本文整理匯總了C#中BigInteger.signum方法的典型用法代碼示例。如果您正苦於以下問題：C# BigInteger.signum方法的具體用法？C# BigInteger.signum怎麽用？C# BigInteger.signum使用的例子？那麽, 這裏精選的方法代碼示例或許可以為您提供幫助。您也可以進一步了解該方法所在類BigInteger的用法示例。
在下文中一共展示了BigInteger.signum方法的2個代碼示例，這些例子默認根據受歡迎程度排序。您可以為喜歡或者感覺有用的代碼點讚，您的評價將有助於係統推薦出更棒的C#代碼示例。
示例1: gcdBinary
/**
* @param m a positive modulus
* Return the greatest common divisor of op1 and op2,
*
* @param op1
* must be greater than zero
* @param op2
* must be greater than zero
* @see BigInteger#gcd(BigInteger)
* @return {@code GCD(op1, op2)}
*/
internal static BigInteger gcdBinary(BigInteger op1, BigInteger op2)
{
// PRE: (op1 > 0) and (op2 > 0)
/*
* Divide both number the maximal possible times by 2 without rounding
* gcd(2*a, 2*b) = 2 * gcd(a,b)
*/
int lsb1 = op1.getLowestSetBit();
int lsb2 = op2.getLowestSetBit();
int pow2Count = java.lang.Math.min(lsb1, lsb2);
BitLevel.inplaceShiftRight(op1, lsb1);
BitLevel.inplaceShiftRight(op2, lsb2);
BigInteger swap;
// I want op2 > op1
if (op1.compareTo(op2) == BigInteger.GREATER) {
swap = op1;
op1 = op2;
op2 = swap;
}
do { // INV: op2 >= op1 && both are odd unless op1 = 0
// Optimization for small operands
// (op2.bitLength() < 64) implies by INV (op1.bitLength() < 64)
if (( op2.numberLength == 1 )
|| ( ( op2.numberLength == 2 ) && ( op2.digits[1] > 0 ) )) {
op2 = BigInteger.valueOf(Division.gcdBinary(op1.longValue(),
op2.longValue()));
break;
}
// Implements one step of the Euclidean algorithm
// To reduce one operand if it's much smaller than the other one
if (op2.numberLength > op1.numberLength * 1.2) {
op2 = op2.remainder(op1);
if (op2.signum() != 0) {
BitLevel.inplaceShiftRight(op2, op2.getLowestSetBit());
}
} else {
// Use Knuth's algorithm of successive subtract and shifting
do {
Elementary.inplaceSubtract(op2, op1); // both are odd
BitLevel.inplaceShiftRight(op2, op2.getLowestSetBit()); // op2 is even
} while (op2.compareTo(op1) >= BigInteger.EQUALS);
}
// now op1 >= op2
swap = op2;
op2 = op1;
op1 = swap;
} while (op1.sign != 0);
return op2.shiftLeft(pow2Count);
}示例2: inplaceShiftRight
/**
* Performs {@code val >>= count} where {@code val} is a positive number.
*/
internal static void inplaceShiftRight(BigInteger val, int count)
{
int sign = val.signum();
if (count == 0 || val.signum() == 0)
return;
int intCount = count >> 5; // count of integers
val.numberLength -= intCount;
if (!shiftRight(val.digits, val.numberLength, val.digits, intCount,
count & 31)
&& sign < 0) {
// remainder not zero: add one to the result
int i;
for (i = 0; ( i < val.numberLength ) && ( val.digits[i] == -1 ); i++) {
val.digits[i] = 0;
}
if (i == val.numberLength) {
val.numberLength++;
}
val.digits[i]++;
}
val.cutOffLeadingZeroes();
val.unCache();
}