Countdown (Editorial)

i = 0 \sum 9 c n t [i] \cdot se g m e n t [i], se g m e n t = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6} .

\sum_{i=0}^{9} cnt[i]\cdot segment[i], \qquad segment=\{6,2,5,5,4,5,6,3,7,6\}.

p o w = 1 0^{j}, high = ⌊ n /10 p o w ⌋, current = ⌊ n / p o w ⌋ mod 10, low = n mod p o w .

pow=10^{j},\quad \mathrm{high}=\bigl\lfloor n/10\,pow\bigr\rfloor,\quad \mathrm{current}=\bigl\lfloor n/pow\bigr\rfloor\bmod10,\quad \mathrm{low}=n\bmod pow.

c n t [i] [j] = high \cdot p o w + ⎩ ⎨ ⎧ p o w, low + 1, 0, current > i, current = i, current < i .

cnt[i][j]= \mathrm{high}\cdot pow+ \begin{cases} pow, & \mathrm{current}>i,\\ \mathrm{low}+1, & \mathrm{current}=i,\\ 0, & \mathrm{current}<i. \end{cases}

c n t [0] [j] = {0, (high - 1) \cdot p o w, high = 0, high \geq 1, + ⎩ ⎨ ⎧ p o w, low + 1, 0, current > 0, current = 0, current < 0.

cnt[0][j]= \begin{cases} 0,&\mathrm{high}=0,\\ (\mathrm{high}-1)\cdot pow,&\mathrm{high}\ge1, \end{cases} + \begin{cases} pow,&\mathrm{current}>0,\\ \mathrm{low}+1,&\mathrm{current}=0,\\ 0,&\mathrm{current}<0. \end{cases}