Well, there's lots to improve about your program.

1. Consider adopting any of the common styles for code-formatting.  
As-is, your current formatting seriously impedes readability.

1. `#include <bits/stdc++.h>` is a bad idea, sharply limiting portability ad increasing compile-times. See: [How does #include `<bits/stdc++.h>` work in C++?](http://stackoverflow.com/questions/25311011/how-does-include-bits-stdc-h-work-in-c)  
Just include those headers you need, which are `<vector>` and `<iostream>`.

1. You are courting conflicting symbols and general bafflement with any minor change of your toolchain. See: [Why is `using namespace std;` considered bad practice?](http://stackoverflow.com/q/1452721)

1. You are using the popular for-if-antipattern. See [Introducing the for-if anti-pattern](http://blogs.msdn.com/b/oldnewthing/archive/2011/12/27/10251210.aspx)  
Why don't you just enumerate the ones you are actually interested in?

1. A for-range-loop is simpler than explicitly using iterators/indices. Unless you actually need them.

1. In the end you don't actually want that whole list, only whether one of them divides your input-number. So why store them at all, and why also those bigger than the input-number?

1. `return 0;` is implicit for `main` in C++.

Doing it as it should be done, [with recursion on coliru](http://coliru.stacked-crooked.com/a/509560d255c0f478):

    #include <stdio.h>
    
    int luckydiv_helper(long in, long num, int free) {
        return !free
            ? !(in % num)
            : luckydiv_helper(in, 10 * num + 4, free - 1)
                || luckydiv_helper(in, 10 * num + 7, free - 1);
    }
    
    int luckydiv(long in) {
        long abort = in;
        for(int digits = 1; abort; digits++, abort /= 10)
            if(luckydiv_helper(in, 0, digits))
                return 1;
        return !in;
    }
    
    int main() {
        long in;
        if(scanf("%ld", &in) != 1 || in < -999999999 || in > 999999999) {
            fprintf(stderr, "You didn't enter a number between -999999999 and "
                "+999999999. Aborting.\n");
            // A long can represent all 9-digit decimal numbers.
            return 1;
        }
        printf("%ld ", in);
        if(in < 0) in = -in;
        puts(luckydiv(in) ? "YES" : "NO");
    }

A way to efficiently get all "lucky" numbers without recursion:

    #include <limits>
    #include <vector>

    template<class T> std::vector<T> luckyvector() {
        std::vector<T> v = {4, 7};
        const auto limit4 = (std::numeric_limits<T>::max() - 4) / 10;
        const auto limit7 = (std::numeric_limits<T>::max() - 7) / 10;
        T x = 0;
        for(size_t i = 0; (x = v[i]) <= limit7; i++) {
            v.push_back(x * 10 + 4);
            v.push_back(x * 10 + 7);
        }
        if(x > limit7 && x <= limit4)
            v.push_pack(x * 10 + 4);
        return v;
    }