四项一起对得上吗

运行下面这段程序：

def greedy(cs, t):
    n = 0
    for c in cs:
        while t >= c:
            t -= c
            n += 1
    return n

def best(cs, t):
    INF = 10 ** 9
    dp = [0] + [INF] * t
    for a in range(1, t + 1):
        for c in cs:
            if c <= a and dp[a - c] + 1 < dp[a]:
                dp[a] = dp[a - c] + 1
    return dp[t]

def sched(iv, key):
    end = -1
    n = 0
    for s, e in sorted(iv, key=key):
        if s >= end:
            n += 1
            end = e
    return n

def wait(order):
    t = 0
    s = 0
    for x in order:
        t += x
        s += t
    return s

ok = (greedy([25, 10, 5, 1], 63) == 6
      and best([4, 3, 1], 6) == 2
      and sched([(1, 4), (2, 3), (3, 5), (0, 7), (5, 6), (6, 8)], lambda t: t[1]) == 4
      and wait(sorted([4, 1, 3, 2])) == 20)
print(ok)