/*-
 * Copyright (c) 1992, 1993
 *	The Regents of the University of California.  All rights reserved.
 *
 * This software was developed by the Computer Systems Engineering group
 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
 * contributed to Berkeley.
 *
 * Redistribution and use in source and binary forms, with or without
 * modification, are permitted provided that the following conditions
 * are met:
 * 1. Redistributions of source code must retain the above copyright
 *    notice, this list of conditions and the following disclaimer.
 * 2. Redistributions in binary form must reproduce the above copyright
 *    notice, this list of conditions and the following disclaimer in the
 *    documentation and/or other materials provided with the distribution.
 * 3. Neither the name of the University nor the names of its contributors
 *    may be used to endorse or promote products derived from this software
 *    without specific prior written permission.
 *
 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
 * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
 * SUCH DAMAGE.
 *
 * From:
 * 	@(#)qdivrem.c	8.1 (Berkeley) 6/4/93
 *	NetBSD: qdivrem.c,v 1.1 2005/12/20 19:28:51 christos Exp
 */

/*
 * Multiprecision divide.  This algorithm is from Knuth vol. 2 (2nd ed),
 * section 4.3.1, pp. 257--259.
 */

#include "longlong.h"

#define	B	((int)1 << HALF_BITS)	/* digit base */

/* Combine two `digits' to make a single two-digit number. */
#define	COMBINE(a, b) (((unsigned int)(a) << HALF_BITS) | (b))

/* select a type for digits in base B: use unsigned short if they fit */
#if UINT_MAX == 0xffffffffU && USHRT_MAX >= 0xffff
typedef unsigned short digit;
#else
typedef unsigned int digit;
#endif

static void shl(digit *p, int len, int sh);

/*
 * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
 *
 * We do this in base 2-sup-HALF_BITS, so that all intermediate
 * products fit within unsigned int.  As a consequence, the maximum
 * length dividend and divisor are 4 `digits' in this base (they are
 * shorter if they have leading zeros).
 */
unsigned long long
__qdivrem(unsigned long long ull, unsigned long long vll,
	  unsigned long long *arq)
{
	union uu tmp;
	digit *u, *v, *q;
	digit v1, v2;
	unsigned int qhat, rhat, t;
	int m, n, d, j, i;
	digit uspace[5], vspace[5], qspace[5];

	/*
	 * Take care of special cases: divide by zero, and u < v.
	 */
	if (vll == 0) {
		/* divide by zero. */
		static volatile const unsigned int zero = 0;

		tmp.ui[H] = tmp.ui[L] = 1 / zero;
		if (arq)
			*arq = ull;
		return (tmp.ll);
	}
	if (ull < vll) {
		if (arq)
			*arq = ull;
		return (0);
	}
	u = &uspace[0];
	v = &vspace[0];
	q = &qspace[0];

	/*
	 * Break dividend and divisor into digits in base B, then
	 * count leading zeros to determine m and n.  When done, we
	 * will have:
	 *	u = (u[1]u[2]...u[m+n]) sub B
	 *	v = (v[1]v[2]...v[n]) sub B
	 *	v[1] != 0
	 *	1 < n <= 4 (if n = 1, we use a different division algorithm)
	 *	m >= 0 (otherwise u < v, which we already checked)
	 *	m + n = 4
	 * and thus
	 *	m = 4 - n <= 2
	 */
	tmp.ull = ull;
	u[0] = 0;
	u[1] = (digit)HHALF(tmp.ui[H]);
	u[2] = (digit)LHALF(tmp.ui[H]);
	u[3] = (digit)HHALF(tmp.ui[L]);
	u[4] = (digit)LHALF(tmp.ui[L]);
	tmp.ull = vll;
	v[1] = (digit)HHALF(tmp.ui[H]);
	v[2] = (digit)LHALF(tmp.ui[H]);
	v[3] = (digit)HHALF(tmp.ui[L]);
	v[4] = (digit)LHALF(tmp.ui[L]);
	for (n = 4; v[1] == 0; v++) {
		if (--n == 1) {
			unsigned int rbj;  /* r*B+u[j] (not root boy jim) */
			digit q1, q2, q3, q4;

			/*
			 * Change of plan, per exercise 16.
			 *	r = 0;
			 *	for j = 1..4:
			 *		q[j] = floor((r*B + u[j]) / v),
			 *		r = (r*B + u[j]) % v;
			 * We unroll this completely here.
			 */
			t = v[2];	/* nonzero, by definition */
			q1 = (digit)(u[1] / t);
			rbj = COMBINE(u[1] % t, u[2]);
			q2 = (digit)(rbj / t);
			rbj = COMBINE(rbj % t, u[3]);
			q3 = (digit)(rbj / t);
			rbj = COMBINE(rbj % t, u[4]);
			q4 = (digit)(rbj / t);
			if (arq)
				*arq = rbj % t;
			tmp.ui[H] = COMBINE(q1, q2);
			tmp.ui[L] = COMBINE(q3, q4);
			return (tmp.ll);
		}
	}

