# nayuki/Project-Euler-solutions

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 # # Solution to Project Euler problem 113 # Copyright (c) Project Nayuki. All rights reserved. # # https://www.nayuki.io/page/project-euler-solutions # https://github.com/nayuki/Project-Euler-solutions # import eulerlib # Let n be the number of digits. To count the number of increasing or decreasing numbers using combinatorics, # let's view each number as a sequence of digit readout slots and operations. For example, suppose n=5 and # we examine the increasing number 23667. We can express it as the sequence "+ + # + # + + + # # + # + +", # where # is a digit and + means increment. This way of thinking will be useful, as we will see. # # For the set of increasing numbers, each number has n readout slots and 9 increments, positioned arbitrarily. # Using this construction, the number is guaranteed to be increasing. Note that leading zeros can be produced. # Conversely, for each increasing number, we can generate a (unique) sequence of slots and increments that represents it # (putting all unused increments after the rightmost digit). Hence there are n+9 objects to arrange in sequence, # so there are binomial(n + 9, 9) ways to arrange them. Finally we subtract 1 because 0 can be formed with this scheme, # which must be excluded from the set of increasing numbers. # # For the set of decreasing numbers, each number has n readout slots and 10 operations. Of the 10 operations, # the leading one must be "increment to 9", and the rest must be decrements. Similar to the increasing case, # each sequence of slots and decrements produces a decreasing number, and conversely each decreasing number # corresponds to a unique sequence of slots and decrements. However, 0 can be formed in n+1 ways, by concentrating # all 10 operations between some pair of slots, e.g. "+9 -9 # # # #", "# +9 -9 # # #", ..., "# # # # +9 -9". # # There are 9n "flat" numbers, for example: 1, 2, ..., 9; 11, 22, ..., 99; 111, 222, ..., 999; ... (note that 0 is excluded). # Since they are double-counted in the increasing and decreasing numbers, we subtract the size of this set. # # In conclusion, the number of non-bouncy numbers is (binomial(n+9,9) - 1) + (binomial(n+10,10) - (n+1)) - 9n. # # (Technically, in the problem statement and this solution, "increasing" actually means "nondecreasing" and "decreasing" means "nonincreasing".) def compute(): DIGITS = 100 increasing = eulerlib.binomial(DIGITS + 9, 9) - 1 decreasing = eulerlib.binomial(DIGITS + 10, 10) - (DIGITS + 1) flat = DIGITS * 9 ans = increasing + decreasing - flat return str(ans) if __name__ == "__main__": print(compute())