Count Numbers with N digits which consists of even number of 0's in C++

We are given a number N as input. The goal is to find all N digit numbers that have an even number of 0’s as digits. The number also could have preceding zeros like in case of N=3 numbers included will be 001,002,003….010….so on.

Let us understand with examples.

Input − N=4

Output − Count of no. with N digits which consists of even number of 0's are − 7047

Explanation − All 4 digits numbers would be like −

Smallest will be 0000, then 0011,0012,0013,0014…..Highest will be 9900.

Input − N=5

Output − Count of no. with N digits which consists of even number of 0's are − 66383

Explanation − All 5 digit numbers would be like −

Smallest will be 00001, then 00002,00003,00004…..Highest will be 99900.

The approach used in the below program is as follows

We will first calculate the total N digit numbers that are T=10N-1. Then calculate all N digit numbers with odd 0’s as digits, that is O=10N-8N . The remaining numbers with Even 0’s in digits will be T-O/2.

• Take an integer N as input.

• Function count_even(int N) takes N and returns the count of N digit numbers with even 0’s.

• Total N digit numbers are total=pow(10,N)-1

• Total N digit numbers with odd 0’s in digit are odd=pow(10,N)-pow(8,N).

• Leftover even 0’s in digit are even=total-odd/2.

• Return even as a count of N digit numbers with even numbers of 0’s.

Example

 Live Demo

#include <bits/stdc++.h> using namespace std; int count_even(int N){    int total = pow(10, N) - 1;    int odd = pow(10, N) - pow(8, N);    int even = total - odd / 2;    return even; } int main(){    int N = 3;    cout<<"Count of Numbers with N digits which consists of even number of 0's are: "<<count_even(N);    return 0; }

Output

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

Count of Numbers with N digits which consists of even number of 0's are: 755