In the above example, the numbers are still small enough to be converted to double
, so in this case you can say
double result = (double)x / (double)y;
If x
and y
are too huge for a double
but still comparable, maybe this great trick is helpful:
double result = Math.Exp(BigInteger.Log(x) - BigInteger.Log(y));
But in general, when the BigInteger
are huge, and their quotient is huge too, this is hard to do without importing a third-party library.
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…