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Count Numbers with N digits which consists of odd 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 odd number of 0’s as digits. The number also could have preceding zeros like in case of N=3 numbers included will be 000,011,012….990.
Let us understand with examples.
Input − N=3
Output − Count of no. with N digits which consists of even number of 0's are − 244
Explanation − All 3 digit numbers would be like −
Smallest will be 000, then 011,012,013,0014…..Highest will be 990.
Input − N=5
Output − Count of no. with N digits which consists of even number of 0's are − 33616
Explanation − All 5 digit numbers would be like −
Smallest will be 00000, then 00011,00012,00013,0014…..Highest will be 99990.
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 even 0’s as digits, that is E=10N-8N . The remaining numbers with Odd0’s in digits will be (T-E)/2.
Take an integer N as input.
Function count_dd(int N) takes N and returns the count of N digit numbers with odd 0’s.
Total N digit numbers are total=pow(10,N)-1
Total N digit numbers with even 0’s in digit are even=pow(10,N)-pow(8,N).
Leftover odd 0’s in digit are odd=(total-even)/2.
Return odd as a count of N digit numbers with odd numbers of 0’s.
Example
#include <bits/stdc++.h>
using namespace std;
int count_odd(int N){
int total = pow(10, N);
int even = pow(8, N);
int odd = (total - even) / 2;
return odd;
}
int main(){
int N = 4;
cout<<"Count of Numbers with N digits which consists of odd number of 0's are: "<<count_odd(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 odd number of 0's are: 2952
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