------
## 最终结论
```python
def freq_count(lst):
    frequency = {}
    for element in lst:
        if element in frequency:
            frequency[element] += 1
        else:
            frequency[element] = 1
    return frequency

# Test cases
assert freq_count([10,10,10,10,20,20,20,20,40,40,50,50,30])==({10: 4, 20: 4, 40: 2, 50: 2, 30: 1})
assert freq_count([1,2,3,4,3,2,4,1,3,1,4])==({1:3, 2:2,3:3,4:3})
assert freq_count([5,6,7,4,9,10,4,5,6,7,9,5])==({10:1,5:3,6:2,7:2,4:2,9:2})
```