E - Σ[k=0..10^100]floor(X/10^k) Editorial by Kiri8128
We will explain an easy (and short) implementation with arbitrary-precision integer calculation (like in Python).
Let \(X_i\) be the \(i\) th smallest digit of \(X\). Let \(\displaystyle s = \sum_{i}{X_i}\) be the sum of all \(X_i\) ’s. Now what we need is
\(\displaystyle \sum_{k=0}^{\infty} \left \lfloor \frac{X}{10^k} \right \rfloor\)
\(=\displaystyle \sum_{k=0}^{\infty} \frac{X}{10^k} - \sum_{k=0}^{\infty} \frac{X\bmod 10^k}{10^k}\)
\(=\displaystyle \sum_{k=0}^{\infty} \frac{X}{10^k} - \sum_{i}\sum_{k=1}^{\infty} \frac{X_i}{10^k}\)
\(=\displaystyle \frac{10X}{9} - \sum_{i}\frac{X_i}{9}\)
\(=\displaystyle \frac{10X-s}{9}\)
AC code(PyPy3)
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