关于Java：从BigDecimal获取减少的分数

Get reduced fraction from BigDecimal

我正在做一些非常精确的十进制计算，最后我将其转换为减少的分数。小数位数需要精确到96个小数位数。

由于精度非常重要，因此我使用BigDecimal和BigInteger。

BigDecimal的计算总是返回正确的十进制值，但是在某些情况下，我无法将此十进制转换为小数的函数

假设我有一个BigDecimal d

1	d.toString() = 32.222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222223

当我的函数试图将其转换为分数时，它输出

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Decimal from BigDecimal is:
32.222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222223

// Run the BigDecimal into getFraction
Denominator before reducing:
1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
Numerator before reducing:
32222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222223

// Reduced fraction turns into:
-1/0

// But should output
290/9

这是我的将小数减为小数的函数：

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static int[] getFraction(BigDecimal x) {
BigDecimal x1 = x.stripTrailingZeros();
//System.out.println(x.toString() +" stripped from zeroes");
//System.out.println(x1.scale());

// If scale is 0 or under we got a whole number fraction something/1
if(x1.scale() <= 0) {
//System.out.println("Whole number");
int[] rf = { x.intValue(), 1 };
return rf;
}

// If the decimal is
if(x.compareTo(BigDecimal.ZERO) < 0) {
// Add"-" to fraction when printing from main function
// Flip boolean to indicate negative decimal number
negative = true;

// Flip the BigDecimal
x = x.negate();
// Perform same function on flipped
return getFraction(x);
}

// Split BigDecimal into the intval and fractional val as strings
String[] parts = x.toString().split("\\\\.");

// Get starting numerator and denominator
BigDecimal denominator = BigDecimal.TEN.pow(parts[1].length());
System.out.println("Denominator :" + denominator.toString());

BigDecimal numerator = (new BigDecimal(parts[0]).multiply(denominator)).add(new BigDecimal(parts[1]));
System.out.println("Numerator :" + numerator.toString());

// Now we reduce
return reduceFraction(numerator.intValue(), denominator.intValue());
}

static int[] reduceFraction(int numerator, int denominator) {
// First find gcd
int gcd = BigInteger.valueOf(numerator).gcd(BigInteger.valueOf(denominator)).intValue();
//System.out.println(gcd);

// Then divide numerator and denominator by gcd
int[] reduced = { numerator / gcd, denominator / gcd };

// Return the fraction
return reduced;
}

如果有人能澄清我是否犯了任何错误，我将不胜感激！

**更新**

更改了reduceFraction函数：
现在返回一个String []而不是int []

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static String[] reduceFraction(BigDecimal numerator, BigDecimal denominator) {
// First find gcd
BigInteger nu = new BigInteger(numerator.toString());
BigInteger de = new BigInteger(denominator.toString());
BigInteger gcd = nu.gcd(de);

// Then divide numerator and denominator by gcd
nu = nu.divide(gcd);
de = de.divide(gcd);
String[] reduced = { nu.toString(), de.toString() };

// Return the fraction
return reduced;
}

getFraction返回：

1 2	// Now we reduce, send BigDecimals for numerator and denominator return reduceFraction(num, den);

而不是

1 2	// Now we reduce return reduceFraction(numerator.intValue(), denominator.intValue());

仍然从函数得到错误答案

输出分数现在为

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// Gcd value
gcd = 1

// Fraction is then:
32222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222223/1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

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//gcd Value should be:
gcd = 111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111

// Whit this gcd the fraction reduces to:
290/9

1 2	// Now we reduce return reduceFraction(numerator.intValue(), denominator.intValue());

好吧，在这种情况下这一定会失败，因为这里的分子或分母都不适合int。

分子变为-1908874353，对它们调用intValue()后分母变为0。您必须携带BigIntegers直到计算结束。

在将它们转换为int或long之前，如果必须这样做，可以通过将它们与Integer.MIN_VALUE，Integer.MAX_VALUE，和Long.MAX_VALUE。