------
## 最终结论
```python
def div_even_odd(lst):
    first_even = None
    first_odd = None
    
    for num in lst:
        if num % 2 == 0 and first_even is None:
            first_even = num
        elif num % 2 != 0 and first_odd is None:
            first_odd = num
        
        if first_even is not None and first_odd is not None:
            break
            
    return first_even / first_odd

# Test cases
assert div_even_odd([1,3,5,7,4,1,6,8]) == 4
assert div_even_odd([1,2,3,4,5,6,7,8,9,10]) == 2
assert div_even_odd([1,5,7,9,10]) == 10
```