C++ 大整數類（BigInteger類）實現
C++ BigInteger類實現，儲存任意大小的非負整數
參考
演算法競賽入門經典第二版 5.3 節——大整數類
程式碼（先放程式碼，後續追加註釋與可以為負數的實現）
#include <string>
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
/**
* 儲存任意大小的非負整數
*/
class BigInteger
{
friend ostream& operator << (ostream &out, const BigInteger &num);
friend istream& operator << (istream &in, BigInteger &num);
public:
BigInteger(long long num = 0)
{
*this = num;
}
// 利用表示非負整數的字串構造物件
BigInteger(const string &numStr)
{
*this = numStr;
}
BigInteger& operator = (long long num)
{
m_bits.clear();
do
{
m_bits.push_back(num % BASE);
num /= BASE;
} while (num > 0);
return *this;
}
// 利用表示非負整數的字串賦值給物件
BigInteger& operator = (const string &numStr)
{ m_bits.clear();
int bit, len = (numStr.length() - 1) / WIDTH + 1;
for (int i = 0; i < len; i++)
{
int end = numStr.length() - i * WIDTH;
int start = max(0, end - WIDTH);
sscanf(numStr.substr(start, end - start).c_str(), "%d", &bit);
m_bits.push_back(bit);
}
return *this;
}
// 加法運算
BigInteger operator + (const BigInteger &rhs) const
{
BigInteger result;
result.m_bits.clear();
for (int i = 0, carry = 0; ; i++)
{
if (i >= m_bits.size() && i >= rhs.m_bits.size() && carry == 0)
break;
int num = carry;
if (i < m_bits.size())
num += m_bits[i];
if (i < rhs.m_bits.size())
num += rhs.m_bits[i];
result.m_bits.push_back(num % BASE);
carry = num / BASE;
}
return result;
}
BigInteger& operator += (const BigInteger &rhs)
{
*this = *this + rhs;
return *this;
}
// 減法運算，因為BigInteger只能表示非負整數，所以返回值為減法運算結果的絕對值
BigInteger operator - (const BigInteger &rhs) const
{
// lhs > rhs， 則結果為 lhs - rhs；否則結果為 rhs - lhs
if (*this > rhs)
return sub(*this, rhs);
else
return sub(rhs, *this);
}
BigInteger& operator -= (const BigInteger &rhs)
{
*this = *this - rhs;
return *this;
}
// 乘法運算
BigInteger operator * (const BigInteger &rhs) const
{
BigInteger result;
result.m_bits.clear();
result.m_bits.resize(m_bits.size() + rhs.m_bits.size(), 0);
for (size_t i = 0; i < m_bits.size(); i++)
{
for (size_t j = 0; j < rhs.m_bits.size(); j++)
{
// 由於int的精度限制，兩者的乘積不能溢位，所以BASE的值不要超過10^4
int num = m_bits[i] * rhs.m_bits[j];
result.m_bits[i + j] += num;
}
}
formatBits(result);
removeHighestZero(result);
return result;
}
BigInteger& operator *= (const BigInteger &rhs)
{
*this = *this * rhs;
return *this;
}
// 整除運算
BigInteger operator / (const BigInteger &rhs) const
{
if (rhs == 0)
throw runtime_error("Divisor cannot be 0!");
BigInteger temp = *this;
return divide(temp, rhs);
}
BigInteger& operator /= (const BigInteger &rhs)
{
*this = *this / rhs;
return *this;
}
// 求餘運算
BigInteger operator % (const BigInteger &rhs) const
{
BigInteger result = *this;
divide(result, rhs);
return result;
}
BigInteger& operator %= (const BigInteger &rhs)
{
divide(*this, rhs);
return *this;
}
bool operator > (const BigInteger &rhs) const
{
// 如果數位不相等
if (m_bits.size() != rhs.m_bits.size())
return m_bits.size() > rhs.m_bits.size();
// 數位相等則逐位進行比較
for (int i = m_bits.size() - 1; i >= 0; i--)
{
if (m_bits[i] != rhs.m_bits[i])
return m_bits[i] > rhs.m_bits[i];
}
return false;
}
bool operator >= (const BigInteger &rhs) const
{
return (*this > rhs || *this == rhs);
}
bool operator < (const BigInteger &rhs) const
{
return !(*this >= rhs);
}
bool operator <= (const BigInteger &rhs) const
{
return !(*this > rhs);
