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APCS Recursion 1, Problem 1. Given a non-negative int n, return the count of the occurrences of 7 as a digit, so for example 717 yields 2. (no loops). 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). R11_count7(717) → 2 R11_count7(7) → 1 R11_count7(123) → 0 ...Save, Compile, Run (ctrl-enter) |
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Difficulty: 300 Post-solution available
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