Hello, you can observe that 1 <= n <= 1e9. This means that the number "n" has atmost 9 digits. Now let us assume that you need to generate all lucky numbers which have k digits. Now you can see that these k digits must comprise of either 4 or 7. So what you can do is just iterate each number from 0 to (2**k-1) and for each number check if current bit is 0/1. If the current bit is 0 it means 4 else it is 7. As an example let us take k = 2. So you have following cases :
00
01
10
11
Now mapping 0 with 4, 1 with 7, you can see that :
00 : 44
01 : 47
10 : 74
11 : 77
This means that you have generated all lucky numbers with digits : 2. Like this you can iterate k from 1 to 9, and generate the numbers. You can see my solution, it's easy to understand : http://codeforces.me/contest/535/submission/10713472