本文整理汇总了Java中com.google.common.math.BigIntegerMath.sqrt方法的典型用法代码示例。如果您正苦于以下问题：Java BigIntegerMath.sqrt方法的具体用法？Java BigIntegerMath.sqrt怎么用？Java BigIntegerMath.sqrt使用的例子？那么, 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com.google.common.math.BigIntegerMath的用法示例。
在下文中一共展示了BigIntegerMath.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Java代码示例。
示例1: countPrimes32749
import com.google.common.math.BigIntegerMath; //导入方法依赖的package包/类
/**
* Factor the given value into primes less than <code>Short.MAX_VALUE</code> .
*
* @param val
* a BigInteger value which should be factored by all primes less equal than 1021
* @param map
* a map which counts the prime integer factors less equal than 32749
* @return the rest factor or zero, if the number could be factored completely into primes less equal then 1021
*/
public static BigInteger countPrimes32749(final BigInteger val, Map<Integer, Integer> map) {
BigInteger[] divRem;
BigInteger result = val;
BigInteger sqrt = BigIntegerMath.sqrt(result, RoundingMode.DOWN);
Integer count;
for (int i = 0; i < primes.length; i++) {
if (sqrt.compareTo(BIprimes[i]) < 0) {
break;
}
divRem = result.divideAndRemainder(BIprimes[i]);
while (divRem[1].equals(BigInteger.ZERO)) {
count = map.get(primes[i]);
if (count == null) {
map.put(primes[i], 1);
} else {
map.put(primes[i], count + 1);
}
result = divRem[0];// quotient
sqrt = BigIntegerMath.sqrt(result, RoundingMode.DOWN);
if (sqrt.compareTo(BIprimes[i]) < 0) {
break;
}
divRem = result.divideAndRemainder(BIprimes[i]);
}
}
BigInteger b;
int prime;
sqrt = BigIntegerMath.sqrt(result, RoundingMode.DOWN);
for (int i = 0; i < SHORT_PRIMES.length; i++) {
prime = SHORT_PRIMES[i];
b = BigInteger.valueOf(prime);
if (sqrt.compareTo(b) < 0) {
break;
}
divRem = result.divideAndRemainder(b);
while (divRem[1].equals(BigInteger.ZERO)) {
count = map.get(prime);
if (count == null) {
map.put(prime, 1);
} else {
map.put(prime, count + 1);
}
result = divRem[0];// quotient
sqrt = BigIntegerMath.sqrt(result, RoundingMode.DOWN);
if (sqrt.compareTo(b) < 0) {
break;
}
divRem = result.divideAndRemainder(b);
}
}
return result;
}
示例2: countRoot1021
import com.google.common.math.BigIntegerMath; //导入方法依赖的package包/类
/**
* Determine the n-th root from the prime decomposition of the primes[] array.
*
* @param val
* a BigInteger value which should be factored by all primes less equal than 1021
* @param root
* th n-th root which should be determined
* @return <code>(result[0] ^ root ) * result[1]</code>
*/
public static BigInteger[] countRoot1021(final BigInteger val, int root) {
BigInteger[] result = new BigInteger[2];
result[1] = val;
result[0] = BigInteger.ONE;
BigInteger[] divRem;
BigInteger sqrt = BigIntegerMath.sqrt(val, RoundingMode.DOWN);
int count = 0;
BigInteger temp = val;
// handle even values
while (temp.and(BigInteger.ONE).equals(BigInteger.ZERO) && !temp.equals(BigInteger.ZERO)) {
temp = temp.shiftRight(1);
count++;
if (count == root) {
count = 0;
result[1] = result[1].shiftRight(root);
result[0] = result[0].shiftLeft(1);
}
}
for (int i = 1; i < primes.length; i++) {
if (sqrt.compareTo(BIprimes[i]) < 0) {
break;
}
count = 0;
divRem = temp.divideAndRemainder(BIprimes[i]);
while (divRem[1].equals(BigInteger.ZERO)) {
count++;
if (count == root) {
count = 0;
result[1] = result[1].divide(BIprimes[i].pow(root));
result[0] = result[0].multiply(BIprimes[i]);
}
temp = divRem[0];// quotient
if (temp.compareTo(BIprimes[i]) < 0) {
break;
}
divRem = temp.divideAndRemainder(BIprimes[i]);
}
}
return result;
}