1372 lines
41 KiB
Java
1372 lines
41 KiB
Java
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/*
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* Copyright 1999-2007 Sun Microsystems, Inc. All Rights Reserved.
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* DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
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*
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* This code is free software; you can redistribute it and/or modify it
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* under the terms of the GNU General Public License version 2 only, as
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* published by the Free Software Foundation. Sun designates this
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* particular file as subject to the "Classpath" exception as provided
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* by Sun in the LICENSE file that accompanied this code.
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*
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* This code is distributed in the hope that it will be useful, but WITHOUT
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* ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
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* FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License
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* version 2 for more details (a copy is included in the LICENSE file that
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* accompanied this code).
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*
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* You should have received a copy of the GNU General Public License version
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* 2 along with this work; if not, write to the Free Software Foundation,
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* Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
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*
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* Please contact Sun Microsystems, Inc., 4150 Network Circle, Santa Clara,
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* CA 95054 USA or visit www.sun.com if you need additional information or
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* have any questions.
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*/
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package java.math;
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/**
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* A class used to represent multiprecision integers that makes efficient
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* use of allocated space by allowing a number to occupy only part of
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* an array so that the arrays do not have to be reallocated as often.
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* When performing an operation with many iterations the array used to
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* hold a number is only reallocated when necessary and does not have to
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* be the same size as the number it represents. A mutable number allows
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* calculations to occur on the same number without having to create
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* a new number for every step of the calculation as occurs with
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* BigIntegers.
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*
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* @see BigInteger
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* @author Michael McCloskey
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* @since 1.3
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*/
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class MutableBigInteger {
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/**
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* Holds the magnitude of this MutableBigInteger in big endian order.
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* The magnitude may start at an offset into the value array, and it may
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* end before the length of the value array.
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*/
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int[] value;
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/**
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* The number of ints of the value array that are currently used
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* to hold the magnitude of this MutableBigInteger. The magnitude starts
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* at an offset and offset + intLen may be less than value.length.
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*/
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int intLen;
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/**
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* The offset into the value array where the magnitude of this
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* MutableBigInteger begins.
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*/
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int offset = 0;
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/**
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* This mask is used to obtain the value of an int as if it were unsigned.
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*/
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private final static long LONG_MASK = 0xffffffffL;
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// Constructors
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/**
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* The default constructor. An empty MutableBigInteger is created with
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* a one word capacity.
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*/
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MutableBigInteger() {
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value = new int[1];
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intLen = 0;
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}
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/**
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* Construct a new MutableBigInteger with a magnitude specified by
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* the int val.
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*/
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MutableBigInteger(int val) {
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value = new int[1];
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intLen = 1;
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value[0] = val;
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}
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/**
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* Construct a new MutableBigInteger with the specified value array
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* up to the specified length.
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*/
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MutableBigInteger(int[] val, int len) {
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value = val;
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intLen = len;
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}
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/**
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* Construct a new MutableBigInteger with the specified value array
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* up to the length of the array supplied.
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*/
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MutableBigInteger(int[] val) {
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value = val;
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intLen = val.length;
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}
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/**
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* Construct a new MutableBigInteger with a magnitude equal to the
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* specified BigInteger.
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*/
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MutableBigInteger(BigInteger b) {
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value = b.mag.clone();
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intLen = value.length;
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}
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/**
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* Construct a new MutableBigInteger with a magnitude equal to the
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* specified MutableBigInteger.
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*/
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MutableBigInteger(MutableBigInteger val) {
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intLen = val.intLen;
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value = new int[intLen];
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for(int i=0; i<intLen; i++)
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value[i] = val.value[val.offset+i];
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}
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/**
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* Clear out a MutableBigInteger for reuse.
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*/
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void clear() {
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offset = intLen = 0;
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for (int index=0, n=value.length; index < n; index++)
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value[index] = 0;
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}
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/**
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* Set a MutableBigInteger to zero, removing its offset.
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*/
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void reset() {
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offset = intLen = 0;
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}
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/**
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* Compare the magnitude of two MutableBigIntegers. Returns -1, 0 or 1
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* as this MutableBigInteger is numerically less than, equal to, or
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* greater than {@code b}.
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*/
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final int compare(MutableBigInteger b) {
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if (intLen < b.intLen)
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return -1;
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if (intLen > b.intLen)
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return 1;
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for (int i=0; i<intLen; i++) {
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int b1 = value[offset+i] + 0x80000000;
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int b2 = b.value[b.offset+i] + 0x80000000;
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if (b1 < b2)
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return -1;
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if (b1 > b2)
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return 1;
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}
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return 0;
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}
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/**
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* Return the index of the lowest set bit in this MutableBigInteger. If the
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* magnitude of this MutableBigInteger is zero, -1 is returned.
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*/
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private final int getLowestSetBit() {
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if (intLen == 0)
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return -1;
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int j, b;
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for (j=intLen-1; (j>0) && (value[j+offset]==0); j--)
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;
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b = value[j+offset];
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if (b==0)
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return -1;
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return ((intLen-1-j)<<5) + BigInteger.trailingZeroCnt(b);
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}
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/**
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* Return the int in use in this MutableBigInteger at the specified
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* index. This method is not used because it is not inlined on all
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* platforms.
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*/
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private final int getInt(int index) {
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return value[offset+index];
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}
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/**
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* Return a long which is equal to the unsigned value of the int in
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* use in this MutableBigInteger at the specified index. This method is
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* not used because it is not inlined on all platforms.
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*/
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private final long getLong(int index) {
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return value[offset+index] & LONG_MASK;
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}
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/**
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* Ensure that the MutableBigInteger is in normal form, specifically
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* making sure that there are no leading zeros, and that if the
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* magnitude is zero, then intLen is zero.
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*/
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final void normalize() {
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if (intLen == 0) {
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offset = 0;
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return;
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}
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int index = offset;
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if (value[index] != 0)
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return;
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int indexBound = index+intLen;
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do {
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index++;
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} while(index < indexBound && value[index]==0);
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int numZeros = index - offset;
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intLen -= numZeros;
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offset = (intLen==0 ? 0 : offset+numZeros);
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}
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/**
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* If this MutableBigInteger cannot hold len words, increase the size
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* of the value array to len words.
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*/
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private final void ensureCapacity(int len) {
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if (value.length < len) {
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value = new int[len];
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offset = 0;
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intLen = len;
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}
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}
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/**
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* Convert this MutableBigInteger into an int array with no leading
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* zeros, of a length that is equal to this MutableBigInteger's intLen.
