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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

//5桁だけ
char isKaibun(int n){
    if(n/10000 == n%10 &&
       (n/1000)%10 == (n%100)/10) return 1;
    else return 0;
}
int main(){
    int A, B;
    scanf("%d %d",&A,&B);
    int count = 0;
    for(int n = A; n <=B; n++){
        if(isKaibun(n)) count++;
    }
    printf("%d\n",count);
}