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## Looking at the below code, write down the final values of A0, A1, …An.

```A0 = dict(zip(('a','b','c','d','e'),(1,2,3,4,5)))

A1 = range(10)

A2 = sorted([i for i in A1 if i in A0])

A3 = sorted([A0[s] for s in A0])

A4 = [i for i in A1 if i in A3]

A5 = {i:i*i for i in A1}

A6 = [[i,i*i] for i in A1]```

The function zip() returns a list of tuples, where each tuple contains the i-th element from each of the argument sequences.  The returned list is truncated in length to the length of the shortest argument sequence.

zip(…)

zip(seq1 [, seq2 […]]) -> [(seq1[0], seq2[0] …), (…)]

The solution of the above code is as follows:

```A0 = {'a': 1, 'c': 3, 'b': 2, 'e': 5, 'd': 4}  # the order may vary

A1 = range(0, 10) # or [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] in python 2

A2 = []

A3 = [1, 2, 3, 4, 5]

A4 = [1, 2, 3, 4, 5]

A5 = {0: 0, 1: 1, 2: 4, 3: 9, 4: 16, 5: 25, 6: 36, 7: 49, 8: 64, 9: 81}

A6 = [[0, 0], [1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81]]```