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*/
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int[] toIntArray() {
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int[] result = new int[intLen];
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for(int i=0; i<intLen; i++)
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result[i] = value[offset+i];
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return result;
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}
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/**
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* Sets the int at index+offset in this MutableBigInteger to val.
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* This does not get inlined on all platforms so it is not used
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* as often as originally intended.
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*/
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void setInt(int index, int val) {
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value[offset + index] = val;
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}
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/**
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* Sets this MutableBigInteger's value array to the specified array.
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* The intLen is set to the specified length.
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*/
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void setValue(int[] val, int length) {
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value = val;
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intLen = length;
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offset = 0;
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}
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/**
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* Sets this MutableBigInteger's value array to a copy of the specified
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* array. The intLen is set to the length of the new array.
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*/
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void copyValue(MutableBigInteger val) {
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int len = val.intLen;
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if (value.length < len)
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value = new int[len];
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for(int i=0; i<len; i++)
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value[i] = val.value[val.offset+i];
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intLen = len;
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offset = 0;
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}
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/**
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* Sets this MutableBigInteger's value array to a copy of the specified
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* array. The intLen is set to the length of the specified array.
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*/
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void copyValue(int[] val) {
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int len = val.length;
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if (value.length < len)
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value = new int[len];
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for(int i=0; i<len; i++)
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value[i] = val[i];
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intLen = len;
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offset = 0;
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}
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/**
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* Returns true iff this MutableBigInteger has a value of one.
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*/
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boolean isOne() {
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return (intLen == 1) && (value[offset] == 1);
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}
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/**
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* Returns true iff this MutableBigInteger has a value of zero.
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*/
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boolean isZero() {
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return (intLen == 0);
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}
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/**
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* Returns true iff this MutableBigInteger is even.
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*/
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boolean isEven() {
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return (intLen == 0) || ((value[offset + intLen - 1] & 1) == 0);
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}
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/**
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* Returns true iff this MutableBigInteger is odd.
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*/
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boolean isOdd() {
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return ((value[offset + intLen - 1] & 1) == 1);
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}
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/**
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* Returns true iff this MutableBigInteger is in normal form. A
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* MutableBigInteger is in normal form if it has no leading zeros
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* after the offset, and intLen + offset <= value.length.
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*/
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boolean isNormal() {
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if (intLen + offset > value.length)
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return false;
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if (intLen ==0)
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return true;
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return (value[offset] != 0);
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}
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/**
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* Returns a String representation of this MutableBigInteger in radix 10.
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*/
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public String toString() {
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BigInteger b = new BigInteger(this, 1);
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return b.toString();
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}
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/**
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* Right shift this MutableBigInteger n bits. The MutableBigInteger is left
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* in normal form.
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*/
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void rightShift(int n) {
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if (intLen == 0)
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return;
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int nInts = n >>> 5;
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int nBits = n & 0x1F;
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this.intLen -= nInts;
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if (nBits == 0)
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return;
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int bitsInHighWord = BigInteger.bitLen(value[offset]);
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if (nBits >= bitsInHighWord) {
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this.primitiveLeftShift(32 - nBits);
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this.intLen--;
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} else {
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primitiveRightShift(nBits);
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}
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}
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/**
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* Left shift this MutableBigInteger n bits.
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*/
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void leftShift(int n) {
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/*
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* If there is enough storage space in this MutableBigInteger already
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* the available space will be used. Space to the right of the used
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* ints in the value array is faster to utilize, so the extra space
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* will be taken from the right if possible.
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*/
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if (intLen == 0)
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return;
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int nInts = n >>> 5;
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int nBits = n&0x1F;
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int bitsInHighWord = BigInteger.bitLen(value[offset]);
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// If shift can be done without moving words, do so
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if (n <= (32-bitsInHighWord)) {
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primitiveLeftShift(nBits);
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return;
|
||
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}
|
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|
|
||
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int newLen = intLen + nInts +1;
|
||
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if (nBits <= (32-bitsInHighWord))
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newLen--;
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if (value.length < newLen) {
|
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// The array must grow
|
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int[] result = new int[newLen];
|
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for (int i=0; i<intLen; i++)
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result[i] = value[offset+i];
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setValue(result, newLen);
|
||
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} else if (value.length - offset >= newLen) {
|
||
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// Use space on right
|
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for(int i=0; i<newLen - intLen; i++)
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value[offset+intLen+i] = 0;
|
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} else {
|
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// Must use space on left
|
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for (int i=0; i<intLen; i++)
|
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value[i] = value[offset+i];
|
||
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for (int i=intLen; i<newLen; i++)
|
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value[i] = 0;
|
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offset = 0;
|
||
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}
|
||
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intLen = newLen;
|
||
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if (nBits == 0)
|
||
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return;
|
||
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if (nBits <= (32-bitsInHighWord))
|
||
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primitiveLeftShift(nBits);
|
||
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else
|
||
|
primitiveRightShift(32 -nBits);
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* A primitive used for division. This method adds in one multiple of the
|
||
|
* divisor a back to the dividend result at a specified offset. It is used
|
||
|
* when qhat was estimated too large, and must be adjusted.
|
||
|
*/
|
||
|
private int divadd(int[] a, int[] result, int offset) {
|
||
|
long carry = 0;
|
||
|
|
||
|
for (int j=a.length-1; j >= 0; j--) {
|
||
|
long sum = (a[j] & LONG_MASK) +
|
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(result[j+offset] & LONG_MASK) + carry;
|
||
|
result[j+offset] = (int)sum;
|
||
|
carry = sum >>> 32;
|
||
|
}
|
||
|
return (int)carry;
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* This method is used for division. It multiplies an n word input a by one
|
||
|
* word input x, and subtracts the n word product from q. This is needed
|
||
|
* when subtracting qhat*divisor from dividend.
|
||
|
*/
|
||
|
private int mulsub(int[] q, int[] a, int x, int len, int offset) {
|
||
|
long xLong = x & LONG_MASK;
|
||
|
long carry = 0;
|
||
|
offset += len;
|
||
|
|
||
|
for (int j=len-1; j >= 0; j--) {
|
||
|
long product = (a[j] & LONG_MASK) * xLong + carry;
|
||
|
long difference = q[offset] - product;
|
||
|
q[offset--] = (int)difference;
|
||
|
carry = (product >>> 32)
|
||
|
+ (((difference & LONG_MASK) >
|
||
|
(((~(int)product) & LONG_MASK))) ? 1:0);
|
||
|
}
|
||
|
return (int)carry;
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* Right shift this MutableBigInteger n bits, where n is
|
||
|
* less than 32.
