在C++中处理大数字？

Question

在C++中处理大型数字输入的最佳方法是什么(例如10^100)？

对于算法,我通常切换到ruby,有时我会使用字符串.

还有其他好方法吗？

Answer 1

听起来你正在寻找一种输入任意精度数字的方法.这里有两个你可以使用的库:GMP和MAPM

Answer 2

查看C++中的大整数案例研究.pdf by Owen Astrachan.我发现这个文件对于详细介绍和代码实现非常有用.它不使用任何第三方库.我用它来处理大量数据(只要你有足够的内存来存储vector<char>)没有问题.

想法:它通过在a中存储big int来实现任意精度整数类vector<char>.

vector<char> myDigits; // stores all digits of number

然后,所有与big int相关的操作(包括<<, >>, +, -, *, ==, <, !=, >, etc.)都可以基于此操作完成char array.

代码的味道:这是头文件,您可以在pdf文件中找到带有代码的cpp.

#include <iostream>
#include <string> // for strings
#include <vector> // for sequence of digits
using namespace std;

class BigInt
{
public:
    BigInt(); // default constructor, value = 0
    BigInt(int); // assign an integer value
    BigInt(const string &); // assign a string
    // may need these in alternative implementation
    // BigInt(const BigInt &); // copy constructor
    // ~BigInt(); // destructor
    // const BigInt & operator = (const BigInt &);
    // assignment operator
    // operators: arithmetic, relational
    const BigInt & operator += (const BigInt &);
    const BigInt & operator -= (const BigInt &);
    const BigInt & operator *= (const BigInt &);
    const BigInt & operator *= (int num);
    string ToString() const; // convert to string
    int ToInt() const; // convert to int
    double ToDouble() const; // convert to double
    // facilitate operators ==, <, << without friends
    bool Equal(const BigInt & rhs) const;
    bool LessThan(const BigInt & rhs) const;
    void Print(ostream & os) const;
private:
    // other helper functions
    bool IsNegative() const; // return true iff number is negative
    bool IsPositive() const; // return true iff number is positive
    int NumDigits() const; // return # digits in number
    int GetDigit(int k) const;
    void AddSigDigit(int value);
    void ChangeDigit(int k, int value);
    void Normalize();
    // private state/instance variables
    enum Sign{positive,negative};
    Sign mySign; // is number positive or negative
    vector<char> myDigits; // stores all digits of number
    int myNumDigits; // stores # of digits of number
};

// free functions
ostream & operator <<(ostream &, const BigInt &);
istream & operator >>(istream &, BigInt &);
BigInt operator +(const BigInt & lhs, const BigInt & rhs);
BigInt operator -(const BigInt & lhs, const BigInt & rhs);
BigInt operator *(const BigInt & lhs, const BigInt & rhs);
BigInt operator *(const BigInt & lhs, int num);
BigInt operator *(int num, const BigInt & rhs);
bool operator == (const BigInt & lhs, const BigInt & rhs);
bool operator < (const BigInt & lhs, const BigInt & rhs);
bool operator != (const BigInt & lhs, const BigInt & rhs);
bool operator > (const BigInt & lhs, const BigInt & rhs);
bool operator >= (const BigInt & lhs, const BigInt & rhs);
bool operator <= (const BigInt & lhs, const BigInt & rhs);

Answer 3

您在寻找如何对您收到的大型输入执行操作吗？有一个大整数C++库(类似于Java),允许您执行算术运算......

Answer 4

如果您希望为此编写自己的代码，请尝试使用字符串存储大数字...然后，您可以在其上创建+-/ *之类的基本操作，例如-

#include <iostream>

using namespace std;

string add (string &s1, string &s2){
    int carry=0,sum,i;

    string  min=s1,
    max=s2,
    result = "";

    if (s1.length()>s2.length()){
        max = s1;
        min = s2;
    } else {
        max = s2;
        min = s1;
    }

    for (i = min.length()-1; i>=0; i--){
        sum = min[i] + max[i + max.length() - min.length()] + carry - 2*'0';

        carry = sum/10;
        sum %=10;

        result = (char)(sum + '0') + result;
    }

    i = max.length() - min.length()-1;

    while (i>=0){
        sum = max[i] + carry - '0';
        carry = sum/10;
        sum%=10;

        result = (char)(sum + '0') + result;
        i--;
    }

    if (carry!=0){
        result = (char)(carry + '0') + result;
    }       

    return result;
}

int main (){
    string a,b;

    cin >> a >> b;

    cout << add (a,b)<<endl;

    return 0;
}

Answer 5

假设您正在谈论输入数字，双精度将使您达到 1.7976931348623157 x 10^308