29 lines
1.4 KiB
Markdown
29 lines
1.4 KiB
Markdown
To calculate the repeating-digit-count we use an algorithm based on the idea of a long division. Here on the example of `1/7`
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| Position | Value | Remainder | Note |
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|----------|-------------- |---------------|----------------------------------------|
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| 0 | 0, | 10/7 | `10/7 = 1` **&** `(10%7)*10 = 30` |
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| 1 | 0,1 | 30/7 | `30/7 = 1` **&** `(30%7)*10 = 20` |
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| 2 | 0,13 | 20/7 | `20/7 = 1` **&** `(20%7)*10 = 60` |
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| 3 | 0,132 | 60/7 | `60/7 = 1` **&** `(60%7)*10 = 40` |
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| 4 | 0,1328 | 40/7 | `40/7 = 1` **&** `(40%7)*10 = 50` |
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| 5 | 0,13285 | 50/7 | `50/7 = 1` **&** `(50%7)*10 = 10` |
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| 6 | 0,132857 | 10/7 | **duplicate remainder -> loop closed** |
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**=>** RepeatingDigitCount := `6 - 0` = `6`
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We use a 1000-field "array" to remember every remainder we already had - as soon as we reach one that is already in use the digits start repeating itself.
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For better understanding here the **FindRepeatingDigitCount(int divisor)** algorithm in pseudo-code:
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```
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int current = 1;
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int position = 0;
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while(true) {
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current = (current*10) % divisor;
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position++;
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if (grid[current] != 0) return position - grid[current];
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else grid[current] = position;
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}
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``` |