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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
int pan(int x)
{
    int a,b;
    a=x/10000+(x/1000)%10*10;
    b=x%100;
    if(a!=b)
        return 0;
    else
        return 1;
}
int main()
{
    int x,y,sum,a,b;
    scanf("%d %d",&x,&y);
    if(x==y)
    {
        if(pan(x)==0)
            sum=0;
        else
            sum=1;
    }
    else
    {
           a=x/10000+(x/1000)%10*10;
    b=x%100;
    if(a<b)
    {
        x=x/100*100+(a%10)*10+(a/10);
        if((x/100%10)==9)
        {
            x=x+100;
            a=x/1000;
            x=x/100*100+(a%10)*10+(a/10);
        }
        else
            x=x+100;
    }
    else if(a=b)
    {
        x=x;
    }
    else
    {

        x=x/100*100+(a%10)*10+(a/10);
    }
    a=(y/10000)+(y/1000)%10*10;
    b=y%100;
    if(a<=b)
    {
        y=y/100*100+(a%10)*10+(a/10);
    }
    else if(a=b)
    {
        y=y;
    }
    else
    {
        y=y/100*100+(a%10)*10+(a/10);
        if((y/100)%10==0)
        {
            y=y-100;
            a=y/1000;
            y=y/100*100+(a%10)*10+(a/10);
        }
    }
    sum=(y-x)/100+1;
    }
    printf("%d\n",sum);
    return 0;
}