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1. Why can the program answer no straight away when the total of the numbers is odd?

Odd totals mean every number in the list is oddEach group would need half the total, and half of an odd whole number is not a whole numberThe table has zero cells when the total is oddThe bitset can hold only even sums

2. The list is five ones, 1, 1, 1, 1, 1, and the target is 3. Which value is p, the sum the chosen subset must reach?

3548

3. The bitset version of the subset sum table needs no backwards loop, because reach << x reads the old value before the | stores the new one.

True, because reach << x reads the old value, and the new value is stored only after the | finishesFalse, because the bitset version still needs a backwards loop so that no number is used twice

4. A mask built as (1 << target) - 1 instead of (1 << (target + 1)) - 1 is just as correct for the bitset method.

True, because the answer is never stored at bit targetFalse, because the shorter mask deletes bit target, which is the bit that holds the answer

5. What does this program print? ``

python
def count_ways(nums, target):
    total = sum(nums)
    if abs(target) > total or (total + target) % 2 != 0:
        return 0
    p = (total + target) // 2
    ways = [0] * (p + 1)
    ways[0] = 1
    for x in nums:
        for s in range(p, x - 1, -1):
            ways[s] += ways[s - x]
    return ways[p]


print(count_ways([2, 3, 5], 1))

``

12010

Subset sums