1. 大整数的存储：用数组倒着存（方便进位：因为数组在末尾添加元素更方便）
2. 高精度加法：Ai + Bi + t（进位）

vector<int> add(vector<int> &A, vector<int> &B)
{
    vector<int> C;
    int t = 0;

    for (int i = 0; i < A.size() || i < B.size(); i ++)
    {
        if (i < A.size()) t += A[i];
        if (i < B.size()) t += B[i];

        C.push_back(t % 10);
        t /= 10;
    }
    if (t) C.push_back(1);
    return C;
}

3. 高精度减法

vector<int> sub(vector<int> &A, vector<int> &B)
{
    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size(); i ++)
    {
        t = A[i] - t;
        if (i < B.size()) t -= B[i];
        C.push_back((t + 10) % 10);
        if (t < 0) t = 1;
        else t = 0;
    }
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

4. 高精度乘法

vector<int> mul(vector<int> &A, int b)
{
    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size() || t; i ++)
    {
        if (i < A.size()) t += A[i] * b;
        C.push_back(t % 10);
        t = t / 10;
    }
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

5. 高精度除法

vector<int> div(vector<int> &A, int b, int &r)
{
    vector<int> C;
    r = 0;
    for (int i = A.size() - 1; i >= 0; i --)
    {
        r = r * 10 + A[i];
        C.push_back(r / b);
        r %= b;
    }
    reverse(C.begin(), C.end());
    while (C.size() > 1 && C.back() == 0) C.pop_back();

    return C;
}