from math import sqrt
sample = {
    "time":     [7, 15,  30],
    "distance": [9, 40, 200],
}

input = {
    "time":     [   48,     87,     69,     81,],
    "distance": [  255,   1288,   1117,   1623,],
}


def solve(data):
    time = data["time"]
    dist = data["distance"]

    part1 = 1
    for t, best in zip(time, dist):
        ways = 0
        for i in range(t):
            v = i
            d = (t-i)*v
            if d > best:
                ways += 1

        part1 *= ways

    return part1


def fancy_solve(data):
    time = data["time"]
    dist = data["distance"]

    for t, best in zip(time, dist):
        # the distance traveled if the boat is released at time i
        # is equal to (t-i)*i
        # we want to know the range of values of i for which this
        # exceeds the best time
        #   (t-i)*i > best
        # a little rearranging gets us the quadratic equation
        #   i^2 - ti + best <= 0
        # which we can determine the crossing points for
        # using the quadratic formula (the good one, not the
        # one you learned in school)
        h = t/2
        s = sqrt(h*h - best)
        #print(h-s, h+s)

        # in general these will not be at integer values,
        # so we probe the floor and ceiling of each crossing
        # point to determine exactly where the condition is met
        a = int(h-s)
        b = int(h+s)
        if (t-a)*a <= best:
            a += 1
        if (t-b)*b > best:
            b += 1

        ways = b - a
        return ways


print(solve(sample))
print(solve(input))

def part2(data):
    return {"time":     [int("".join(str(x) for x in data["time"]))],
            "distance": [int("".join(str(x) for x in data["distance"]))]}

print(solve(part2(sample)))
print(fancy_solve(part2(sample)))
print(fancy_solve(part2(input)))