Problem:
Score : 200 points

Problem Statement

Find the number of palindromic numbers among the integers between A and B (inclusive).
Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.

Constraints

10000 \leq A \leq B \leq 99999

All input values are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the number of palindromic numbers among the integers between A and B (inclusive).

Sample Input 1

11009 11332

Sample Output 1

4

There are four integers that satisfy the conditions: 11011, 11111, 11211 and 11311.

Sample Input 2

31415 92653

Sample Output 2

612

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

int main(){
    int A,B;
    cin >> A >> B;
    int a,b,c,d,e;
    int count = 0;
    a = A / 10000;
    b = (A -a*10000)/1000;
    c = (A - a*10000 - b*1000)/100;
    d = (A - a*10000 - b*1000 - c*100)/10;
    e = (A - a*10000 - b*1000 - c*100 - d*10);
    int a1,b1,c1,d1,e1;
    a1 = B/ 10000;
    b1 = (B -a1*10000)/1000;
    c1 = (B - a1*10000 - b1*1000)/100;
    d1 = (B - a1*10000 - b1*1000 - c1*100)/10;
    e1 = (B - a1*10000 - b1*1000 - c1*100 - d1*10);   
    if (10*d + e <= 10*b + a){
        count += 1;
        count += (9 - c +1);

        if(10*d1 + e1 <= 10*b1 + a1){
            count += 1;
            count += (9 -c1 + 1);
        }
    }
    count += (((10*a1 + b1 -1) - (10*a + b +1) + 1)*10);//真ん中の数は0-9のなかのどれでもいいので10倍
    cout << count << endl;

}