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 <stdio.h>
#include <string.h>
#include <math.h>
#include <stdbool.h>
#include <stdlib.h>

int kaibun_num(int x){
    int ans = 0;
    int a,b,c,d,e;
    a = x/10000;
    b = x/1000-10*a;
    c = (x%100-x%10)/10;
    d = x%10;
    e = x/100-10*b-100*a;

    ans+=(a-1)*100+(b-1)*10;
    if (a <=d);{
        if (b <=c){
            ans +=e+1;
        }
    }
    return ans;

}

int main(){
    int a,b,A,B,p,q,r,s;
    scanf("%d %d",&a,&b);
    A = kaibun_num(a);
    B = kaibun_num(b);

    p = a/10000;
    q = a/1000-10*p;
    r = (a%100-a%10)/10;
    s = a%10;
    if (p ==s){
        if (q ==r){
            printf("%d\n",B-A+1);
        }
        else{
            printf("%d\n",B-A);
        }
    }
    else{
        printf("%d\n",B-A);
    }

    return 0;
}