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: Accepted)

//~ while (clock()<=69*CLOCKS_PER_SEC)
//~ #pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("O3")
#pragma GCC optimize("-Ofast","-funroll-all-loops","-ffast-math")
//~ #pragma GCC target ("avx2")
//~ #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <algorithm>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <random>
#include <cmath>
#include <chrono>
#include <cstring>
#include <string>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cassert>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
using namespace std;
using LL = long long;
#define eps 1e-8
#define fi first
#define se second
#define endl '\12'
#define eb emplace_back
#define SZ(a) int(a.size())
#define ALL(x) (x).begin(),(x).end()
#define trav(a,x) for (auto& a: x)
#define LOG(FMT...) fprintf(stderr, FMT)
#define close std::ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define FOR(i, x, y) for (LL i = (x), _##i = (y); i < _##i; ++i)
#define FORD(i, x, y) for (LL i = (x), _##i = (y); i > _##i; --i)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define CASET int ___T; cin>>___T; for(int __CS=1;__CS<=___T;__CS++)
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef pair<int,int> pii;
constexpr int mod = 1e9+7;
constexpr int Erica = 998244353;
mt19937 dlsrand(random_device{}());
mt19937 mrand(std::chrono::system_clock::now().time_since_epoch().count()); 
int rnd(int x) { return mrand() % x;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
ll ex_gcd(ll a, ll b, ll& x, ll& y){if(!b){x=1;y=0;return a;}int ret=ex_gcd(b,a%b,y,x);y-=a/b*x;return ret;}
LL bin(LL x, LL n, LL MOD) {LL ret = MOD != 1;for (x %= MOD; n; n >>= 1, x = x * x % MOD)if (n & 1) ret = ret * x % MOD;return ret;}
int norm(int x) { return x >= mod ? (x - mod) : x; }
inline LL get_inv(LL x, LL p) { return bin(x, p - 2, p); }
inline ll get_inv(ll a) { ll x, y; ex_gcd(a, mod, x, y); return norm(x + mod);}
template<class T>inline void umin(T &x, T y) {x = x > y ? y : x;}
template<class T>inline void umax(T &x, T y) {x = x < y ? y : x;}
template<class T>inline void dec(T &x, T y) {x -= y; if(x < 0) x += mod;}
template<class T>inline void add(T &x, T y) {x += y; if(x >= mod) x -= mod;}
const double PI = acos(-1.0);
constexpr int maxn = 1e6+10;
constexpr int INF = 0x3f3f3f3f;
constexpr ll linf = 0x3f3f3f3f3f3f3f3f;
constexpr ull base=2333, P_1=19260817, P_2=999998639;
bool chk(int x){
    string s = to_string(x);
    for(int i = 0; i < s.size() && 2*i <= s.size(); i++){
        if(s[i] != s[s.size() - 1 - i]) return 0;
    }
    return 1;
}
int main()
{
    int a , b , cnt = 0;
    cin >> a >> b;
    for(int i = a; i <= b; i++) if(chk(i)) cnt++;
    cout << cnt << endl;
    return 0;
}