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A lucky number is defined as a positive integer whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. I need to check if a given number is evenly divisible by any lucky number or not.

Now, suppose I want to add all Lucky Numbers under a given integer [N] to a vector, without using recursion. For the sake of simplicity, let N = 1000.

A lucky number is defined as a positive integer whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not. I need to check if a given number is evenly divisible by any lucky number or not.

Now, suppose I want to add all Lucky Numbers under a given integer [N] to a vector, without using recursion. For the sake of simplicity, let N = 1000.

EDIT:

#include <iostream>
#include <vector>
bool is_lucky(int num)
{
while(num!=0)
 {
 if((num%10!=4)&&(num%10!=7))
 {
 return false;
 }
 num/=10;
 }
return true;
}
int main()
{
std::vector <long long> lucky;
for(int in_num=1;in_num<1000;in_num++)
 {
 if(is_lucky(in_num))
 {
 lucky.push_back(in_num);
 }
 }
}

Even though I am still using the for-if anti-pattern as Deduplicator pointed out, is the is_lucky function comparable in efficiency as the ones below?

Even though I am still using the for-if anti-pattern as Deduplicator pointed out, is the above function (almost) as efficient as the ones below?

Even though I am still using the for-if anti-pattern as Deduplicator pointed out, is the is_lucky function comparable in efficiency as the ones below?