本文整理汇总了C#中Granados.BigInteger.sqrt方法的典型用法代码示例。如果您正苦于以下问题：C# BigInteger.sqrt方法的具体用法？C# BigInteger.sqrt怎么用？C# BigInteger.sqrt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.sqrt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: SqrtTest
//***********************************************************************
// Tests the correct implementation of sqrt() method.
//***********************************************************************
public static void SqrtTest(int rounds) {
Random rand = new Random();
for (int count = 0; count < rounds; count++) {
// generate data of random length
int t1 = 0;
while (t1 == 0)
t1 = (int)(rand.NextDouble() * 1024);
Console.Write("Round = " + count);
BigInteger a = new BigInteger();
a.genRandomBits(t1, rand);
BigInteger b = a.sqrt();
BigInteger c = (b + 1) * (b + 1);
// check that b is the largest integer such that b*b <= a
if (c <= a) {
Console.WriteLine("\nError at round " + count);
Console.WriteLine(a + "\n");
return;
}
Console.WriteLine(" <PASSED>.");
}
}示例2: LucasStrongTestHelper
private bool LucasStrongTestHelper(BigInteger thisVal) {
// Do the test (selects D based on Selfridge)
// Let D be the first element of the sequence
// 5, -7, 9, -11, 13, ... for which J(D,n) = -1
// Let P = 1, Q = (1-D) / 4
long D = 5, sign = -1, dCount = 0;
bool done = false;
while (!done) {
int Jresult = BigInteger.Jacobi(D, thisVal);
if (Jresult == -1)
done = true; // J(D, this) = 1
else {
if (Jresult == 0 && Math.Abs(D) < thisVal) // divisor found
return false;
if (dCount == 20) {
// check for square
BigInteger root = thisVal.sqrt();
if (root * root == thisVal)
return false;
}
//Console.WriteLine(D);
D = (Math.Abs(D) + 2) * sign;
sign = -sign;
}
dCount++;
}
long Q = (1 - D) >> 2;
/*
Console.WriteLine("D = " + D);
Console.WriteLine("Q = " + Q);
Console.WriteLine("(n,D) = " + thisVal.gcd(D));
Console.WriteLine("(n,Q) = " + thisVal.gcd(Q));
Console.WriteLine("J(D|n) = " + BigInteger.Jacobi(D, thisVal));
*/
BigInteger p_add1 = thisVal + 1;
int s = 0;
for (int index = 0; index < p_add1.dataLength; index++) {
uint mask = 0x01;
for (int i = 0; i < 32; i++) {
if ((p_add1.data[index] & mask) != 0) {
index = p_add1.dataLength; // to break the outer loop
break;
}
mask <<= 1;
s++;
}
}
BigInteger t = p_add1 >> s;
// calculate constant = b^(2k) / m
// for Barrett Reduction
BigInteger constant = new BigInteger();
int nLen = thisVal.dataLength << 1;
constant.data[nLen] = 0x00000001;
constant.dataLength = nLen + 1;
constant = constant / thisVal;
BigInteger[] lucas = LucasSequenceHelper(1, Q, t, thisVal, constant, 0);
bool isPrime = false;
if ((lucas[0].dataLength == 1 && lucas[0].data[0] == 0) ||
(lucas[1].dataLength == 1 && lucas[1].data[0] == 0)) {
// u(t) = 0 or V(t) = 0
isPrime = true;
}
for (int i = 1; i < s; i++) {
if (!isPrime) {
// doubling of index
lucas[1] = thisVal.BarrettReduction(lucas[1] * lucas[1], thisVal, constant);
lucas[1] = (lucas[1] - (lucas[2] << 1)) % thisVal;
//lucas[1] = ((lucas[1] * lucas[1]) - (lucas[2] << 1)) % thisVal;
if ((lucas[1].dataLength == 1 && lucas[1].data[0] == 0))
isPrime = true;
}
lucas[2] = thisVal.BarrettReduction(lucas[2] * lucas[2], thisVal, constant); //Q^k
}
if (isPrime) { // additional checks for composite numbers
// If n is prime and gcd(n, Q) == 1, then
// Q^((n+1)/2) = Q * Q^((n-1)/2) is congruent to (Q * J(Q, n)) mod n
BigInteger g = thisVal.gcd(Q);
//.........这里部分代码省略.........