急求c++代码。。。大数运算，包括（+ - * / %），用类实现。。。

Posted 2023-03-23

输入描述：
输入数据由若干组数据，每组数据由一个运算符（+ - * / %）和两个整数构成，整数范围在-10150~10150之间，如果运算符晕倒‘@’，则表示运行结束。
输出描述
对于每组数据，输出其运算符序号和运算结果，其中的‘/’为整除运算。每个结果单独成行。
如果运算结果超过200位，则应输出“Too Large Number。”
如果输入整数空，或者有前导0，则应输出“Illegal Number。”
如果输入符号不是上述的物种运算符之一，则输出”Illegal Opeartor”，并在输入操作上，应跳过后续的两个操作符。
如果除0，则应输出“Divide By Zero.：”

参考技术A 如果你想学习大数运算的编程的话，你应该学习GMP库的用法，这个是非常优秀的大数运算程序库，GMP是用C语言写成的。学会怎么用之后，利用C++的运算符重载技术，把这些C的函数封装成类，这样就完成了你的要求。
如何封装，这个在GMP的官方网站上有详细的说明，自己去看吧 gmplib.org追问

可是我想要个代码先看看，用c++的。。。

追答

这个是我从gmplib.org 直接copy下来的例子，1分钟就能看明白，自己看吧。要是连搜索引擎都懒得用，还是别学编程了吧

#include
int main (void)

mpz_class a, b, c;

a = 1234123412341234;
b = "-5678567856785678";
c = a+b;
cout << "sum is " << c << "\n";
cout << "absolute value is " << abs(c) << "\n";

return 0;

追问

我想问一下，你编程怎么样啊？其实每次有不懂的都是google的。。。所以搜索引擎都懒得用这不是问题。。。

本回答被提问者采纳

c++实现大数运算

刷上交大的题遇到大数运算的问题（权当记录）
题目描述如下：
Today, facing the rapid development of business, SJTU recognizes that more powerful calculator should be studied, developed and appeared in future market shortly. SJTU now invites you attending such amazing research and development work. In most business applications, the top three useful calculation operators are Addition (+), Subtraction (-) and Multiplication (×) between two given integers. Normally, you may think it is just a piece of cake. However, since some integers for calculation in business application may be very big, such as the GDP of the whole world, the calculator becomes harder to develop. For example, if we have two integers 20 000 000 000 000 000 and 4 000 000 000 000 000, the exact results of addition, subtraction and multiplication are: 20000000000000000 + 4000000000000000 = 24 000 000 000 000 000 20000000000000000 - 4000000000000000 = 16 000 000 000 000 000 20000000000000000 × 4000000000000000 = 80 000 000 000 000 000 000 000 000 000 000 Note: SJTU prefers the exact format of the results rather than the float format or scientific remark format. For instance, we need “24000000000000000” rather than 2.4×10^16. As a programmer in SJTU, your current task is to develop a program to obtain the exact results of the addition (a + b), subtraction (a - b) and multiplication (a × b) between two given integers a and b.
输入描述:
Each case consists of two separate lines where the first line gives the integer a and the second gives b (|a| <10^400 and |b| < 10^400).
输出描述:
For each case, output three separate lines showing the exact results of addition (a + b), subtraction (a - b) and multiplication (a × b) of that case, one result per lines.

c++实现：

#include<iostream>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;

#define Max 401

string Get_sum(int a[], int b[], int length) 
    string result = "";
    int s[Max] = 0;
    int to_high = 0, l = length, i;
    for (i = 0; i < l; i++) 
        s[i] = (a[i]+b[i]+to_high)%10;
        to_high = (a[i]+b[i]+to_high)/10;
    
    if (to_high >0) s[l++] = 1;
    for (i = 0; i < l; i++) result.insert(0, to_string(s[i]));
    return result;


