本文整理汇总了Java中org.apache.commons.math3.util.ArithmeticUtils.mulAndCheck方法的典型用法代码示例。如果您正苦于以下问题：Java ArithmeticUtils.mulAndCheck方法的具体用法？Java ArithmeticUtils.mulAndCheck怎么用？Java ArithmeticUtils.mulAndCheck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.util.ArithmeticUtils的用法示例。
在下文中一共展示了ArithmeticUtils.mulAndCheck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: getEntry
import org.apache.commons.math3.util.ArithmeticUtils; //导入方法依赖的package包/类
/**
* Returns the {@code (i, j)} entry of the inverse Hilbert matrix. Exact
* arithmetic is used; in case of overflow, an exception is thrown.
*
* @param i Row index (starts at 0).
* @param j Column index (starts at 0).
* @return The coefficient of the inverse Hilbert matrix.
*/
public long getEntry(final int i, final int j) {
long val = i + j + 1;
long aux = CombinatoricsUtils.binomialCoefficient(n + i, n - j - 1);
val = ArithmeticUtils.mulAndCheck(val, aux);
aux = CombinatoricsUtils.binomialCoefficient(n + j, n - i - 1);
val = ArithmeticUtils.mulAndCheck(val, aux);
aux = CombinatoricsUtils.binomialCoefficient(i + j, i);
val = ArithmeticUtils.mulAndCheck(val, aux);
val = ArithmeticUtils.mulAndCheck(val, aux);
return ((i + j) & 1) == 0 ? val : -val;
}
示例2: getEntry
import org.apache.commons.math3.util.ArithmeticUtils; //导入方法依赖的package包/类
/**
* Returns the {@code (i, j)} entry of the inverse Hilbert matrix. Exact
* arithmetic is used; in case of overflow, an exception is thrown.
*
* @param i Row index (starts at 0).
* @param j Column index (starts at 0).
* @return The coefficient of the inverse Hilbert matrix.
*/
public long getEntry(final int i, final int j) {
long val = i + j + 1;
long aux = ArithmeticUtils.binomialCoefficient(n + i, n - j - 1);
val = ArithmeticUtils.mulAndCheck(val, aux);
aux = ArithmeticUtils.binomialCoefficient(n + j, n - i - 1);
val = ArithmeticUtils.mulAndCheck(val, aux);
aux = ArithmeticUtils.binomialCoefficient(i + j, i);
val = ArithmeticUtils.mulAndCheck(val, aux);
val = ArithmeticUtils.mulAndCheck(val, aux);
return ((i + j) & 1) == 0 ? val : -val;
}