about | help | code help+videos | done | prefs |
Given a non-negative int n, compute recursively (no loops) the count of the occurrences of 8 as a digit, except that an 8 with another 8 immediately to its left counts double, so 8818 yields 4. Note that mod (%) by 10 yields the rightmost digit (126 % 10 is 6), while divide (/) by 10 removes the rightmost digit (126 / 10 is 12). count8(8) → 1 count8(818) → 2 count8(8818) → 4 ...Save, Compile, Run (ctrl-enter) |
Progress graphs:
Your progress graph for this problem
Random user progress graph for this problem
Random Epic Progress Graph
Difficulty: 324.0
Copyright Nick Parlante 2017 - privacy