本文整理汇总了C#中Granados.BigInteger.bitCount方法的典型用法代码示例。如果您正苦于以下问题：C# BigInteger.bitCount方法的具体用法？C# BigInteger.bitCount怎么用？C# BigInteger.bitCount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Granados.BigInteger的用法示例。
在下文中一共展示了BigInteger.bitCount方法的2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: LucasSequenceHelper
//***********************************************************************
// Performs the calculation of the kth term in the Lucas Sequence.
// For details of the algorithm, see reference [9].
//
// k must be odd. i.e LSB == 1
//***********************************************************************
private static BigInteger[] LucasSequenceHelper(BigInteger P, BigInteger Q,
BigInteger k, BigInteger n,
BigInteger constant, int s) {
BigInteger[] result = new BigInteger[3];
if ((k.data[0] & 0x00000001) == 0)
throw (new ArgumentException("Argument k must be odd."));
int numbits = k.bitCount();
uint mask = (uint)0x1 << ((numbits & 0x1F) - 1);
// v = v0, v1 = v1, u1 = u1, Q_k = Q^0
BigInteger v = 2 % n, Q_k = 1 % n,
v1 = P % n, u1 = Q_k;
bool flag = true;
for (int i = k.dataLength - 1; i >= 0; i--) { // iterate on the binary expansion of k
//Console.WriteLine("round");
while (mask != 0) {
if (i == 0 && mask == 0x00000001) // last bit
break;
if ((k.data[i] & mask) != 0) { // bit is set
// index doubling with addition
u1 = (u1 * v1) % n;
v = ((v * v1) - (P * Q_k)) % n;
v1 = n.BarrettReduction(v1 * v1, n, constant);
v1 = (v1 - ((Q_k * Q) << 1)) % n;
if (flag)
flag = false;
else
Q_k = n.BarrettReduction(Q_k * Q_k, n, constant);
Q_k = (Q_k * Q) % n;
}
else {
// index doubling
u1 = ((u1 * v) - Q_k) % n;
v1 = ((v * v1) - (P * Q_k)) % n;
v = n.BarrettReduction(v * v, n, constant);
v = (v - (Q_k << 1)) % n;
if (flag) {
Q_k = Q % n;
flag = false;
}
else
Q_k = n.BarrettReduction(Q_k * Q_k, n, constant);
}
mask >>= 1;
}
mask = 0x80000000;
}
// at this point u1 = u(n+1) and v = v(n)
// since the last bit always 1, we need to transform u1 to u(2n+1) and v to v(2n+1)
u1 = ((u1 * v) - Q_k) % n;
v = ((v * v1) - (P * Q_k)) % n;
if (flag)
flag = false;
else
Q_k = n.BarrettReduction(Q_k * Q_k, n, constant);
Q_k = (Q_k * Q) % n;
for (int i = 0; i < s; i++) {
// index doubling
u1 = (u1 * v) % n;
v = ((v * v) - (Q_k << 1)) % n;
if (flag) {
Q_k = Q % n;
flag = false;
}
else
Q_k = n.BarrettReduction(Q_k * Q_k, n, constant);
}
result[0] = u1;
result[1] = v;
result[2] = Q_k;
return result;
}示例2: modPow
//***********************************************************************
// Modulo Exponentiation
//***********************************************************************
public BigInteger modPow(BigInteger exp, BigInteger n) {
if ((exp.data[maxLength - 1] & 0x80000000) != 0)
throw (new ArithmeticException("Positive exponents only."));
BigInteger resultNum = 1;
BigInteger tempNum;
bool thisNegative = false;
if ((this.data[maxLength - 1] & 0x80000000) != 0) { // negative this
tempNum = -this % n;
thisNegative = true;
}
else
tempNum = this % n; // ensures (tempNum * tempNum) < b^(2k)
if ((n.data[maxLength - 1] & 0x80000000) != 0) // negative n
n = -n;
// calculate constant = b^(2k) / m
BigInteger constant = new BigInteger();
int i = n.dataLength << 1;
constant.data[i] = 0x00000001;
constant.dataLength = i + 1;
constant = constant / n;
int totalBits = exp.bitCount();
int count = 0;
// perform squaring and multiply exponentiation
for (int pos = 0; pos < exp.dataLength; pos++) {
uint mask = 0x01;
//Console.WriteLine("pos = " + pos);
for (int index = 0; index < 32; index++) {
if ((exp.data[pos] & mask) != 0)
resultNum = BarrettReduction(resultNum * tempNum, n, constant);
mask <<= 1;
tempNum = BarrettReduction(tempNum * tempNum, n, constant);
if (tempNum.dataLength == 1 && tempNum.data[0] == 1) {
if (thisNegative && (exp.data[0] & 0x1) != 0) //odd exp
return -resultNum;
return resultNum;
}
count++;
if (count == totalBits)
break;
}
}
if (thisNegative && (exp.data[0] & 0x1) != 0) //odd exp
return -resultNum;
return resultNum;
}