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We are given with an integer range, a variable m which is used as divisor and a variable d which is used to check whether the digit 'd' is at even position or not and the task is to calculate the count of those numbers in a range which are divisible by the variable m and have digit d in even positions.

**Input -** int start = 20, end = 50, d = 8 and m = 4

**Output -** Count of Numbers in a Range divisible by m and having digit d in even positions are: 2

**Explanation -** The range is starting from 20 to 50. So the possible numbers with digit d i.e. 8 are 28, 38 and 48 having 8 at even positions i.e. 2 and the numbers 24 and 48 are divisible by m i.e. 4 therefore the count is 2.

**Input -** int start = 10, end = 100, d = 6 and m = 2

**Output -** Count of Numbers in a Range divisible by m and having digit d in even positions are: 8

**Explanation -** The range is starting from 10 to 100. So the possible numbers with digit d i.e. 6, 16, 26, 36, 46, 56, 66, 76, 86 and 96 having 6 at even positions except the number 6 and 66 which are at odd positions so we will not include this and now we will check the numbers from the list that are divisible by 2 so all the numbers i.e. 16, 26, 36, 46, 56, 76, 86 and 96 are divisible by 2 therefore the count is 8.

- Create a range of integer numbers starting from the variable start till the variable end and declare the variable d and m and input the values. Pass the data to the function for further processing.
- Create a variable of type vector let's say vec.
- Start loop while till the value which is the value inside the variable start. Now, inside the while push the value as val % 10 to the vector and set val as val / 10.
- Call the reverse function in STL by passing vec.begin() and vec.end() as an argument to it.
- Set the values in the array as -1 using memset.
- Return the set_total(0, 0, 0, vec) which is a function that will check whether the numbers with even position d and are divisible by m
- Inside the set_totalfunction-:
- Check IF place equals to size of a vector then check IF temp = 0 then return 1 or return 0.
- Check IF arr[place][temp][val] not equals to -1 then return the value at arr[place][temp][val].
- Check IF place % 2 = 1 then check IF val = 0 then check IF d greater than vec[place] then return 0
- Declare the variable temp_2 and set it to 0.
- Check IF d less than vec[place] then set temp_2 to 1.
- Declare variable temp_3 and make a recursive call to the set_total() and return arr[place][temp][val] = temp_3
- Declare a variable count to store the result.
- Declare a variable set_limit and set it to 9 IF val equals 1 ELSE set it with vec[place].
- Start loop FOR from i to 0 till the set_limit and check IF i equals d then continue
- Declare a variable as temp_2 and set it to val
- Check IF i less than vec[place] then set temp_2 as 1
- Set count with the recursive call to the function set_total
- Return arr[place][temp][val] = count.

#include <bits/stdc++.h> using namespace std; int arr[20][20][2]; int d, m; int set_total(int place, int temp, int val, vector < int > vec) { if (place == vec.size()) { if (temp == 0) { return 1; } return 0; } if (arr[place][temp][val] != -1) { return arr[place][temp][val]; } if (place % 2) { if (val == 0) { if (d > vec[place]) { return 0; } } int temp_2 = val; if (d < vec[place]) { temp_2 = 1; } int temp_3 = set_total(place + 1, (10 * temp + d) % m, temp_2, vec); return arr[place][temp][val] = temp_3; } int count = 0; int set_limit = (val ? 9 : vec[place]); for (int i = 0; i <= set_limit; i++) { if (i == d) { continue; } int temp_2 = val; if (i < vec[place]) { temp_2 = 1; } count += set_total(place + 1, (10 * temp + i) % m, temp_2, vec); } return arr[place][temp][val] = count; } int divisible(int val) { vector < int > vec; while (val) { vec.push_back(val % 10); val = val / 10; } reverse(vec.begin(), vec.end()); memset(arr, -1, sizeof(arr)); return set_total(0, 0, 0, vec); } int main() { int start = 20, end = 50; d = 8, m = 4; int count = divisible(end) - divisible(start); cout << "Count of Numbers in a Range divisible by m and having digit d in even positions are: " << count; return 0; }

If we run the above code it will generate the following output −

Count of Numbers in a Range divisible by m and having digit d in even positions are: 2

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