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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

// 標準入力から改行かスペースまで読み取り、数字として返す
int get_num() {
    // 10文字+区切り文字までしか受け付けられない
    char s[11];
    int i, n;
    for (i = 0; i < sizeof(s); i++) {
        s[i] = getchar();
        if (s[i] == ' ' || s[i] == '\n') {
            s[i] = '\0';
            break;
        }
    }
    sscanf(s, "%d", &n);
    return n;
}

int is_loop_num(int x) {
    int y0, y1;
    y0 = x;
    y1 = 0;
    
    while(x > 0) {
        y1 = y1 * 10 + x % 10;
        x /= 10;
    }

    if (y0 == y1) {
        return 1;
    } else {
        return 0;
    }
}

int main(void) {
    int A, B, i, count = 0;

    A = get_num();
    B = get_num();
    
    for (i = A; i <= B; i++) {
        count += is_loop_num(i);
    }
    
    printf("%d\n", count);
    
    return 0;
}