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Count all possible N digit numbers that satisfy the given condition in C++
In this tutorial, we will be discussing a program to find the number of possible N digit numbers that satisfy the given condition.
For this we will be provided with an integer. Our task is to check which one of number having N digits follow
Number + Reverse(Number) = 10N -1
Example
#include <bits/stdc++.h>
using namespace std;
//returning the count of numbers
string count_num(int N){
if (N % 2 == 1)
return 0;
string result = "9";
for (int i = 1; i <= N / 2 - 1; i++)
result += "0";
return result;
}
int main(){
int N = 4;
cout << count_num(N);
return 0;
}Output
90
Updated on: 10-Feb-2020
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