}
bool operator == (const BigInteger &rhs) const
{
return m_bits == rhs.m_bits;
}
bool operator != (const BigInteger &rhs) const
{
return m_bits != rhs.m_bits;
}
private:
// 該函式假設 lhs >= rhs
static BigInteger sub(const BigInteger &lhs, const BigInteger &rhs)
{
BigInteger result;
result.m_bits.clear();
for (int i = 0, carry = 0; ; i++)
{
if (i >= lhs.m_bits.size() && i >= rhs.m_bits.size() && carry == 0)
break;
int num = carry;
if (i < lhs.m_bits.size())
num += lhs.m_bits[i];
if (i < rhs.m_bits.size())
num -= rhs.m_bits[i];
num += BASE;
result.m_bits.push_back(num % BASE);
carry = num / BASE - 1;
}
removeHighestZero(result);
return result;
}
// 利用長除法進行除法運算
static BigInteger divide(BigInteger &lhs, const BigInteger &rhs)
{
BigInteger result;
result.m_bits.clear();
result.m_bits.resize(lhs.m_bits.size() - rhs.m_bits.size() + 1, 0);
BigInteger temp;
int pos = 0, quotient = 0;
while (lhs >= rhs)
{
if (lhs.m_bits.back() >= rhs.m_bits.back())
{
quotient = lhs.m_bits.back() / rhs.m_bits.back();
pos = lhs.m_bits.size() - rhs.m_bits.size();
}
else
{
// 由於int的精度限制，所以quotient的運算不能產生溢位，因而BASE的值不能太大
quotient = lhs.m_bits.back() * BASE / rhs.m_bits.back();
pos = lhs.m_bits.size() - rhs.m_bits.size() - 1;
}
temp.m_bits.clear();
temp.m_bits.resize(pos + 1, 0);
temp.m_bits[pos] = quotient;
result.m_bits[pos] += quotient;
lhs = lhs - rhs * temp;
}
removeHighestZero(result);
return result;
}
// 去掉最高位的0
static void removeHighestZero(BigInteger &num)
{
while (num.m_bits.size() > 1 && num.m_bits.back() == 0)
num.m_bits.pop_back();
}
// 格式化各數位，使其符合進位制
static void formatBits(BigInteger &num)
{
int carry = 0;
for (size_t i = 0; i < num.m_bits.size(); i++)
{
int temp = num.m_bits[i] + carry;
num.m_bits[i] = temp % BASE;
carry = temp / BASE;
}
while (carry != 0)
{
num.m_bits.push_back(carry % BASE);
carry /= BASE;
}
}
private:
// 採用BASE進位制記錄任意大小的非負整數
// BASE為10的指數，因乘法運算具體實現的限制，BASE的值不要超過10^4
static const int BASE = 10000;
static const int WIDTH = 4;
// BASE進位制各數位，從低位到高位對應陣列從左到右
vector<int> m_bits;
};
ostream& operator << (ostream &out, const BigInteger &num)
{
static string format = "%0" + to_string(BigInteger::WIDTH) +"d";
out << num.m_bits.back();
for (int i = num.m_bits.size() - 2; i >= 0; i--)
{
char buf[20];
sprintf(buf, format.c_str(), num.m_bits[i]);
out << buf;
}
return out;
}
istream& operator >> (istream &in, BigInteger &num)
{
string str;
if (!(in >> str))
return in;
num = str;
return in;
}
int main()
{
cout << "-----------------------Test procedure for BigInteger-----------------------" << endl;
BigInteger num1, num2;
char instruction;
while (1)
{
cout << "Enter q to exit and other to start" << endl;
cin >> instruction;
if (instruction == 'q')
break;
cout << "please input num1:" << endl;
cin >> num1;
cout << "please input num2:" << endl;
cin >> num2;
cout << "the result of (num1 + num2) is:" << endl;
cout << num1 + num2 << endl;
cout << "the result of (num1 - num2):" << endl;
cout << num1 - num2 << endl;
cout << "the result of (num1 * num2):" << endl;
cout << num1 * num2 << endl;
cout << "the result of (num1 / num2):" << endl;
cout << num1 / num2 << endl;
cout << "the result of (num1 % num2):" << endl;
cout << num1 % num2 << endl;
cout << endl;
}
return 0;
}相關文章
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