|
||
|
* Assumes that intLen > 0, n > 0 for speed
|
||
|
*/
|
||
|
private final void primitiveRightShift(int n) {
|
||
|
int[] val = value;
|
||
|
int n2 = 32 - n;
|
||
|
for (int i=offset+intLen-1, c=val[i]; i>offset; i--) {
|
||
|
int b = c;
|
||
|
c = val[i-1];
|
||
|
val[i] = (c << n2) | (b >>> n);
|
||
|
}
|
||
|
val[offset] >>>= n;
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* Left shift this MutableBigInteger n bits, where n is
|
||
|
* less than 32.
|
||
|
* Assumes that intLen > 0, n > 0 for speed
|
||
|
*/
|
||
|
private final void primitiveLeftShift(int n) {
|
||
|
int[] val = value;
|
||
|
int n2 = 32 - n;
|
||
|
for (int i=offset, c=val[i], m=i+intLen-1; i<m; i++) {
|
||
|
int b = c;
|
||
|
c = val[i+1];
|
||
|
val[i] = (b << n) | (c >>> n2);
|
||
|
}
|
||
|
val[offset+intLen-1] <<= n;
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* Adds the contents of two MutableBigInteger objects.The result
|
||
|
* is placed within this MutableBigInteger.
|
||
|
* The contents of the addend are not changed.
|
||
|
*/
|
||
|
void add(MutableBigInteger addend) {
|
||
|
int x = intLen;
|
||
|
int y = addend.intLen;
|
||
|
int resultLen = (intLen > addend.intLen ? intLen : addend.intLen);
|
||
|
int[] result = (value.length < resultLen ? new int[resultLen] : value);
|
||
|
|
||
|
int rstart = result.length-1;
|
||
|
long sum = 0;
|
||
|
|
||
|
// Add common parts of both numbers
|
||
|
while(x>0 && y>0) {
|
||
|
x--; y--;
|
||
|
sum = (value[x+offset] & LONG_MASK) +
|
||
|
(addend.value[y+addend.offset] & LONG_MASK) + (sum >>> 32);
|
||
|
result[rstart--] = (int)sum;
|
||
|
}
|
||
|
|
||
|
// Add remainder of the longer number
|
||
|
while(x>0) {
|
||
|
x--;
|
||
|
sum = (value[x+offset] & LONG_MASK) + (sum >>> 32);
|
||
|
result[rstart--] = (int)sum;
|
||
|
}
|
||
|
while(y>0) {
|
||
|
y--;
|
||
|
sum = (addend.value[y+addend.offset] & LONG_MASK) + (sum >>> 32);
|
||
|
result[rstart--] = (int)sum;
|
||
|
}
|
||
|
|
||
|
if ((sum >>> 32) > 0) { // Result must grow in length
|
||
|
resultLen++;
|
||
|
if (result.length < resultLen) {
|
||
|
int temp[] = new int[resultLen];
|
||
|
for (int i=resultLen-1; i>0; i--)
|
||
|
temp[i] = result[i-1];
|
||
|
temp[0] = 1;
|
||
|
result = temp;
|
||
|
} else {
|
||
|
result[rstart--] = 1;
|
||
|
}
|
||
|
}
|
||
|
|
||
|
value = result;
|
||
|
intLen = resultLen;
|
||
|
offset = result.length - resultLen;
|
||
|
}
|
||
|
|
||
|
|
||
|
/**
|
||
|
* Subtracts the smaller of this and b from the larger and places the
|
||
|
* result into this MutableBigInteger.
|
||
|
*/
|
||
|
int subtract(MutableBigInteger b) {
|
||
|
MutableBigInteger a = this;
|
||
|
|
||
|
int[] result = value;
|
||
|
int sign = a.compare(b);
|
||
|
|
||
|
if (sign == 0) {
|
||
|
reset();
|
||
|
return 0;
|
||
|
}
|
||
|
if (sign < 0) {
|
||
|
MutableBigInteger tmp = a;
|
||
|
a = b;
|
||
|
b = tmp;
|
||
|
}
|
||
|
|
||
|
int resultLen = a.intLen;
|
||
|
if (result.length < resultLen)
|
||
|
result = new int[resultLen];
|
||
|
|
||
|
long diff = 0;
|
||
|
int x = a.intLen;
|
||
|
int y = b.intLen;
|
||
|
int rstart = result.length - 1;
|
||
|
|
||
|
// Subtract common parts of both numbers
|
||
|
while (y>0) {
|
||
|
x--; y--;
|
||
|
|
||
|
diff = (a.value[x+a.offset] & LONG_MASK) -
|
||
|
(b.value[y+b.offset] & LONG_MASK) - ((int)-(diff>>32));
|
||
|
result[rstart--] = (int)diff;
|
||
|
}
|
||
|
// Subtract remainder of longer number
|
||
|
while (x>0) {
|
||
|
x--;
|
||
|
diff = (a.value[x+a.offset] & LONG_MASK) - ((int)-(diff>>32));
|
||
|
result[rstart--] = (int)diff;
|
||
|
}
|
||
|
|
||
|
value = result;
|
||
|
intLen = resultLen;
|
||
|
offset = value.length - resultLen;
|
||
|
normalize();
|
||
|
return sign;
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* Subtracts the smaller of a and b from the larger and places the result
|
||
|
* into the larger. Returns 1 if the answer is in a, -1 if in b, 0 if no
|
||
|
* operation was performed.