string Get_diff(int a[], int b[], int length) 
    string result = "";
    int s[Max] = 0;
    int need_high = 0, l = length, i;
    for (i = 0; i < l; i++) 
        if ((a[i]-need_high) < b[i]) 
            s[i] = a[i]+10-b[i]-need_high;
            need_high = 1;
         else 
            s[i] = a[i]-b[i]-need_high;
            need_high = 0;
        
    
    for (i = 0; i < l; i++) result.insert(0, to_string(s[i]));
    while(result.length() >0) 
        if (result[0] == '0') result.erase(0,1);
        else break;
    
    return result;


string Get_mul(int a[], int b[], int a_length, int b_length) 
    string result = "";
    int s[Max*2] = 0;
    for (int i = 0; i < a_length; i++) 
        for (int j = 0; j < b_length; j++) 
            s[i+j] += a[i]*b[j];
        
    
    int to_high = 0, total = a_length+b_length;
    for (int i = 0; i < total; i++) 
        int temp = s[i];
        s[i] = (temp+to_high)%10;
        to_high  = (temp+to_high)/10;
    
    if (s[total-1] == 0) total -= 1;
    for (int i = 0; i < total; i++) result.insert(0, to_string(s[i]));
    return result;


int main() 
    string a, b;
    while (cin >> a >> b) 
        int s1[Max] = 0, s2[Max] = 0;
        bool a_flag = false, b_flag = false;
        if (a[0] == '-') 
            a.erase(0,1);
            a_flag = true;
        
        if (b[0] == '-') 
            b.erase(0,1);
            b_flag = true;
        
        reverse(a.begin(), a.end());
        reverse(b.begin(), b.end());
        unsigned int k;
        for (k = 0; k < a.length(); k++) s1[k] = a[k] - 48;
        for (k = 0; k < b.length(); k++) s2[k] = b[k] - 48;
        string result;
        int length = max(a.length(), b.length());
        reverse(a.begin(), a.end());
        reverse(b.begin(), b.end());
        if ((!a_flag) && (!b_flag)) 
            cout << Get_sum(s1, s2, length) << endl;

            if (a.length() > b.length() || (a.length() == b.length() && a >= b)) // a >= b
                cout << Get_diff(s1, s2, length) << endl;
                // cout << "YES" <<endl;
            else cout << "-" << Get_diff(s2, s1, length) << endl;

            cout << Get_mul(s1, s2, a.length(), b.length()) << endl;
         else if (a_flag && b_flag) 
            cout << "-" << Get_sum(s1, s2, length) << endl;

            if (b.length() > a.length() || (a.length() == b.length() && b >= a)) cout << Get_diff(s2, s1, length) << endl; // b >= a
            else cout << "-" << Get_diff(s1, s2, length) << endl;

            cout << Get_mul(s1, s2, a.length(), b.length()) << endl;
         else 
            if ((!a_flag) && b_flag) 
                if (a.length() > b.length() || (a.length() == b.length() && a >= b)) // a >= b
                    cout << Get_diff(s1, s2, length) << endl;
                else cout << "-" << Get_diff(s2, s1, length) << endl;

                cout << Get_sum(s1, s2, length) << endl;
             else 
                if (b.length() > a.length() || (a.length() == b.length() && b >= a)) cout << Get_diff(s2, s1, length) << endl; // b >= a
                else cout << "-" << Get_diff(s1, s2, length) << endl;

                cout << "-" << Get_sum(s1, s2, length) << endl;
            
            cout << "-" << Get_mul(s1, s2, a.length(), b.length()) << endl;
        
    
    return 0;

关键点：
主要思想是利用字符串储存需要运算的大数，运算时按位运算并按位储存在int数组中，最后每位考虑进位的情况。即模拟人工手算过程。
大数运算乘法实现的关键在于要理解：一个数的第i 位和另一个数的第j 位相乘所得的数，一定是要累加到结果的第i+j 位上。这里i， j 都是从右往左，从0 开始数（反向储存在数组后则是按从左到右）。
即：result[i+j] += a[i]*b[j];