怎么用C + +编高精度

使用C++编写高精度可以通过自定义大整数类来实现。主要思路是将大整数按照数组的形式存储，通过重载运算符来实现加减乘除等运算。

以下是一个简单的大整数类的示例代码：

#include <iostream>
#include <cstring>
using namespace std;

const int MAXN = 1005;

class BigInteger {
private:
    int num[MAXN];
    bool sign;
public:
    BigInteger() {
        memset(num, 0, sizeof(num));
        sign = true;
    }

    BigInteger(string s) {
        memset(num, 0, sizeof(num));
        if (s[0] == '-') {
            sign = false;
            s = s.substr(1);
        } else {
            sign = true;
        }
        int len = s.length();
        for (int i = 0; i < len; i++) {
            num[len - i - 1] = s[i] - '0';
        }
    }

    BigInteger operator + (const BigInteger& b) const {
        BigInteger res;
        res.sign = this->sign;
        int carry = 0;
        for (int i = 0; i < MAXN; i++) {
            res.num[i] = this->num[i] + b.num[i] + carry;
            carry = res.num[i] / 10;
            res.num[i] %= 10;
        }
        return res;
    }

    BigInteger operator - (const BigInteger& b) const {
        BigInteger res;
        res.sign = this->sign;
        int borrow = 0;
        for (int i = 0; i < MAXN; i++) {
            res.num[i] = this->num[i] - b.num[i] - borrow;
            if (res.num[i] < 0) {
                res.num[i] += 10;
                borrow = 1;
            } else {
                borrow = 0;
            }
        }
        return res;
    }

    BigInteger operator * (const BigInteger& b) const {
        BigInteger res;
        res.sign = (this->sign == b.sign);
        for (int i = 0; i < MAXN - 1; i++) {
            for (int j = 0; j < MAXN - 1; j++) {
                int pos = i + j;
                res.num[pos] += this->num[i] * b.num[j];
                res.num[pos + 1] += res.num[pos] / 10;
                res.num[pos] %= 10;
            }
        }
        return res;
    }

    BigInteger operator / (const BigInteger& b) const {
        BigInteger res, tmp;
        res.sign = (this->sign == b.sign);
        tmp.sign = true;
        int len = this->getLen() - b.getLen();
        for (int i = len; i >= 0; i--) {
            tmp = tmp * 10;
            tmp.num[0] = this->num[i + b.getLen() - 1];
            int l = 0, r = 9, mid;
            while (l < r) {
                mid = (l + r + 1) / 2;
                if (tmp >= b * mid) {
                    l = mid;
                } else {
                    r = mid - 1;
                }
            }
            res.num[i] = l;
            tmp = tmp - b * l;
        }
        return res;
    }

    bool operator < (const BigInteger& b) const {
        if (this->sign != b.sign) {
            return !this->sign;
        }
        for (int i = MAXN - 1; i >= 0; i--) {
            if (this->num[i] != b.num[i]) {
                return (this->sign) ? (this->num[i] < b.num[i]) : (this->num[i] > b.num[i]);
            }
        }
        return false;
    }

    int getLen() const {
        for (int i = MAXN - 1; i >= 0; i--) {
            if (this->num[i] != 0) {
                return i + 1;
            }
        }
        return 1;
    }

    friend ostream& operator << (ostream& out, const BigInteger& b) {
        if (!b.sign) {
            out << '-';
        }
        int len = b.getLen();
        for (int i = len - 1; i >= 0; i--) {
            out << b.num[i];
        }
        return out;
    }
};

int main() {
    BigInteger a("12345678901234567890");
    BigInteger b("98765432109876543210");
    cout << a + b << endl;
    cout << a - b << endl;
    cout << a * b << endl;
    cout << a / b << endl;
    return 0;
}

该示例代码可以实现大整数的加减乘除等基本运算，可以通过自定义更多的运算符来实现其他运算。