i具有一个数字数字的向量,向量代表基本2^32的系统中的大整数。例如:
vector <unsigned> vec = {453860625, 469837947, 3503557200, 40}
此向量表示这个大整数:
base = 2 ^ 32
3233755723588593872632005090577 = 40 * base ^ 3 + 3503557200 * base ^ 2 + 469837947 * base + 453860625
如何在字符串中获得此小数点表示?
这是一种做您想做的事情的低效方法,从代表任意大小的整数的单词值的向量中获取小数字符串。
我宁愿将其作为类实现,以更好地封装,因此可以添加数学运算符,但是为了更好地遵守问题,这只是用于操纵std::vector<unsigned>
对象的一堆免费功能。这确实使用Typedef BiType
作为std::vector<unsigned>
的别名。
进行二进制部门的功能构成了大部分此代码。它的大部分内容复制了std::bitset
可以做的事情,但对于任意大小的位,作为unsigned
单词的向量。如果您想提高效率,请插入执行人均操作而不是每位操作的部门算法。此外,当仅用于除以10的情况时,分区代码是通用的,因此您可以用特殊用途的分区代码替换。
代码通常假定unsigned
单词的向量,并且基础是最大unsigned
值,另外一个值。我在较小的基地或不是2个权力的较小基地或基地出错的地方留下了评论
另外,我仅测试了1个情况,即您在OP中提供的情况 - 这是新的未验证代码,因此您可能需要进行更多测试。如果您发现问题案例,我会很乐意在此处修复错误。
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
namespace bigint {
using BiType = std::vector<unsigned>;
// cmp compares a with b, returning 1:a>b, 0:a==b, -1:a<b
int cmp(const BiType& a, const BiType& b) {
const auto max_size = std::max(a.size(), b.size());
for(auto i=max_size-1; i+1; --i) {
const auto wa = i < a.size() ? a[i] : 0;
const auto wb = i < b.size() ? b[i] : 0;
if(wa != wb) { return wa > wb ? 1 : -1; }
}
return 0;
}
bool is_zero(BiType& bi) {
for(auto w : bi) { if(w) return false; }
return true;
}
// canonize removes leading zero words
void canonize(BiType& bi) {
const auto size = bi.size();
if(!size || bi[size-1]) return;
for(auto i=size-2; i+1; --i) {
if(bi[i]) {
bi.resize(i + 1);
return;
}
}
bi.clear();
}
// subfrom subtracts b from a, modifying a
// a >= b must be guaranteed by caller
void subfrom(BiType& a, const BiType& b) {
unsigned borrow = 0;
for(std::size_t i=0; i<b.size(); ++i) {
if(b[i] || borrow) {
// TODO: handle error if i >= a.size()
const auto w = a[i] - b[i] - borrow;
// this relies on the automatic w = w (mod base),
// assuming unsigned max is base-1
// if this is not the case, w must be set to w % base here
borrow = w >= a[i];
a[i] = w;
}
}
for(auto i=b.size(); borrow; ++i) {
// TODO: handle error if i >= a.size()
borrow = !a[i];
--a[i];
// a[i] must be set modulo base here too
// (this is automatic when base is unsigned max + 1)
}
}
// binary division and its helpers: these require base to be a power of 2
// hi_bit_set is base/2
// the definition assumes CHAR_BIT == 8
const auto hi_bit_set = unsigned(1) << (sizeof(unsigned) * 8 - 1);
// shift_right_1 divides bi by 2, truncating any fraction
void shift_right_1(BiType& bi) {
unsigned carry = 0;
for(auto i=bi.size()-1; i+1; --i) {
const auto next_carry = (bi[i] & 1) ? hi_bit_set : 0;
bi[i] >>= 1;
bi[i] |= carry;
carry = next_carry;
}
// if carry is nonzero here, 1/2 was truncated from the result
canonize(bi);
}
// shift_left_1 multiplies bi by 2
void shift_left_1(BiType& bi) {
unsigned carry = 0;
for(std::size_t i=0; i<bi.size(); ++i) {
const unsigned next_carry = !!(bi[i] & hi_bit_set);
bi[i] <<= 1; // assumes high bit is lost, i.e. base is unsigned max + 1
bi[i] |= carry;
carry = next_carry;
}
if(carry) { bi.push_back(1); }
}
// sets an indexed bit in bi, growing the vector when required
void set_bit_at(BiType& bi, std::size_t index, bool set=true) {
std::size_t widx = index / (sizeof(unsigned) * 8);
std::size_t bidx = index % (sizeof(unsigned) * 8);
if(bi.size() < widx + 1) { bi.resize(widx + 1); }
if(set) { bi[widx] |= unsigned(1) << bidx; }
else { bi[widx] &= ~(unsigned(1) << bidx); }
}
// divide divides n by d, returning the result and leaving the remainder in n
// this is implemented using binary division
BiType divide(BiType& n, BiType d) {
if(is_zero(d)) {
// TODO: handle divide by zero
return {};
}
std::size_t shift = 0;
while(cmp(n, d) == 1) {
shift_left_1(d);
++shift;
}
BiType result;
do {
if(cmp(n, d) >= 0) {
set_bit_at(result, shift);
subfrom(n, d);
}
shift_right_1(d);
} while(shift--);
canonize(result);
canonize(n);
return result;
}
std::string get_decimal(BiType bi) {
std::string dec_string;
// repeat division by 10, using the remainder as a decimal digit
// this will build a string with digits in reverse order, so
// before returning, it will be reversed to correct this.
do {
const auto next_bi = divide(bi, {10});
const char digit_value = static_cast<char>(bi.size() ? bi[0] : 0);
dec_string.push_back('0' + digit_value);
bi = next_bi;
} while(!is_zero(bi));
std::reverse(dec_string.begin(), dec_string.end());
return dec_string;
}
}
int main() {
bigint::BiType my_big_int = {453860625, 469837947, 3503557200, 40};
auto dec_string = bigint::get_decimal(my_big_int);
std::cout << dec_string << 'n';
}
输出:
3233755723588593872632005090577