将加数与被加数存储在数组A、B中
模拟人工加法的过程，结果使用数组C存储


方法一：同位相加同时进位存余
//实现1
vector<int> add(vector<int> &A, vector<int> &B)
{
    if (A.size() < B.size()) return add(B, A);

    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size(); i ++ )
    {
        t += A[i];    //同位相加
        if (i < B.size()) t += B[i];

        C.push_back(t % 10);    //存余
        t /= 10;    //进位
    }
    if (t) C.push_back(t);

    return C;
}
//实现2
vector<int> add(vector<int> &A, vector<int> &B)
{
    if (A.size() < B.size()) return add(B, A);

    vector<int> C;
    C.push_back(0);
    for (int i = 0; i < A.size(); i ++ )
    {
        C.push_back(0);

        if (i < B.size()) C[i] = A[i] + B[i] + C[i];    //同位相加
        else C[i] = A[i] + C[i];

        C[i + 1] = C[i] / 10;    //存余
        C[i] %= 10;    //进位
    }
    if (!C.back()) C.pop_back();

    return C;
}

方法二：先同位相加，再进位存余
vector<int> add(vector<int> &A, vector<int> &B)
{
    if (A.size() < B.size()) return add(B, A);

    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size(); i ++ )    //同位相加
    {
        if (i < B.size()) t = A[i] + B[i];
        else t = A[i];

        C.push_back(t);
    }
    C.push_back(0);
    for (int i = 0; i < A.size(); i ++ )
    {
        C[i + 1] += C[i] / 10;    //进位
        C[i] %= 10;    //存余
    }
    if(!C.back()) C.pop_back();

    return C;
}