检查BigInteger是不是一个完美的正方形 [英] Check if BigInteger is not a perfect square
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问题描述
我有一个BigInteger值,假设它是282并且在变量x中。我现在想写一个while循环,声明:
I have a BigInteger value, let's say it is 282 and is inside the variable x. I now want to write a while loop that states:
while b2 isn't a perfect square:
a ← a + 1
b2 ← a*a - N
endwhile
我将如何使用BigInteger做这样的事情?
How would I do such a thing using BigInteger?
编辑:这样做的目的是为了写一个此方法。正如文章所述,必须检查b2是否不是正方形。
The purpose for this is so I can write this method. As the article states one must check if b2 is not square.
推荐答案
计算整数平方根,然后检查它的正方形是你的号码。这是我使用 Heron方法计算平方根的方法:
Compute the integer square root, then check that its square is your number. Here is my method of computing the square root using Heron's method:
private static final BigInteger TWO = BigInteger.valueOf(2);
/**
* Computes the integer square root of a number.
*
* @param n The number.
*
* @return The integer square root, i.e. the largest number whose square
* doesn't exceed n.
*/
public static BigInteger sqrt(BigInteger n)
{
if (n.signum() >= 0)
{
final int bitLength = n.bitLength();
BigInteger root = BigInteger.ONE.shiftLeft(bitLength / 2);
while (!isSqrt(n, root))
{
root = root.add(n.divide(root)).divide(TWO);
}
return root;
}
else
{
throw new ArithmeticException("square root of negative number");
}
}
private static boolean isSqrt(BigInteger n, BigInteger root)
{
final BigInteger lowerBound = root.pow(2);
final BigInteger upperBound = root.add(BigInteger.ONE).pow(2);
return lowerBound.compareTo(n) <= 0
&& n.compareTo(upperBound) < 0;
}
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