|
||
|
*/
|
||
|
private int difference(MutableBigInteger b) {
|
||
|
MutableBigInteger a = this;
|
||
|
int sign = a.compare(b);
|
||
|
if (sign ==0)
|
||
|
return 0;
|
||
|
if (sign < 0) {
|
||
|
MutableBigInteger tmp = a;
|
||
|
a = b;
|
||
|
b = tmp;
|
||
|
}
|
||
|
|
||
|
long diff = 0;
|
||
|
int x = a.intLen;
|
||
|
int y = b.intLen;
|
||
|
|
||
|
// Subtract common parts of both numbers
|
||
|
while (y>0) {
|
||
|
x--; y--;
|
||
|
diff = (a.value[a.offset+ x] & LONG_MASK) -
|
||
|
(b.value[b.offset+ y] & LONG_MASK) - ((int)-(diff>>32));
|
||
|
a.value[a.offset+x] = (int)diff;
|
||
|
}
|
||
|
// Subtract remainder of longer number
|
||
|
while (x>0) {
|
||
|
x--;
|
||
|
diff = (a.value[a.offset+ x] & LONG_MASK) - ((int)-(diff>>32));
|
||
|
a.value[a.offset+x] = (int)diff;
|
||
|
}
|
||
|
|
||
|
a.normalize();
|
||
|
return sign;
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* Multiply the contents of two MutableBigInteger objects. The result is
|
||
|
* placed into MutableBigInteger z. The contents of y are not changed.
|
||
|
*/
|
||
|
void multiply(MutableBigInteger y, MutableBigInteger z) {
|
||
|
int xLen = intLen;
|
||
|
int yLen = y.intLen;
|
||
|
int newLen = xLen + yLen;
|
||
|
|
||
|
// Put z into an appropriate state to receive product
|
||
|
if (z.value.length < newLen)
|
||
|
z.value = new int[newLen];
|
||
|
z.offset = 0;
|
||
|
z.intLen = newLen;
|
||
|
|
||
|
// The first iteration is hoisted out of the loop to avoid extra add
|
||
|
long carry = 0;
|
||
|
for (int j=yLen-1, k=yLen+xLen-1; j >= 0; j--, k--) {
|
||
|
long product = (y.value[j+y.offset] & LONG_MASK) *
|
||
|
(value[xLen-1+offset] & LONG_MASK) + carry;
|
||
|
z.value[k] = (int)product;
|
||
|
carry = product >>> 32;
|
||
|
}
|
||
|
z.value[xLen-1] = (int)carry;
|
||
|
|
||
|
// Perform the multiplication word by word
|
||
|
for (int i = xLen-2; i >= 0; i--) {
|
||
|
carry = 0;
|
||
|
for (int j=yLen-1, k=yLen+i; j >= 0; j--, k--) {
|
||
|
long product = (y.value[j+y.offset] & LONG_MASK) *
|
||
|
(value[i+offset] & LONG_MASK) +
|
||
|
(z.value[k] & LONG_MASK) + carry;
|
||
|
z.value[k] = (int)product;
|
||
|
carry = product >>> 32;
|
||
|
}
|
||
|
z.value[i] = (int)carry;
|
||
|
}
|
||
|
|
||
|
// Remove leading zeros from product
|
||
|
z.normalize();
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* Multiply the contents of this MutableBigInteger by the word y. The
|
||
|
* result is placed into z.
|
||
|
*/
|
||
|
void mul(int y, MutableBigInteger z) {
|
||
|
if (y == 1) {
|
||
|
z.copyValue(this);
|
||
|
return;
|
||
|
}
|
||
|
|
||
|
if (y == 0) {
|
||
|
z.clear();
|
||
|
return;
|
||
|
}
|
||
|
|
||
|
// Perform the multiplication word by word
|
||
|
long ylong = y & LONG_MASK;
|
||
|
int[] zval = (z.value.length<intLen+1 ? new int[intLen + 1]
|
||
|
: z.value);
|
||
|
long carry = 0;
|
||
|
for (int i = intLen-1; i >= 0; i--) {
|
||
|
long product = ylong * (value[i+offset] & LONG_MASK) + carry;
|
||
|
zval[i+1] = (int)product;
|
||
|
carry = product >>> 32;
|
||
|
}
|
||
|
|
||
|
if (carry == 0) {
|
||
|
z.offset = 1;
|
||
|
z.intLen = intLen;
|
||
|
} else {
|
||
|
z.offset = 0;
|
||
|
z.intLen = intLen + 1;
|
||
|
zval[0] = (int)carry;
|
||
|
}
|
||
|
z.value = zval;
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* This method is used for division of an n word dividend by a one word
|
||
|
* divisor. The quotient is placed into quotient. The one word divisor is
|
||
|
* specified by divisor. The value of this MutableBigInteger is the
|
||
|
* dividend at invocation but is replaced by the remainder.
|
||
|
*
|
||
|
* NOTE: The value of this MutableBigInteger is modified by this method.
|
||
|
*/
|
||
|
void divideOneWord(int divisor, MutableBigInteger quotient) {
|
||
|
long divLong = divisor & LONG_MASK;
|
||
|
|
||
|
// Special case of one word dividend
|
||
|
if (intLen == 1) {
|
||
|
long remValue = value[offset] & LONG_MASK;
|
||
|
quotient.value[0] = (int) (remValue / divLong);
|
||
|
quotient.intLen = (quotient.value[0] == 0) ? 0 : 1;
|
||
|
quotient.offset = 0;
|
||
|
|
||
|
value[0] = (int) (remValue - (quotient.value[0] * divLong));
|
||
|
offset = 0;
|
||
|
intLen = (value[0] == 0) ? 0 : 1;
|
||
|
|
||
|
return;
|
||
|
}
|
||
|
|
||
|
if (quotient.value.length < intLen)
|
||
|
quotient.value = new int[intLen];
|
||
|
quotient.offset = 0;
|
||
|
quotient.intLen = intLen;
|
||
|
|
||
|
// Normalize the divisor
|
||
|
int shift = 32 - BigInteger.bitLen(divisor);
|
||
|
|
||
|
int rem = value[offset];
|
||
|
long remLong = rem & LONG_MASK;
|
||
|
if (remLong < divLong) {
|
||
|
quotient.value[0] = 0;
|
||
|
} else {
|
||
|
quotient.value[0] = (int)(remLong/divLong);
|
||
|
rem = (int) (remLong - (quotient.value[0] * divLong));
|
||
|
remLong = rem & LONG_MASK;
|
||
|
}
|
||
|
|
||
|
int xlen = intLen;
|
||
|
int[] qWord = new int[2];
|
||
|
while (--xlen > 0) {
|
||
|
long dividendEstimate = (remLong<<32) |
|
||
|
(value[offset + intLen - xlen] & LONG_MASK);
|
||
|
if (dividendEstimate >= 0) {
|
||
|
qWord[0] = (int) (dividendEstimate/divLong);
|
||
|
qWord[1] = (int) (dividendEstimate - (qWord[0] * divLong));
|
||
|
} else {
|
||
|
divWord(qWord, dividendEstimate, divisor);
|
||
|
}
|
||
|
quotient.value[intLen - xlen] = qWord[0];
|
||
|
rem = qWord[1];
|
||
|
remLong = rem & LONG_MASK;
|
||
|
}
|
||
|
|
||
|
// Unnormalize
|
||
|
if (shift > 0)
|
||
|
value[0] = rem %= divisor;
|
||
|
else
|
||
|
value[0] = rem;
|
||
|
intLen = (value[0] == 0) ? 0 : 1;
|
||
|
|
||
|
quotient.normalize();
|
||
|
}
|
||
|
|
||
|
|
||
|
/**
|
||
|
* Calculates the quotient and remainder of this div b and places them
|
||
|
* in the MutableBigInteger objects provided.
