# Some phone usage rate may be described as follows:

# first minute of a call costs min1 cents,
# each minute from the 2nd up to 10th (inclusive) costs min2_10 cents
# each minute after 10th costs min11 cents.
# You have s cents on your account before the call. What is the duration of the longest call (in minutes rounded down to the nearest integer) you can have?

# Example

# For min1 = 3, min2_10 = 1, min11 = 2, and s = 20, the output should be
# phoneCall(min1, min2_10, min11, s) = 14.

# Here's why:

# the first minute costs 3 cents, which leaves you with 20 - 3 = 17 cents;
# the total cost of minutes 2 through 10 is 1 * 9 = 9, so you can talk 9 more minutes and still have 17 - 9 = 8 cents;
# each next minute costs 2 cents, which means that you can talk 8 / 2 = 4 more minutes.
# Thus, the longest call you can make is 1 + 9 + 4 = 14 minutes long.

def phoneCall(min1, min2_10, min11, s):
    if s < min1:
        return 0
    elif s == min1:
        return 1
    elif s < 11:
        return s - min1 - min2_10
    else:
        a = s - min1
        b = a - min2_10*9
        c = b // min11
    return 1 + 9 + c


print(phoneCall(3, 1, 2, 24))
