long fact[12], ten[12], nine[12], dp[12];
 
int DigitCount(long n)
{
    int cnt = 0;
    while (n)
    {
        cnt++;
        n /= 10;
    }
    return cnt ; // number of digits n has in it's decimal rep
}
 
int Get_ith_Digit(long n, int i) 
{
    n = n % ten[i];
    n /= ten[i - 1];
    i = n;
    return i; // returns ith digit of n , Get_ith_Digit(1234,2) returns 3
}
 
long calc(int len) // counting part , 
{
    long sum = 0;
 
    for (int i = 1; i <= len; i++)
    {
        long temp = fact[len] / (fact[i] * fact[len - i]);
        temp *= nine[len - i];
        sum += i * temp;
    }
    return sum; 
    // lets say, comb(x) = num of possible combination with x zeroes
   // here, sum = 1*comb(1) + 2*comb(2) + ..... + len*comb(len)
}
 
long fun(long n)// let's say n = 5043
{
    if (n < 0)
        return 0;
 
    if (n == 0)
        return 1;
 
    int cnt = DigitCount(n);
    if (cnt == 1) return 1;
 
    long ans = dp[cnt-1]; // we are done with up to 999 , now we've to calculate the ans/ 
    // for 5000 to 5043
 
    long d = n/ten[cnt-1]; // check out for the MSB
    d--;

    // note , ans(1000 to 1999) == ans(2000 to 2999) == ans(3000 to 3999) and so on
   // but , ans(1000 to 1999) = ?.....use counting knowledge

    if(d) ans += d*calc(cnt-1); // d was five, so ans for 1000 to 4999 done 

   // if you understand above part you r done with digits>0....but what about 0??/
//in btwn 5000 to 5045 , zero in second MSB will be in count (45+1) times.....convinced?/
//if not then see , 5000 , 5001 , 5002......5043,5044,5045   
/// now i hope you will understand the what happened rest of the code 

    for(int i=cnt-1; i ; i--){
        d = Get_ith_Digit(n,i);
 
        if(i==1)ans++;
        else if(d)ans += ( (d)*calc(i-1) + ten[i-1] );
        else ans += (n%ten[i]+1);
    }
    return ans ;
}
int main(){
    dp[1] = ten[0] = fact[0] = nine[0] = 1;
 
    for (int i = 1; i <= 11; i++)
    { 
        fact[i] = fact[i - 1] * i;    // computing factorial
        ten[i] = ten[i - 1] * 10;     // ten[i] = 10^i
        nine[i] = nine[i - 1] * 9;    // nine[i] = 9^i
 
        if(i>1) dp[i] = 9*calc(i-1) + dp[i - 1];  
        // dp[i] = ans for the numbers which has <= i digits in their decimal rep
    }
   ans = fun(b) - fun(a - 1));
    return 0;
}