >> a >> b;
    for (int i = nian(a); i <= nian(b); i++)
    {
        if (i == nian(a))
        {
            if (yue(i * 1000 + dao(i)) == yue(a))
            {
                if (ri(i * 1000 + dao(i)) > ri(a))
                {
                    cnt ++;
                }
            }
            else if (yue(i * 1000 + dao(i)) > yue(a))
            {
                if (yue(i * 1000 + dao(i)) <= 12)
                {
                    cnt++;
                }
            }
        }
        else if (i == nian(b))
        {
            if(yue(i * 1000 + dao(i)) == yue(b))
            {
                if (ri(i * 1000 + dao(i)) < ri(b))
                {
                    cnt++;
                }
            }
            else if (yue(i * 1000 + dao(i)) < yue(b))
            {
                if (yue(i * 1000 + dao(i)) <= 12)
                {
                    cnt++;
                }
            }

        }
        else
        {
            if (yue(i * 1000 + dao(i)) <= 12 && ri(i * 1000 + dao(i)) <= s[yue(i * 1000 + dao(i))])
            {
                cnt++;
            }
        }

    }
    cout << cnt;
}

吾乃四年级小学生，即蒟蒻也，帮帮我呀