|
||
|
*
|
||
|
* Uses Algorithm D in Knuth section 4.3.1.
|
||
|
* Many optimizations to that algorithm have been adapted from the Colin
|
||
|
* Plumb C library.
|
||
|
* It special cases one word divisors for speed.
|
||
|
* The contents of a and b are not changed.
|
||
|
*
|
||
|
*/
|
||
|
void divide(MutableBigInteger b,
|
||
|
MutableBigInteger quotient, MutableBigInteger rem) {
|
||
|
if (b.intLen == 0)
|
||
|
throw new ArithmeticException("BigInteger divide by zero");
|
||
|
|
||
|
// Dividend is zero
|
||
|
if (intLen == 0) {
|
||
|
quotient.intLen = quotient.offset = rem.intLen = rem.offset = 0;
|
||
|
return;
|
||
|
}
|
||
|
|
||
|
int cmp = compare(b);
|
||
|
|
||
|
// Dividend less than divisor
|
||
|
if (cmp < 0) {
|
||
|
quotient.intLen = quotient.offset = 0;
|
||
|
rem.copyValue(this);
|
||
|
return;
|
||
|
}
|
||
|
// Dividend equal to divisor
|
||
|
if (cmp == 0) {
|
||
|
quotient.value[0] = quotient.intLen = 1;
|
||
|
quotient.offset = rem.intLen = rem.offset = 0;
|
||
|
return;
|
||
|
}
|
||
|
|
||
|
quotient.clear();
|
||
|
|
||
|
// Special case one word divisor
|
||
|
if (b.intLen == 1) {
|
||
|
rem.copyValue(this);
|
||
|
rem.divideOneWord(b.value[b.offset], quotient);
|
||
|
return;
|
||
|
}
|
||
|
|
||
|
// Copy divisor value to protect divisor
|
||
|
int[] d = new int[b.intLen];
|
||
|
for(int i=0; i<b.intLen; i++)
|
||
|
d[i] = b.value[b.offset+i];
|
||
|
int dlen = b.intLen;
|
||
|
|
||
|
// Remainder starts as dividend with space for a leading zero
|
||
|
if (rem.value.length < intLen +1)
|
||
|
rem.value = new int[intLen+1];
|
||
|
|
||
|
for (int i=0; i<intLen; i++)
|
||
|
rem.value[i+1] = value[i+offset];
|
||
|
rem.intLen = intLen;
|
||
|
rem.offset = 1;
|
||
|
|
||
|
int nlen = rem.intLen;
|
||
|
|
||
|
// Set the quotient size
|
||
|
int limit = nlen - dlen + 1;
|
||
|
if (quotient.value.length < limit) {
|
||
|
quotient.value = new int[limit];
|
||
|
quotient.offset = 0;
|
||
|
}
|
||
|
quotient.intLen = limit;
|
||
|
int[] q = quotient.value;
|
||
|
|
||
|
// D1 normalize the divisor
|
||
|
int shift = 32 - BigInteger.bitLen(d[0]);
|
||
|
if (shift > 0) {
|
||
|
// First shift will not grow array
|
||
|
BigInteger.primitiveLeftShift(d, dlen, shift);
|
||
|
// But this one might
|
||
|
rem.leftShift(shift);
|
||
|
}
|
||
|
|
||
|
// Must insert leading 0 in rem if its length did not change
|
||
|
if (rem.intLen == nlen) {
|
||
|
rem.offset = 0;
|
||
|
rem.value[0] = 0;
|
||
|
rem.intLen++;
|
||
|
}
|
||
|
|
||
|
int dh = d[0];
|
||
|
long dhLong = dh & LONG_MASK;
|
||
|
int dl = d[1];
|
||
|
int[] qWord = new int[2];
|
||
|
|
||
|
// D2 Initialize j
|
||
|
for(int j=0; j<limit; j++) {
|
||
|
// D3 Calculate qhat
|
||
|
// estimate qhat
|
||
|
int qhat = 0;
|
||
|
int qrem = 0;
|
||
|
boolean skipCorrection = false;
|
||
|
int nh = rem.value[j+rem.offset];
|
||
|
int nh2 = nh + 0x80000000;
|
||
|
int nm = rem.value[j+1+rem.offset];
|
||
|
|
||
|
if (nh == dh) {
|
||
|
qhat = ~0;
|
||
|
qrem = nh + nm;
|
||
|
skipCorrection = qrem + 0x80000000 < nh2;
|
||
|
} else {
|
||
|
long nChunk = (((long)nh) << 32) | (nm & LONG_MASK);
|
||
|
if (nChunk >= 0) {
|
||
|
qhat = (int) (nChunk / dhLong);
|
||
|
qrem = (int) (nChunk - (qhat * dhLong));
|
||
|
} else {
|
||
|
divWord(qWord, nChunk, dh);
|
||
|
qhat = qWord[0];
|
||
|
qrem = qWord[1];
|
||
|
}
|
||
|
}
|
||
|
|
||
|
if (qhat == 0)
|
||
|
continue;
|
||
|
|
||
|
if (!skipCorrection) { // Correct qhat
|
||
|
long nl = rem.value[j+2+rem.offset] & LONG_MASK;
|
||
|
long rs = ((qrem & LONG_MASK) << 32) | nl;
|
||
|
long estProduct = (dl & LONG_MASK) * (qhat & LONG_MASK);
|
||
|
|
||
|
if (unsignedLongCompare(estProduct, rs)) {
|
||
|
qhat--;
|
||
|
qrem = (int)((qrem & LONG_MASK) + dhLong);
|
||
|
if ((qrem & LONG_MASK) >= dhLong) {
|
||
|
estProduct = (dl & LONG_MASK) * (qhat & LONG_MASK);
|
||
|
rs = ((qrem & LONG_MASK) << 32) | nl;
|
||
|
if (unsignedLongCompare(estProduct, rs))
|
||
|
qhat--;
|
||
|
}
|
||
|
}
|
||
|
}
|
||
|
|
||
|
// D4 Multiply and subtract
|
||
|
rem.value[j+rem.offset] = 0;
|
||
|
int borrow = mulsub(rem.value, d, qhat, dlen, j+rem.offset);
|
||
|
|
||
|
// D5 Test remainder
|
||
|
if (borrow + 0x80000000 > nh2) {
|
||
|
// D6 Add back
|
||
|
divadd(d, rem.value, j+1+rem.offset);
|
||
|
qhat--;
|
||
|
}
|
||
|
|
||
|
// Store the quotient digit
|
||
|
q[j] = qhat;
|
||
|
} // D7 loop on j
|
||
|
|
||
|
// D8 Unnormalize
|
||
|
if (shift > 0)
|
||
|
rem.rightShift(shift);
|
||
|
|
||
|
rem.normalize();
|
||
|
quotient.normalize();
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* Compare two longs as if they were unsigned.
