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大数计算：e^1000/300！

news 2025/7/10 9:26:32

1.问题：大数计算可能超出数据类型范围

当单独计算 $e^{1000}$ ，因为 $e^{1000} > e^{700} \approx 1.01432*e304$ ，double的最大取值为1.79769e+308，所以 $e^{1000}$ 肯定超过了double的表示范围。

同样，对于300！也是如此。

那我们应该怎么去计算和存储结果呢？

2.解决方案

从数学角度出发，对于超级大的数，运算方式、运算规律等肯定保持不变的。

很多时候，我们主要是利用相关的定理、公式进行简化或者极限处理。

由于我项目里的精度要求就 e-10，于是，可以采用相对宽松的方式解决这个问题：

科学计数法！

3.代码实现

#ifndef __BigNumeric_h
#define __BigNumeric_h
#include <cmath>template<typename _T>
class BigNumeric
{
private:_T mantissa;        //基数int exponent;       //指数public:BigNumeric(_T num){if (num == 0) {exponent = 0;mantissa = 0;}else{exponent = std::floor(std::log10(std::abs(num)));mantissa = num / std::pow(10, exponent);}}~BigNumeric(){}BigNumeric(const BigNumeric& other){if (this == &other) return;this->exponent = other.exponent;this->mantissa = other.mantissa;}BigNumeric& operator=(const BigNumeric& other){/*if(this == &other) return *this;*/this->exponent = other.exponent;this->mantissa = other.mantissa;return *this;}public:_T value(){if (this->exponent >= 308) return 0.0;return this->mantissa * std::pow(10.0, this->exponent);}//乘法BigNumeric& operator*(const BigNumeric& opr){BigNumeric<_T> resmnt(this->mantissa * opr.mantissa);this->exponent = resmnt.exponent + this->exponent + opr.exponent;this->mantissa = resmnt.mantissa;return *this;}BigNumeric& operator*(const _T opr){BigNumeric<_T> oprbgn(opr);*this = *this * oprbgn;return *this;}friend    BigNumeric operator*(const _T opr1, BigNumeric& opr2){return opr2 * opr1;}//除法BigNumeric& operator/(const BigNumeric& opr){BigNumeric<_T> resmnt(this->mantissa / opr.mantissa);this->exponent = resmnt.exponent + this->exponent - opr.exponent;this->mantissa = resmnt.mantissa;return *this;}BigNumeric& operator/(const _T opr){BigNumeric<_T> oprbgn(opr);*this = *this / oprbgn;return *this;}friend     BigNumeric operator/(const _T opr, const BigNumeric& opr1){BigNumeric<_T> oprbgn(opr);return oprbgn / opr1;}//加法BigNumeric& operator+(const BigNumeric& opr){if (this->exponent - opr.exponent > 15) return *this;else if (this->exponent - opr.exponent < -15){*this = opr;return *this;}   int min = this->exponent > opr.exponent ? opr.exponent : this->exponent;BigNumeric<_T> resmnt(this->mantissa * std::pow(10.0, this->exponent - min) + opr.mantissa * std::pow(10.0, opr.exponent - min));this->exponent = resmnt.exponent + min;this->mantissa = resmnt.mantissa;return *this;}BigNumeric& operator+(const _T opr){BigNumeric<_T> oprbgn(opr);*this = *this + oprbgn;return *this;}friend    BigNumeric operator+(const _T opr1, BigNumeric& opr2){return opr2 + opr1;}//减法BigNumeric& operator-(const BigNumeric& opr){BigNumeric temp(opr);*this = *this + temp * (-1.0);return *this;}BigNumeric& operator-(const _T opr){BigNumeric oprbgn(opr);*this = *this - oprbgn;return *this;}friend    BigNumeric operator-(const _T opr1, BigNumeric& opr2){return opr2 - opr1;}//开方BigNumeric& Sqrt(){_T bgnmant = std::sqrt(this->mantissa);int bgnexp = this->exponent;if (bgnexp % 2 == 0){this->mantissa = bgnmant;this->exponent = bgnexp / 2;}else{BigNumeric temp(bgnmant * std::sqrt(10.0));this->mantissa = temp.mantissa;this->exponent = temp.exponent + bgnexp / 2;}return *this;}//幂BigNumeric& Pow(_T exp){BigNumeric temp(Vpow(this->mantissa, exp));this->mantissa = temp.mantissa;this->exponent = temp.exponent + this->exponent * exp;return *this;}public:static BigNumeric Factorial(int opr){if (opr < 0) return 1.0 / Factorial(-1.0 * opr + 1);else if (opr == 0) return BigNumeric(1.0);return Factorial(opr - 1) * opr;}static BigNumeric Epow(_T exp){BigNumeric res(1.0);double e = 2.71828182845904523536;if (std::abs(exp) <= 700) return BigNumeric(std::pow(e, exp));int count = exp / 700;BigNumeric bgn(std::pow(e, 700.0));for (size_t i = 0; i < count; i++)res = res * bgn;BigNumeric bgn1(std::pow(e, exp - count * 700));res = res * bgn1;return res;}static BigNumeric Vpow(_T e, _T exp){BigNumeric res(1.0);BigNumeric bgnmant(e);int chk = bgnmant.exponent == 0 ? std::abs(exp) : std::abs(exp) * bgnmant.exponent;if (chk <= 300) return BigNumeric(std::pow(e, exp));int count = exp / 300;BigNumeric bgn(std::pow(e, 300.0));for (size_t i = 0; i < count; i++)res = res * bgn;BigNumeric bgn1(std::pow(e, exp - count * 300));res = res * bgn1;return res;}};#endif // !__BigNumeric_h

4.测试

//测试
#include "BigNumeric.hpp"int main() {BigNumeric<double> bignum = BigNumeric<double>::Factorial(300);BigNumeric<double> bignum1 = BigNumeric<double>::Epow(1000);bignum = bignum1 / bignum;return 0;
}

结果：6.4369310844548986e-181，数字部分精度为 e-12，指数部分完全准确。

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