c++實現大數運算

刷上交大的題遇到大數運算的問題（權當記錄）
題目描述如下：
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.
輸入描述:
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).
輸出描述:
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++實現：

#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;
}

關鍵點：
主要思想是利用字符串儲存需要運算的大數，運算時按位運算並按位儲存在int數組中，最後每位考慮進位的情況。即模擬人工手算過程。
大數運算乘法實現的關鍵在於要理解：一個數的第i 位和另一個數的第j 位相乘所得的數，一定是要累加到結果的第i+j 位上。這裏i， j 都是從右往左，從0 開始數（反向儲存在數組後則是按從左到右）。
即：result[i+j] += a[i]*b[j];