|
||
|
* Returns true iff one is bigger than two.
|
||
|
*/
|
||
|
private boolean unsignedLongCompare(long one, long two) {
|
||
|
return (one+Long.MIN_VALUE) > (two+Long.MIN_VALUE);
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* This method divides a long quantity by an int to estimate
|
||
|
* qhat for two multi precision numbers. It is used when
|
||
|
* the signed value of n is less than zero.
|
||
|
*/
|
||
|
private void divWord(int[] result, long n, int d) {
|
||
|
long dLong = d & LONG_MASK;
|
||
|
|
||
|
if (dLong == 1) {
|
||
|
result[0] = (int)n;
|
||
|
result[1] = 0;
|
||
|
return;
|
||
|
}
|
||
|
|
||
|
// Approximate the quotient and remainder
|
||
|
long q = (n >>> 1) / (dLong >>> 1);
|
||
|
long r = n - q*dLong;
|
||
|
|
||
|
// Correct the approximation
|
||
|
while (r < 0) {
|
||
|
r += dLong;
|
||
|
q--;
|
||
|
}
|
||
|
while (r >= dLong) {
|
||
|
r -= dLong;
|
||
|
q++;
|
||
|
}
|
||
|
|
||
|
// n - q*dlong == r && 0 <= r <dLong, hence we're done.
|
||
|
result[0] = (int)q;
|
||
|
result[1] = (int)r;
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* Calculate GCD of this and b. This and b are changed by the computation.
|
||
|
*/
|
||
|
MutableBigInteger hybridGCD(MutableBigInteger b) {
|
||
|
// Use Euclid's algorithm until the numbers are approximately the
|
||
|
// same length, then use the binary GCD algorithm to find the GCD.
|
||
|
MutableBigInteger a = this;
|
||
|
MutableBigInteger q = new MutableBigInteger(),
|
||
|
r = new MutableBigInteger();
|
||
|
|
||
|
while (b.intLen != 0) {
|
||
|
if (Math.abs(a.intLen - b.intLen) < 2)
|
||
|
return a.binaryGCD(b);
|
||
|
|
||
|
a.divide(b, q, r);
|
||
|
MutableBigInteger swapper = a;
|
||
|
a = b; b = r; r = swapper;
|
||
|
}
|
||
|
return a;
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* Calculate GCD of this and v.
|
||
|
* Assumes that this and v are not zero.
|
||
|
*/
|
||
|
private MutableBigInteger binaryGCD(MutableBigInteger v) {
|
||
|
// Algorithm B from Knuth section 4.5.2
|
||
|
MutableBigInteger u = this;
|
||
|
MutableBigInteger r = new MutableBigInteger();
|
||
|
|
||
|
// step B1
|
||
|
int s1 = u.getLowestSetBit();
|
||
|
int s2 = v.getLowestSetBit();
|
||
|
int k = (s1 < s2) ? s1 : s2;
|
||
|
if (k != 0) {
|
||
|
u.rightShift(k);
|
||
|
v.rightShift(k);
|
||
|
}
|
||
|
|
||
|
// step B2
|
||
|
boolean uOdd = (k==s1);
|
||
|
MutableBigInteger t = uOdd ? v: u;
|
||
|
int tsign = uOdd ? -1 : 1;
|
||
|
|
||
|
int lb;
|
||
|
while ((lb = t.getLowestSetBit()) >= 0) {
|
||
|
// steps B3 and B4
|
||
|
t.rightShift(lb);
|
||
|
// step B5
|
||
|
if (tsign > 0)
|
||
|
u = t;
|
||
|
else
|
||
|
v = t;
|
||
|
|
||
|
// Special case one word numbers
|
||
|
if (u.intLen < 2 && v.intLen < 2) {
|
||
|
int x = u.value[u.offset];
|
||
|
int y = v.value[v.offset];
|
||
|
x = binaryGcd(x, y);
|
||
|
r.value[0] = x;
|
||
|
r.intLen = 1;
|
||
|
r.offset = 0;
|
||
|
if (k > 0)
|
||
|
r.leftShift(k);
|
||
|
return r;
|
||
|
}
|
||
|
|
||
|
// step B6
|
||
|
if ((tsign = u.difference(v)) == 0)
|
||
|
break;
|
||
|
t = (tsign >= 0) ? u : v;
|
||
|
}
|
||
|
|
||
|
if (k > 0)
|
||
|
u.leftShift(k);
|
||
|
return u;
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* Calculate GCD of a and b interpreted as unsigned integers.
