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。