[PS] Baekjoon #1461

Baekjoon #1461 Library

Conditions

• Current position is 0, books are also at 0.
• You can carry M books at a time and move a total of N books.
• There's no need to return to 0 for the last move.

Strategy

• Split positive and negative ways.
• Since there's a requirement to return to zero, the sum of travel distances is minimal when rounding as little as possible.
• If the number of books in one direction is divided by M, then the sum of the distance traveled should be minimal.
• If it does not divided, move r books first and move M books is the smallest.
• The last move does not need to return to 0.

po = [i for i in b if i > 0]
ne = sorted(-i for i in b if i < 0)
# Sort in positive ascending order for ease of calculation.

def g(li, m):
    if len(li) == 0: # Case if there's no books in the side.
        return 0
    o = len(li) % m

    if o == 0:
        # Can divide by M
        return sum(li[m - 1::m]) * 2
    else:
        return sum(li[o - 1::m]) * 2

    # Last move does not need to return to zero.
    # Substract more far distance to make the result smaller.
    print(g(po, m) + g(ne, m) - max(abs(b[0]), abs(b[-1])))