操作数大小大致相同的大整数除法
Big integer division with operands aproximately of the same size
我正在尝试实现一个函数，该函数计算两个大小大致相同的数字之间的除法。我使用unsigned int的数组来存储每个数字的值(参见下面的类结构)。我想实现一个有效的除法函数当两者的unsigned int个数之差不大于1时。我研究过长除法算法(如维基百科所述)，但它太慢了。有没有有效的算法来处理这种特殊情况?目前，我正在处理最多100位的数字。
class BigInt {
public:
short sign; //sign of the number
unsigned int size; //number of unsigned int's needed to store the number
unsigned int* values; //values of the number
...
}
谢谢
我相信Knuth算法D (The Art of Computer Programming Vol.2, Section 4.3.1)的修改版本应该可以在线性时间最坏情况下，以恒定的平均时间处理随机输入。重要的简化源于只需要初始迭代产生的第一个商字。
下面是我无耻地改编Hacker’s Delight的通用实现的尝试。该实现假设32位字和64位中间/除法，但可以在更广泛的算术可用的地方自然扩展。
我确实担心这个函数实际上没有经过测试，所以请把它看作是一个想法的萌芽，而不是一个完全成熟的实现。老实说，我甚至不记得的微妙之处为什么算法有效。
// A helper function for finding the position of the most significant set bit.
// Please plug in your intrinsic of choice here
static unsigned int find_first_bit(uint32_t value) {
# ifdef _MSC_VER
unsigned long index;
(void) _BitScanReverse(&index, value);
return index + 1;
# else
unsigned int count = 0;
for(count = 0; value; ++count)
value >>= 1;
return count;
# endif
}
// Multi-word division in 32-bit big-endian segments, returning the most significant
// word of the quotient.
uint32_t divide(const uint32_t *num, const uint32_t *den, size_t len) {
// Skip past leading zero denominator digits. The quotient must fit in a single
// 32-bit word and so only the preceeding numerator digit needs be examined
uint32_t norm_den;
uint64_t norm_num = 0;
size_t top = 0;
while(norm_den = den[top], !norm_den)
norm_num = num[top++];
// Please pad the input to insure at least three denominator words counting from
// the first non-zero digit
assert(len >= top + 3);
// Divide the first two digits of the numerator by the leading digit of the
// denominator as an initial quotient digit guess, yielding an upper bound
// for the quotient at most two steps above the true value.
// Simultaneously normalize the denominator with the MSB in the 31st bit.
unsigned int norm = find_first_bit(norm_den);
norm_num = norm_num << (64 - norm);
norm_num |= ((uint64_t) num[top + 0] << 32 | num[top + 1]) >> norm;
norm_den = ((uint64_t) norm_den << 32 | den[top + 1]) >> norm;
// We are using a straight 64/64 division where 64/32=32 would suffice. The latter
// is available on e.g. x86-32 but difficult to generate short of assembly code.
uint32_t quot = (uint32_t) (norm_num / norm_den);
// Substitute norm_num - quot * norm_den if your optimizer is too thick-headed to
// efficiently extract the remainder
uint32_t rem = norm_num % norm_den;
// Test the next word of the input, reducing the upper bound to within one step
// of the true quotient. See Knuth for proofs of this reduction and the bounds
// of the first guess
norm_num = ((uint64_t) num[top + 1] << 32 | num[top + 2]) >> norm;
norm_num = (uint64_t) rem << 32 | (uint32_t) norm_num;
norm_den = ((uint64_t) den[top + 1] << 32 | den[top + 2]) >> norm;
if((uint64_t) quot * norm_den > norm_num) {
--quot;
// There is no "add-back" step try to avoid and so there is little point
// in looping to refine the guess further since the bound is sufficiently
// tight already
}
// Compare quotient guess multiplied by the denominator to the numerator
// across whole numbers to account for the final quotient step.
// There is no need to bother with normalization here. Furthermore we can
// compare from the most to the least significant and cut off early when the
// intermediate result becomes large, yielding a constant average running
// time for random input
uint64_t accum = 0;
do {
uint64_t prod = (uint64_t) quot * *den++;
accum = accum << 32 | *num++;
// A negative partial result can never recover, so pick the lower
// quotient. A separate test is required to avoid 65-bit arithmetic
if(accum < prod)
return --quot;
accum -= prod;
// Similarly a partial result which spills into the upper 32-bits can't
// recover either, so go with the upper quotient
if((uint64_t) accum >= 0x100000000)
return quot;
} while(--len);
return quot;
}
假设您想要一个整数结果，您应该能够相当快地执行此操作。
最简单的方法可能是用十进制来表示，以稍微小一点的数字开始。
让我们考虑一些较小的数字，如:123456789/54321876。其中一个比另一个多一位，所以最大的可能结果是在10左右。
如果我们把两边同时除，那么(比如说)大数剩下三位数，小数剩下两位数，我们得到123/54，它产生的整数结果与我们在整个过程中做了完全精确的工作一样。
在你的例子中，你可以做几乎相同的事情，除了不使用十进制数字，你将在较大的项目中保留三个精确的单词，在较小的项目中保留两个精确的单词(实际上，即使这是保守的-分别使用2个和1个单词可能就足够了，但我现在懒得证明这一点，如果你这样做，你可能需要更小心舍入)。
所以，是的，你的想法，这是一种特殊的情况下，你可以节省工作基于数字几乎相同的大小是正确的。
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