|
||
|
*/
|
||
|
static int binaryGcd(int a, int b) {
|
||
|
if (b==0)
|
||
|
return a;
|
||
|
if (a==0)
|
||
|
return b;
|
||
|
|
||
|
int x;
|
||
|
int aZeros = 0;
|
||
|
while ((x = a & 0xff) == 0) {
|
||
|
a >>>= 8;
|
||
|
aZeros += 8;
|
||
|
}
|
||
|
int y = BigInteger.trailingZeroTable[x];
|
||
|
aZeros += y;
|
||
|
a >>>= y;
|
||
|
|
||
|
int bZeros = 0;
|
||
|
while ((x = b & 0xff) == 0) {
|
||
|
b >>>= 8;
|
||
|
bZeros += 8;
|
||
|
}
|
||
|
y = BigInteger.trailingZeroTable[x];
|
||
|
bZeros += y;
|
||
|
b >>>= y;
|
||
|
|
||
|
int t = (aZeros < bZeros ? aZeros : bZeros);
|
||
|
|
||
|
while (a != b) {
|
||
|
if ((a+0x80000000) > (b+0x80000000)) { // a > b as unsigned
|
||
|
a -= b;
|
||
|
|
||
|
while ((x = a & 0xff) == 0)
|
||
|
a >>>= 8;
|
||
|
a >>>= BigInteger.trailingZeroTable[x];
|
||
|
} else {
|
||
|
b -= a;
|
||
|
|
||
|
while ((x = b & 0xff) == 0)
|
||
|
b >>>= 8;
|
||
|
b >>>= BigInteger.trailingZeroTable[x];
|
||
|
}
|
||
|
}
|
||
|
return a<<t;
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* Returns the modInverse of this mod p. This and p are not affected by
|
||
|
* the operation.
|
||
|
*/
|
||
|
MutableBigInteger mutableModInverse(MutableBigInteger p) {
|
||
|
// Modulus is odd, use Schroeppel's algorithm
|
||
|
if (p.isOdd())
|
||
|
return modInverse(p);
|
||
|
|
||
|
// Base and modulus are even, throw exception
|
||
|
if (isEven())
|
||
|
throw new ArithmeticException("BigInteger not invertible.");
|
||
|
|
||
|
// Get even part of modulus expressed as a power of 2
|
||
|
int powersOf2 = p.getLowestSetBit();
|
||
|
|
||
|
// Construct odd part of modulus
|
||
|
MutableBigInteger oddMod = new MutableBigInteger(p);
|
||
|
oddMod.rightShift(powersOf2);
|
||
|
|
||
|
if (oddMod.isOne())
|
||
|
return modInverseMP2(powersOf2);
|
||
|
|
||
|
// Calculate 1/a mod oddMod
|
||
|
MutableBigInteger oddPart = modInverse(oddMod);
|
||
|
|
||
|
// Calculate 1/a mod evenMod
|
||
|
MutableBigInteger evenPart = modInverseMP2(powersOf2);
|
||
|
|
||
|
// Combine the results using Chinese Remainder Theorem
|
||
|
MutableBigInteger y1 = modInverseBP2(oddMod, powersOf2);
|
||
|
MutableBigInteger y2 = oddMod.modInverseMP2(powersOf2);
|
||
|
|
||
|
MutableBigInteger temp1 = new MutableBigInteger();
|
||
|
MutableBigInteger temp2 = new MutableBigInteger();
|
||
|
MutableBigInteger result = new MutableBigInteger();
|
||
|
|
||
|
oddPart.leftShift(powersOf2);
|
||
|
oddPart.multiply(y1, result);
|
||
|
|
||
|
evenPart.multiply(oddMod, temp1);
|
||
|
temp1.multiply(y2, temp2);
|
||
|
|
||
|
result.add(temp2);
|
||
|
result.divide(p, temp1, temp2);
|
||
|
return temp2;
|
||
|
}
|
||
|
|
||
|
/*
|
||
|
* Calculate the multiplicative inverse of this mod 2^k.
|
||
|
*/
|
||
|
MutableBigInteger modInverseMP2(int k) {
|
||
|
if (isEven())
|
||
|
throw new ArithmeticException("Non-invertible. (GCD != 1)");
|
||
|
|
||
|
if (k > 64)
|
||
|
return euclidModInverse(k);
|
||
|
|
||
|
int t = inverseMod32(value[offset+intLen-1]);
|
||
|
|
||
|
if (k < 33) {
|
||
|
t = (k == 32 ? t : t & ((1 << k) - 1));
|
||
|
return new MutableBigInteger(t);
|
||
|
}
|
||
|
|
||
|
long pLong = (value[offset+intLen-1] & LONG_MASK);
|
||
|
if (intLen > 1)
|
||
|
pLong |= ((long)value[offset+intLen-2] << 32);
|
||
|
long tLong = t & LONG_MASK;
|
||
|
tLong = tLong * (2 - pLong * tLong); // 1 more Newton iter step
|
||
|
tLong = (k == 64 ? tLong : tLong & ((1L << k) - 1));
|
||
|
|
||
|
MutableBigInteger result = new MutableBigInteger(new int[2]);
|
||
|
result.value[0] = (int)(tLong >>> 32);
|
||
|
result.value[1] = (int)tLong;
|
||
|
result.intLen = 2;
|
||
|
result.normalize();
|
||
|
return result;
|
||
|
}
|
||
|
|
||
|
/*
|
||
|
* Returns the multiplicative inverse of val mod 2^32. Assumes val is odd.
|
||
|
*/
|
||
|
static int inverseMod32(int val) {
|
||
|
// Newton's iteration!
|
||
|
int t = val;
|
||
|
t *= 2 - val*t;
|
||
|
t *= 2 - val*t;
|
||
|
t *= 2 - val*t;
|
||
|
t *= 2 - val*t;
|
||
|
return t;
|
||
|
}
|
||
|
|
||
|
/*
|
||
|
* Calculate the multiplicative inverse of 2^k mod mod, where mod is odd.
|
||
|
*/
|
||
|
static MutableBigInteger modInverseBP2(MutableBigInteger mod, int k) {
|
||
|
// Copy the mod to protect original
|
||
|
return fixup(new MutableBigInteger(1), new MutableBigInteger(mod), k);
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* Calculate the multiplicative inverse of this mod mod, where mod is odd.
|
||
|
* This and mod are not changed by the calculation.
|
||
|
*
|
||
|
* This method implements an algorithm due to Richard Schroeppel, that uses
|
||
|
* the same intermediate representation as Montgomery Reduction
|
||
|
* ("Montgomery Form"). The algorithm is described in an unpublished
|
||
|
* manuscript entitled "Fast Modular Reciprocals."
