AcWing 1912. Odometer (enumeration)

【 Title Description 】
John's cow is traveling on the road .
The odometer on their car shows the whole mileage .
At the beginning of the journey, the mileage displayed on the odometer is $X$, At the end of the trip, the mileage displayed on the odometer is $Y$.
Whenever the odometer shows “ amusing ” Numbers （ Including the numbers displayed at the beginning and end ） when , The cows will make a pleasant cry .
If a number is in all numbers except the leading zero , Except for a different number , All other numbers are the same , So this number is “ amusing ”.
for example ,33323 and 110 It's interesting. , and 9779 and 55555 Not interesting .
Please help John count the number of times the cows cry during the journey .

【 Input format 】
All in one line , Contains two integers $X$ and $Y$.

【 Output format 】
Output the number of times the cows make calls during the journey .

【 Data range 】
$100\le X\le Y\le 10^{16}$

【 sample input 】

110 133

【 sample output 】

13

【 Sample explanation 】
$110\sim 133$ Of all the numbers between , Interesting numbers are ：
$110,112,113,114,115,116,117,118,119,121,122,131,133$.

【 analysis 】

---

Observing this data range, we cannot enumerate $X\sim Y$ And then determine whether it is an interesting number , So we need to change the angle , You can enumerate all the interesting numbers , That is, the number of digits to be satisfied is $3\sim 17$ There is only one digit difference , Other figures are the same . For every interesting number of enumerations , If in $X\sim Y$ In the range of , Add one to the answer .

---

【 Code 】

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long LL;
LL x, y;
int res;

int main()
{
    
    cin >> x >> y;
    for (int i = 3; i <= 17; i++)// Enumerator bits 
        for (int j = 0; j < 10; j++)// Enumerate which number is the same 
            for (int k = 0; k < 10; k++)// Enumerate the only different number 
                if (j != k)// The two numbers must be different 
                    for (int u = 0; u < i; u++)// Enumerate where different numbers appear 
                    {
    
                        string str(i, '0' + j);
                        str[u] = '0' + k;
                        if (str[0] != '0')// There can't be leading zeros 
                        {
    
                            LL num = stoll(str);
                            if (num >= x && num <= y) res++;
                        }
                    }
    cout << res << endl;
    return 0;
}