小数幂次c++高精度代码

小数幂次计算在C++中通常涉及到高精度数学，因为直接对浮点数进行乘方可能会导致精度丢失。如果你需要处理非整数幂，可以使用一种称为“牛顿法”或者“二分法”的算法结合字符串或大数库来进行。这里提供一个简单的基于stringstream的大数乘法和除法的示例：

#include <sstream>
#include <iostream>

std::string multiply(const std::string &a, const std::string &b) {
    int n1 = a.size(), n2 = b.size();
    if (n1 == 0 || n2 == 0)
        return "0";
    std::vector<int> res(n1 + n2);
    for (int i = n1 - 1; i >= 0; --i) {
        long long int carry = 0;
        for (int j = n2 - 1; j >= 0; --j) {
            int sum = (a[i] - '0') * (b[j] - '0') + carry + res[i + j];
            res[i + j] = static_cast<int>(sum % 10);
            carry = sum / 10;
        }
        if (carry > 0)
            res[i] += carry;
    }
    std::reverse(res.begin(), res.end());
    return std::string(res.begin(), res.end());
}

std::string divide(const std::string &dividend, const std::string &divisor) {
    int n1 = dividend.size(), n2 = divisor.size();
    if (n2 == 0)
        return "1";
    bool negative = (dividend[0] - '0') < 0 ^ (divisor[0] - '0' < 0);
    dividend = std::abs(dividend);
    divisor = std::abs(divisor);

    std::string result = "0.";
    while (!dividend.empty()) {
        int shift = 1;
        for (int i = 1; i < divisor.size(); ++i) {
            shift *= 10;
            dividend.pop_back();
        }
        int quotient = 0;
        for (int i = 0; i < divisor.size(); ++i) {
            quotient += ((dividend.back() - '0') * shift) / (divisor[i] - '0');
            dividend.pop_back();
        }
        result.push_back(quotient + '0');
    }

    return negative ? "-" + result : result;
}

std::string power(std::string base, double exponent) {
    std::string result("1");
    while (exponent != 0) {
        if (exponent > 0) {
            result = multiply(result, base);
            exponent--;
        } else {
            exponent++;
            base = divide(base, "10"); // 这里假设我们只支持小数部分不超过一位的情况
        }
    }
    return result;
}

int main() {
    std::cout << power("12345", 2.5) << std::endl; // 示例：输出以特定精度表示的12345^2.5
    return 0;
}