本文整理汇总了C#中Org.BouncyCastle.Math.BigInteger.QuickPow2Check方法的典型用法代码示例。如果您正苦于以下问题:C# BigInteger.QuickPow2Check方法的具体用法?C# BigInteger.QuickPow2Check怎么用?C# BigInteger.QuickPow2Check使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类Org.BouncyCastle.Math.BigInteger
的用法示例。
在下文中一共展示了BigInteger.QuickPow2Check方法的6个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的C#代码示例。
示例1: Remainder
public BigInteger Remainder(
BigInteger n)
{
if (n.sign == 0)
throw new ArithmeticException("Division by zero error");
if (this.sign == 0)
return Zero;
// For small values, use fast remainder method
if (n.magnitude.Length == 1)
{
int val = n.magnitude[0];
if (val > 0)
{
if (val == 1)
return Zero;
// TODO Make this func work on uint, and handle val == 1?
int rem = Remainder(val);
return rem == 0
? Zero
: new BigInteger(sign, new int[]{ rem }, false);
}
}
if (CompareNoLeadingZeroes(0, magnitude, 0, n.magnitude) < 0)
return this;
int[] result;
if (n.QuickPow2Check()) // n is power of two
{
// TODO Move before small values branch above?
result = LastNBits(n.Abs().BitLength - 1);
}
else
{
result = (int[]) this.magnitude.Clone();
result = Remainder(result, n.magnitude);
}
return new BigInteger(sign, result, true);
}
示例2: Multiply
public BigInteger Multiply(
BigInteger val)
{
if (sign == 0 || val.sign == 0)
return Zero;
if (val.QuickPow2Check()) // val is power of two
{
BigInteger result = this.ShiftLeft(val.Abs().BitLength - 1);
return val.sign > 0 ? result : result.Negate();
}
if (this.QuickPow2Check()) // this is power of two
{
BigInteger result = val.ShiftLeft(this.Abs().BitLength - 1);
return this.sign > 0 ? result : result.Negate();
}
int resLength = (this.BitLength + val.BitLength) / BitsPerInt + 1;
int[] res = new int[resLength];
if (val == this)
{
Square(res, this.magnitude);
}
else
{
Multiply(res, this.magnitude, val.magnitude);
}
return new BigInteger(sign * val.sign, res, true);
}
示例3: DivideAndRemainder
public BigInteger[] DivideAndRemainder(
BigInteger val)
{
if (val.sign == 0)
throw new ArithmeticException("Division by zero error");
BigInteger[] biggies = new BigInteger[2];
if (sign == 0)
{
biggies[0] = Zero;
biggies[1] = Zero;
}
else if (val.QuickPow2Check()) // val is power of two
{
int e = val.Abs().BitLength - 1;
BigInteger quotient = this.Abs().ShiftRight(e);
int[] remainder = this.LastNBits(e);
biggies[0] = val.sign == this.sign ? quotient : quotient.Negate();
biggies[1] = new BigInteger(this.sign, remainder, true);
}
else
{
int[] remainder = (int[]) this.magnitude.Clone();
int[] quotient = Divide(remainder, val.magnitude);
biggies[0] = new BigInteger(this.sign * val.sign, quotient, true);
biggies[1] = new BigInteger(this.sign, remainder, true);
}
return biggies;
}
示例4: Divide
public BigInteger Divide(
BigInteger val)
{
if (val.sign == 0)
throw new ArithmeticException("Division by zero error");
if (sign == 0)
return Zero;
if (val.QuickPow2Check()) // val is power of two
{
BigInteger result = this.Abs().ShiftRight(val.Abs().BitLength - 1);
return val.sign == this.sign ? result : result.Negate();
}
int[] mag = (int[]) this.magnitude.Clone();
return new BigInteger(this.sign * val.sign, Divide(mag, val.magnitude), true);
}
示例5: Multiply
public BigInteger Multiply(
BigInteger val)
{
if (val == this)
return Square();
if ((sign & val.sign) == 0)
return Zero;
if (val.QuickPow2Check()) // val is power of two
{
BigInteger result = this.ShiftLeft(val.Abs().BitLength - 1);
return val.sign > 0 ? result : result.Negate();
}
if (this.QuickPow2Check()) // this is power of two
{
BigInteger result = val.ShiftLeft(this.Abs().BitLength - 1);
return this.sign > 0 ? result : result.Negate();
}
int resLength = magnitude.Length + val.magnitude.Length;
int[] res = new int[resLength];
Multiply(res, this.magnitude, val.magnitude);
int resSign = sign ^ val.sign ^ 1;
return new BigInteger(resSign, res, true);
}
示例6: ModInverse
public BigInteger ModInverse(
BigInteger m)
{
if (m.sign < 1)
throw new ArithmeticException("Modulus must be positive");
// TODO Too slow at the moment
// // "Fast Key Exchange with Elliptic Curve Systems" R.Schoeppel
// if (m.TestBit(0))
// {
// //The Almost Inverse Algorithm
// int k = 0;
// BigInteger B = One, C = Zero, F = this, G = m, tmp;
//
// for (;;)
// {
// // While F is even, do F=F/u, C=C*u, k=k+1.
// int zeroes = F.GetLowestSetBit();
// if (zeroes > 0)
// {
// F = F.ShiftRight(zeroes);
// C = C.ShiftLeft(zeroes);
// k += zeroes;
// }
//
// // If F = 1, then return B,k.
// if (F.Equals(One))
// {
// BigInteger half = m.Add(One).ShiftRight(1);
// BigInteger halfK = half.ModPow(BigInteger.ValueOf(k), m);
// return B.Multiply(halfK).Mod(m);
// }
//
// if (F.CompareTo(G) < 0)
// {
// tmp = G; G = F; F = tmp;
// tmp = B; B = C; C = tmp;
// }
//
// F = F.Add(G);
// B = B.Add(C);
// }
// }
if (m.QuickPow2Check())
{
return ModInversePow2(m);
}
BigInteger d = this.Remainder(m);
BigInteger x;
BigInteger gcd = ExtEuclid(d, m, out x);
if (!gcd.Equals(One))
throw new ArithmeticException("Numbers not relatively prime.");
if (x.sign < 0)
{
x = x.Add(m);
}
return x;
}