本文整理汇总了Java中org.apache.commons.math3.fraction.BigFraction.divide方法的典型用法代码示例。如果您正苦于以下问题:Java BigFraction.divide方法的具体用法?Java BigFraction.divide怎么用?Java BigFraction.divide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.apache.commons.math3.fraction.BigFraction
的用法示例。
在下文中一共展示了BigFraction.divide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: msToSmpteTimecode
import org.apache.commons.math3.fraction.BigFraction; //导入方法依赖的package包/类
/**
* Transforms milliseconds to an SMPTE timecode according to the given edit unit rate.
* <ul>
* <li>An example of edit units is a frame.</li>
* <li>The output timecode has the following format 'hh:mm:ss:ff'.</li>
* </ul>
*
* @param milliseconds milliseconds to be transformed
* @param unitsInSec edit unit rate
* @return timecode as a string in "hh:mm:ss:ff" format.
*/
public static String msToSmpteTimecode(long milliseconds, BigFraction unitsInSec) {
BigFraction ms = new BigFraction(milliseconds);
BigFraction msInMin = new BigFraction(60 * 1000);
BigFraction msInHour = new BigFraction(60 * 60 * 1000);
BigFraction msInSec = new BigFraction(1000);
BigFraction unitsInMs = unitsInSec.divide(msInSec);
int hours = ms
.divide(msInHour)
.intValue();
int minutes = ms
.subtract(msInHour.multiply(hours))
.divide(msInMin)
.intValue();
int seconds = ms
.subtract(msInHour.multiply(hours))
.subtract(msInMin.multiply(minutes))
.divide(msInSec)
.intValue();
int units = ms
.subtract(msInHour.multiply(hours))
.subtract(msInMin.multiply(minutes))
.subtract(msInSec.multiply(seconds))
.multiply(unitsInMs)
.intValue();
return String.format("%02d:%02d:%02d:%02d", hours, minutes, seconds, units);
}
示例2: ballOnSupport
import org.apache.commons.math3.fraction.BigFraction; //导入方法依赖的package包/类
/** {@inheritDoc} */
public EnclosingBall<Euclidean2D, Vector2D> ballOnSupport(final List<Vector2D> support) {
if (support.size() < 1) {
return new EnclosingBall<Euclidean2D, Vector2D>(Vector2D.ZERO, Double.NEGATIVE_INFINITY);
} else {
final Vector2D vA = support.get(0);
if (support.size() < 2) {
return new EnclosingBall<Euclidean2D, Vector2D>(vA, 0, vA);
} else {
final Vector2D vB = support.get(1);
if (support.size() < 3) {
return new EnclosingBall<Euclidean2D, Vector2D>(new Vector2D(0.5, vA, 0.5, vB),
0.5 * vA.distance(vB),
vA, vB);
} else {
final Vector2D vC = support.get(2);
// a disk is 2D can be defined as:
// (1) (x - x_0)^2 + (y - y_0)^2 = r^2
// which can be written:
// (2) (x^2 + y^2) - 2 x_0 x - 2 y_0 y + (x_0^2 + y_0^2 - r^2) = 0
// or simply:
// (3) (x^2 + y^2) + a x + b y + c = 0
// with disk center coordinates -a/2, -b/2
// If the disk exists, a, b and c are a non-zero solution to
// [ (x^2 + y^2 ) x y 1 ] [ 1 ] [ 0 ]
// [ (xA^2 + yA^2) xA yA 1 ] [ a ] [ 0 ]
// [ (xB^2 + yB^2) xB yB 1 ] * [ b ] = [ 0 ]
// [ (xC^2 + yC^2) xC yC 1 ] [ c ] [ 0 ]
// So the determinant of the matrix is zero. Computing this determinant
// by expanding it using the minors m_ij of first row leads to
// (4) m_11 (x^2 + y^2) - m_12 x + m_13 y - m_14 = 0
// So by identifying equations (2) and (4) we get the coordinates
// of center as:
// x_0 = +m_12 / (2 m_11)
// y_0 = -m_13 / (2 m_11)
// Note that the minors m_11, m_12 and m_13 all have the last column
// filled with 1.0, hence simplifying the computation
final BigFraction[] c2 = new BigFraction[] {
new BigFraction(vA.getX()), new BigFraction(vB.getX()), new BigFraction(vC.getX())
};
final BigFraction[] c3 = new BigFraction[] {
new BigFraction(vA.getY()), new BigFraction(vB.getY()), new BigFraction(vC.getY())
};
final BigFraction[] c1 = new BigFraction[] {
c2[0].multiply(c2[0]).add(c3[0].multiply(c3[0])),
c2[1].multiply(c2[1]).add(c3[1].multiply(c3[1])),
c2[2].multiply(c2[2]).add(c3[2].multiply(c3[2]))
};
final BigFraction twoM11 = minor(c2, c3).multiply(2);
final BigFraction m12 = minor(c1, c3);
final BigFraction m13 = minor(c1, c2);
final BigFraction centerX = m12.divide(twoM11);
final BigFraction centerY = m13.divide(twoM11).negate();
final BigFraction dx = c2[0].subtract(centerX);
final BigFraction dy = c3[0].subtract(centerY);
final BigFraction r2 = dx.multiply(dx).add(dy.multiply(dy));
return new EnclosingBall<Euclidean2D, Vector2D>(new Vector2D(centerX.doubleValue(),
centerY.doubleValue()),
FastMath.sqrt(r2.doubleValue()),
vA, vB, vC);
}
}
}
}