/// <summary>减</summary>
public static BigNumber Minus(BigNumber one, BigNumber two)
{
BigNumber oneAbsolute = one.AbsoluteNumber;
BigNumber twoAbsolute = two.AbsoluteNumber;
//正数减正数
if (one.IsPlus && two.IsPlus)
{
if (oneAbsolute.CompareTo(twoAbsolute) == 1)
{
return(PlusMinus(one, two));
}
else if (oneAbsolute.CompareTo(twoAbsolute) == -1)
{
return(PlusMinus(two, one).ReverseNumber);
}
else
{
return(BigNumber.Zero);
}
}
//正数减负数
else if (one.IsPlus && !two.IsPlus)
{
return(PlusAdd(oneAbsolute, twoAbsolute));
}
//负数减正数
else if (!one.IsPlus && two.IsPlus)
{
return(PlusAdd(oneAbsolute, twoAbsolute).ReverseNumber);
}
//负数减负数
else if (!one.IsPlus && !two.IsPlus)
{
return(Add(one, twoAbsolute));
}
throw new Exception("不可能到达的地方");
}
/// <summary>计算value^pow</summary>
static internal BigNumber Power(BigNumber value, BigNumber pow, int precision)
{
if (pow.DecimalPart.Count != 0 && !value.IsPlus) //第一个条件是小数次幂，第二个条件是负数
{
throw new ExpressionException("只有正数才有小数次幂");
}
BigNumber result = new BigNumber("1");
for (BigNumber i = new BigNumber("1"); i.CompareTo(pow) != 1; i++)
{
result = result * value;
}
BigNumber two = new BigNumber("2");
int n = 1;
BigNumber tt = new BigNumber(new List <int>(), pow.DecimalPart, true) * two;
while (true)
{
// 如果整数部分为1，那么就得进行一次2^n开方运算
if (tt.IntPart.Count == 1 && tt.IntPart[0] == 1)
{
tt.IntPart.Clear();
BigNumber r = Root(value, n, precision + 1);
if (r.GetPrecision(1) >= precision / BigNumber.OneCount + 2)
{
break;
}
result = result * r;
}
else if (tt.IsZero())
{
break;
}
tt = tt * two;
n++;
}
result.KeepPrecision(precision);
return(result);
}