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# Project Euler #113: Non-bouncy numbers

# Project Euler #113: Non-bouncy numbers

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I arrived at the formula (k+9)C9+(k+10)C10 - 10*k - 2 non bouncy numbers less than 10^k using pen and paper. (combinatorics)

Also number of increasing and decreasing non bouncy numbers won't be the same, some people might make that mistake.

Seriously, putting two times the same number... just so you know for the ones that come after.

Terminated due to timeout What does this mean???

Input (stdin) 3 3 5 10

Your Output (stdout) 474 4953

Expected Output 474 4953 277032

Compiler Message

Terminated due to timeout

Can it be solved using digit dp? Means states will be like: dp[N][last]=summation(dp[N-1][i]) for all i>=last for increasing? I tried this but its counting lesser no of increasing and decreasing numbers. Can anyone explain whats wrong?

By the definitions given, the number 4444 is trivially both increasing and decreasing and thus not bouncy, correct?