Luntik and Subsequences (Codeforces 1582B)
Tags: Math, combinatorics

Luntik came out for a morning stroll and found an array $$$a$$$ of length $$$n$$$. He calculated the sum $$$s$$$ of the elements of the array ($$$s= \sum_{i=1}^{n} a_i$$$). Luntik calls a subsequence of the array $$$a$$$ nearly full if the sum of the numbers in that subsequence is equal to $$$s-1$$$.Luntik really wants to know the number of nearly full subsequences of the array $$$a$$$. But he needs to come home so he asks you to solve that problem!A sequence $$$x$$$ is a subsequence of a sequence $$$y$$$ if $$$x$$$ can be obtained from $$$y$$$ by deletion of several (possibly, zero or all) elements.

Input: The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The next $$$2 \cdot t$$$ lines contain descriptions of test cases. The description of each test case consists of two lines.The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 60$$$) — the length of the array.The second line contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$0 \le a_i \le 10^9$$$) — the elements of the array $$$a$$$.

Output: For each test case print the number of nearly full subsequences of the array.

Note: In the first test case, $$$s=1+2+3+4+5=15$$$, only $$$(2,3,4,5)$$$ is a nearly full subsequence among all subsequences, the sum in it is equal to $$$2+3+4+5=14=15-1$$$.In the second test case, there are no nearly full subsequences.In the third test case, $$$s=1+0=1$$$, the nearly full subsequences are $$$(0)$$$ and $$$()$$$ (the sum of an empty subsequence is $$$0$$$).

Sample Inputs:
5
5
1 2 3 4 5
2
1000 1000
2
1 0
5
3 0 2 1 1
5
2 1 0 3 0

Sample Outputs:
1
0
2
4
4

Solution (Python):
for s in [*open(0)][2::2]:
    f = s.split().count
    print(f('1') << f('0'))

Solution (Python):
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
for _ in range(int(readline())):
    N = int(readline())
    A = list(map(int, readline().split()))
    (c1, c0) = (A.count(1), A.count(0))
    print(2 ** c0 * c1)

Solution (Python):
for s in [*open(0)][2::2]:
    f = s.split().count
    print(f('1') << f('0'))