|
||
|
*/
|
||
|
private MutableBigInteger modInverse(MutableBigInteger mod) {
|
||
|
MutableBigInteger p = new MutableBigInteger(mod);
|
||
|
MutableBigInteger f = new MutableBigInteger(this);
|
||
|
MutableBigInteger g = new MutableBigInteger(p);
|
||
|
SignedMutableBigInteger c = new SignedMutableBigInteger(1);
|
||
|
SignedMutableBigInteger d = new SignedMutableBigInteger();
|
||
|
MutableBigInteger temp = null;
|
||
|
SignedMutableBigInteger sTemp = null;
|
||
|
|
||
|
int k = 0;
|
||
|
// Right shift f k times until odd, left shift d k times
|
||
|
if (f.isEven()) {
|
||
|
int trailingZeros = f.getLowestSetBit();
|
||
|
f.rightShift(trailingZeros);
|
||
|
d.leftShift(trailingZeros);
|
||
|
k = trailingZeros;
|
||
|
}
|
||
|
|
||
|
// The Almost Inverse Algorithm
|
||
|
while(!f.isOne()) {
|
||
|
// If gcd(f, g) != 1, number is not invertible modulo mod
|
||
|
if (f.isZero())
|
||
|
throw new ArithmeticException("BigInteger not invertible.");
|
||
|
|
||
|
// If f < g exchange f, g and c, d
|
||
|
if (f.compare(g) < 0) {
|
||
|
temp = f; f = g; g = temp;
|
||
|
sTemp = d; d = c; c = sTemp;
|
||
|
}
|
||
|
|
||
|
// If f == g (mod 4)
|
||
|
if (((f.value[f.offset + f.intLen - 1] ^
|
||
|
g.value[g.offset + g.intLen - 1]) & 3) == 0) {
|
||
|
f.subtract(g);
|
||
|
c.signedSubtract(d);
|
||
|
} else { // If f != g (mod 4)
|
||
|
f.add(g);
|
||
|
c.signedAdd(d);
|
||
|
}
|
||
|
|
||
|
// Right shift f k times until odd, left shift d k times
|
||
|
int trailingZeros = f.getLowestSetBit();
|
||
|
f.rightShift(trailingZeros);
|
||
|
d.leftShift(trailingZeros);
|
||
|
k += trailingZeros;
|
||
|
}
|
||
|
|
||
|
while (c.sign < 0)
|
||
|
c.signedAdd(p);
|
||
|
|
||
|
return fixup(c, p, k);
|
||
|
}
|
||
|
|
||
|
/*
|
||
|
* The Fixup Algorithm
|
||
|
* Calculates X such that X = C * 2^(-k) (mod P)
|
||
|
* Assumes C<P and P is odd.
|
||
|
*/
|
||
|
static MutableBigInteger fixup(MutableBigInteger c, MutableBigInteger p,
|
||
|
int k) {
|
||
|
MutableBigInteger temp = new MutableBigInteger();
|
||
|
// Set r to the multiplicative inverse of p mod 2^32
|
||
|
int r = -inverseMod32(p.value[p.offset+p.intLen-1]);
|
||
|
|
||
|
for(int i=0, numWords = k >> 5; i<numWords; i++) {
|
||
|
// V = R * c (mod 2^j)
|
||
|
int v = r * c.value[c.offset + c.intLen-1];
|
||
|
// c = c + (v * p)
|
||
|
p.mul(v, temp);
|
||
|
c.add(temp);
|
||
|
// c = c / 2^j
|
||
|
c.intLen--;
|
||
|
}
|
||
|
int numBits = k & 0x1f;
|
||
|
if (numBits != 0) {
|
||
|
// V = R * c (mod 2^j)
|
||
|
int v = r * c.value[c.offset + c.intLen-1];
|
||
|
v &= ((1<<numBits) - 1);
|
||
|
// c = c + (v * p)
|
||
|
p.mul(v, temp);
|
||
|
c.add(temp);
|
||
|
// c = c / 2^j
|
||
|
c.rightShift(numBits);
|
||
|
}
|
||
|
|
||
|
// In theory, c may be greater than p at this point (Very rare!)
|
||
|
while (c.compare(p) >= 0)
|
||
|
c.subtract(p);
|
||
|
|
||
|
return c;
|
||
|
}
|
||
|
|
||
|
/**
|
||
|
* Uses the extended Euclidean algorithm to compute the modInverse of base
|
||
|
* mod a modulus that is a power of 2. The modulus is 2^k.
|
||
|
*/
|
||
|
MutableBigInteger euclidModInverse(int k) {
|
||
|
MutableBigInteger b = new MutableBigInteger(1);
|
||
|
b.leftShift(k);
|
||
|
MutableBigInteger mod = new MutableBigInteger(b);
|
||
|
|
||
|
MutableBigInteger a = new MutableBigInteger(this);
|
||
|
MutableBigInteger q = new MutableBigInteger();
|
||
|
MutableBigInteger r = new MutableBigInteger();
|
||
|
|
||
|
b.divide(a, q, r);
|
||
|
MutableBigInteger swapper = b; b = r; r = swapper;
|
||
|
|
||
|
MutableBigInteger t1 = new MutableBigInteger(q);
|
||
|
MutableBigInteger t0 = new MutableBigInteger(1);
|
||
|
MutableBigInteger temp = new MutableBigInteger();
|
||
|
|
||
|
while (!b.isOne()) {
|
||
|
a.divide(b, q, r);
|
||
|
|
||
|
if (r.intLen == 0)
|
||
|
throw new ArithmeticException("BigInteger not invertible.");
|
||
|
|
||
|
swapper = r; r = a; a = swapper;
|
||
|
|
||
|
if (q.intLen == 1)
|
||
|
t1.mul(q.value[q.offset], temp);
|
||
|
else
|
||
|
q.multiply(t1, temp);
|
||
|
swapper = q; q = temp; temp = swapper;
|
||
|
|
||
|
t0.add(q);
|
||
|
|
||
|
if (a.isOne())
|
||
|
return t0;
|
||
|
|
||
|
b.divide(a, q, r);
|
||
|
|
||
|
if (r.intLen == 0)
|
||
|
throw new ArithmeticException("BigInteger not invertible.");
|
||
|
|
||
|
swapper = b; b = r; r = swapper;
|
||
|
|
||
|
if (q.intLen == 1)
|
||
|
t0.mul(q.value[q.offset], temp);
|
||
|
else
|
||
|
q.multiply(t0, temp);
|
||
|
swapper = q; q = temp; temp = swapper;
|
||
|
|
||
|
t1.add(q);
|
||
|
}
|
||
|
mod.subtract(t1);
|
||
|
return mod;
|
||
|
}
|
||
|
|
||
|
}
|