如何知道一小部分的重复小数？

问题描述

我已经知道，当一个分数循环小数。这里是功能

I already know when a fraction is repeating decimals. Here is the function.

public bool IsRepeatingDecimal
{
    get
    {
        if (Numerator % Denominator == 0)
            return false;

        var primes = MathAlgorithms.Primes(Denominator);

        foreach (int n in primes)
        {
            if (n != 2 && n != 5)
                return true;
        }

        return false;
    }
}

现在，我试图让重复的次数。我检查这个网站：http://en.wikipedia.org/wiki/Repeating_decimal

Now, I'm trying to get the repeated number. I'm checking this web site: http://en.wikipedia.org/wiki/Repeating_decimal

public decimal RepeatingDecimal()
{
    if (!IsRepeatingDecimal) throw new InvalidOperationException("The fraction is not producing repeating decimals");

    int digitsToTake;
    switch (Denominator)
    {
        case 3:
        case 9: digitsToTake = 1; break;
        case 11: digitsToTake = 2; break;
        case 13: digitsToTake = 6; break;
        default: digitsToTake = Denominator - 1; break;
    }

    return MathExtensions.TruncateAt((decimal)Numerator / Denominator, digitsToTake);
}

不过，我真正意识到，有些数字有部分十进制有限，后来无限。例如：1/28

But I really realized, that some numbers has a partial decimal finite and later infinite. For example: 1/28

你知道一个更好的方式来做到这一点？或算法？

Do you know a better way to do this? Or an Algorithm?

推荐答案

一个很简单的算法是这样的：执行长除法。记录你做的每一个中间划分。当你看到一个分裂等同于一个你以前做过，你有什么东西被重复。

A very simple algorithm is this: implement long division. Record every intermediate division you do. As soon as you see a division identical to the one you've done before, you have what's being repeated.

例如：7/13

1. 13 goes into   7 0 times with remainder  7; bring down a 0.
2. 13 goes into  70 5 times with remainder  5; bring down a 0.
3. 13 goes into  50 3 times with remainder 11; bring down a 0.
4. 13 goes into 110 8 times with remainder  6; bring down a 0.
5. 13 goes into  60 4 times with remainder  8; bring down a 0.
6. 13 goes into  80 6 times with remainder  2; bring down a 0.
7. 13 goes into  20 1 time  with remainder  7; bring down a 0.
8. We have already seen 13/70 on line 2; so lines 2-7 have the repeating part

该算法为我们提供了538461作为重复的部分。我的计算器说7/13是0.538461538。看起来我的权利！这一切仍然是实现细节，还是找一个更好的算法！

The algorithm gives us 538461 as the repeating part. My calculator says 7/13 is 0.538461538. Looks right to me! All that remains are implementation details, or to find a better algorithm!

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