	/*
	 * By adjusting q once we determine m, we can guarantee that
	 * there is a complete four-digit quotient at &qspace[1] when
	 * we finally stop.
	 */
	for (m = 4 - n; u[1] == 0; u++)
		m--;
	for (i = 4 - m; --i >= 0;)
		q[i] = 0;
	q += 4 - m;

	/*
	 * Here we run Program D, translated from MIX to C and acquiring
	 * a few minor changes.
	 *
	 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
	 */
	d = 0;
	for (t = v[1]; t < B / 2; t <<= 1)
		d++;
	if (d > 0) {
		shl(&u[0], m + n, d);		/* u <<= d */
		shl(&v[1], n - 1, d);		/* v <<= d */
	}
	/*
	 * D2: j = 0.
	 */
	j = 0;
	v1 = v[1];	/* for D3 -- note that v[1..n] are constant */
	v2 = v[2];	/* for D3 */
	do {
		digit uj0, uj1, uj2;

		/*
		 * D3: Calculate qhat (\^q, in TeX notation).
		 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
		 * let rhat = (u[j]*B + u[j+1]) mod v[1].
		 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
		 * decrement qhat and increase rhat correspondingly.
		 * Note that if rhat >= B, v[2]*qhat < rhat*B.
		 */
		uj0 = u[j + 0];	/* for D3 only -- note that u[j+...] change */
		uj1 = u[j + 1];	/* for D3 only */
		uj2 = u[j + 2];	/* for D3 only */
		if (uj0 == v1) {
			qhat = B;
			rhat = uj1;
			goto qhat_too_big;
		} else {
			unsigned int nn = COMBINE(uj0, uj1);
			qhat = nn / v1;
			rhat = nn % v1;
		}
		while (v2 * qhat > COMBINE(rhat, uj2)) {
	qhat_too_big:
			qhat--;
			if ((rhat += v1) >= B)
				break;
		}
		/*
		 * D4: Multiply and subtract.
		 * The variable `t' holds any borrows across the loop.
		 * We split this up so that we do not require v[0] = 0,
		 * and to eliminate a final special case.
		 */
		for (t = 0, i = n; i > 0; i--) {
			t = u[i + j] - v[i] * qhat - t;
			u[i + j] = (digit)LHALF(t);
			t = (B - HHALF(t)) & (B - 1);
		}
		t = u[j] - t;
		u[j] = (digit)LHALF(t);
		/*
		 * D5: test remainder.
		 * There is a borrow if and only if HHALF(t) is nonzero;
		 * in that (rare) case, qhat was too large (by exactly 1).
		 * Fix it by adding v[1..n] to u[j..j+n].
		 */
		if (HHALF(t)) {
			qhat--;
			for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
				t += u[i + j] + v[i];
				u[i + j] = (digit)LHALF(t);
				t = HHALF(t);
			}
			u[j] = (digit)LHALF(u[j] + t);
		}
		q[j] = (digit)qhat;
	} while (++j <= m);		/* D7: loop on j. */

	/*
	 * If caller wants the remainder, we have to calculate it as
	 * u[m..m+n] >> d (this is at most n digits and thus fits in
	 * u[m+1..m+n], but we may need more source digits).
	 */
	if (arq) {
		if (d) {
			for (i = m + n; i > m; --i)
				u[i] = (digit)(((unsigned int)u[i] >> d) |
				    LHALF((unsigned int)u[i - 1] <<
					  (HALF_BITS - d)));
			u[i] = 0;
		}
		tmp.ui[H] = COMBINE(uspace[1], uspace[2]);
		tmp.ui[L] = COMBINE(uspace[3], uspace[4]);
		*arq = tmp.ll;
	}

	tmp.ui[H] = COMBINE(qspace[1], qspace[2]);
	tmp.ui[L] = COMBINE(qspace[3], qspace[4]);
	return (tmp.ll);
}

/*
 * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
 * `fall out' the left (there never will be any such anyway).
 * We may assume len >= 0.  NOTE THAT THIS WRITES len+1 DIGITS.
 */
static void
shl(digit *p, int len, int sh)
{
	int i;

	for (i = 0; i < len; i++)
		p[i] = (digit)(LHALF((unsigned int)p[i] << sh) |
		    ((unsigned int)p[i + 1] >> (HALF_BITS - sh)));
	p[i] = (digit)(LHALF((unsigned int)p[i] << sh));
}