设计一个求两个大整数四则运算的类。要求如下。 (1)类中私有数据成员为2个运算数及位数、运算结果。 (2)公有成员函数包括:构造函数和析构函数;接收数据的函数;和函数;差函数; 积函数;除函数;显示函数;清零函数等。 程序运行结果如下所示: Please enter one: 1234567 please enter two: 2345678 one+two=3580245 one-two=-1111111 one*two=2895896651426 one/two=52631742743。c++

以下是一个可能的实现：

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

class BigNumArith {
private:
    int num1[100], num2[100], res[200];
    int len1, len2, resLen;
public:
    BigNumArith() {
        memset(num1, 0, sizeof(num1));
        memset(num2, 0, sizeof(num2));
        memset(res, 0, sizeof(res));
        len1 = len2 = resLen = 0;
    }

    ~BigNumArith() {}

    void input() {
        char str1[100], str2[100];
        cout << "Please enter one: ";
        cin >> str1;
        cout << "please enter two: ";
        cin >> str2;

        len1 = strlen(str1);
        len2 = strlen(str2);

        for (int i = 0; i < len1; i++) {
            num1[i] = str1[len1-1-i] - '0';
        }

        for (int i = 0; i < len2; i++) {
            num2[i] = str2[len2-1-i] - '0';
        }
    }

    void add() {
        int carry = 0;
        resLen = max(len1, len2);
        for (int i = 0; i < resLen; i++) {
            int sum = num1[i] + num2[i] + carry;
            res[i] = sum % 10;
            carry = sum / 10;
        }
        if (carry > 0) {
            res[resLen++] = carry;
        }
    }

    void sub() {
        int borrow = 0;
        resLen = max(len1, len2);
        for (int i = 0; i < resLen; i++) {
            int diff = num1[i] - num2[i] - borrow;
            if (diff < 0) {
                diff += 10;
                borrow = 1;
            } else {
                borrow = 0;
            }
            res[i] = diff;
        }
        while (resLen > 0 && res[resLen-1] == 0) {
            resLen--;
        }
    }

    void mul() {
        resLen = len1 + len2;
        for (int i = 0; i < len1; i++) {
            for (int j = 0; j < len2; j++) {
                res[i+j] += num1[i] * num2[j];
                res[i+j+1] += res[i+j] / 10;
                res[i+j] %= 10;
            }
        }
        while (resLen > 0 && res[resLen-1] == 0) {
            resLen--;
        }
    }

    void div() {
        int divisor[100], quotient[100];
        memcpy(divisor, num2, sizeof(num2));
        memset(quotient, 0, sizeof(quotient));
        resLen = len1 - len2 + 1;
        for (int i = len1-1; i >= len2-1; i--) {
            int q = 0;
            while (compare(num1+i-len2+1, divisor, len2) >= 0) {
                substract(num1+i-len2+1, divisor, len2);
                q++;
            }
            quotient[i-len2+1] = q;
        }
        while (resLen > 0 && quotient[resLen-1] == 0) {
            resLen--;
        }
        memcpy(res, quotient, sizeof(quotient));
    }

    void display() {
        cout << "one+two=";
        for (int i = resLen-1; i >= 0; i--) {
            cout << res[i];
        }
        cout << endl;
    }

    void clear() {
        memset(num1, 0, sizeof(num1));
        memset(num2, 0, sizeof(num2));
        memset(res, 0, sizeof(res));
        len1 = len2 = resLen = 0;
    }

private:
    int compare(int* a, int* b, int len) {
        for (int i = len-1; i >= 0; i--) {
            if (a[i] < b[i]) {
                return -1;
            } else if (a[i] > b[i]) {
                return 1;
            }
        }
        return 0;
    }

    void substract(int* a, int* b, int len) {
        int borrow = 0;
        for (int i = 0; i < len; i++) {
            int diff = a[i] - b[i] - borrow;
            if (diff < 0) {
                diff += 10;
                borrow = 1;
            } else {
                borrow = 0;
            }
            a[i] = diff;
        }
    }
};

int main() {
    BigNumArith bna;
    bna.input();
    bna.add();
    bna.display();
    bna.clear();
    bna.input();
    bna.sub();
    bna.display();
    bna.clear();
    bna.input();
    bna.mul();
    bna.display();
    bna.clear();
    bna.input();
    bna.div();
    bna.display();
    bna.clear();
    return 0;
}

说明：

1. 本实现中，使用了三个私有函数：`compare`、`substract`和`clear`。其中，`compare`函数用于比较两个整数数组的大小，返回1表示第一个大于第二个，返回-1表示第一个小于第二个，返回0表示相等；`substract`函数用于计算两个整数数组的差，并将结果保存在第一个数组中；`clear`函数用于将所有成员变量清零，以便进行下一轮计算。

2. 在进行加、减、乘、除运算时，需要考虑进位和借位的情况，需要注意不同位数的处理方式不同。

3. 由于本题中的数据非常大，因此需要使用数组来保存数据，同时需要注意数组的长度需要足够大。

4. 在除法运算中，需要用到两个数组的值，因此需要使用memcpy函数进行拷贝。