高精度 X 低精度

#include <iostream>
#include <vector>

using namespace std;

vector <int> mul(vector <int> & A, int b) {
    vector <int> C;

    int t = 0;
    for (int i = 0; i < A.size(); i ++) {
        t += A[i] * b;       // t + A[i] * b = 7218
        C.push_back(t % 10); // 只取个位 8
        t /= 10;             // 721 看作 进位
    }

    while (t) {            // 处理最后剩余的 t
        C.push_back(t % 10);
        t /= 10;
    }

    while (C.size() > 1 && C.back() == 0) C.pop_back();

    return C;
}

int main() {
    string a;
    int b;
    cin >> a >> b;

    vector <int> A;
    for (int i = a.size() - 1; i >= 0; i --) A.push_back(a[i] - '0');

    auto C = mul(A, b);

    for (int i = C.size() - 1; i >= 0; i --) {
        cout << C[i];
    }

    return 0;
}

高精度 X 高精度

#include <iostream>
#include <vector>

using namespace std;

vector<int> mul(vector<int> &A, vector<int> &B) {
    vector<int> C(A.size() + B.size() + 7, 0); // 初始化为 0，C的size可以大一点

    for (int i = 0; i < A.size(); i++)
        for (int j = 0; j < B.size(); j++)
            C[i + j] += A[i] * B[j];

    int t = 0;
    for (int i = 0; i < C.size(); i++) { // i = C.size() - 1时 t 一定小于 10
        t += C[i];
        C[i] = t % 10;
        t /= 10;
    }

    while (C.size() > 1 && C.back() == 0) C.pop_back(); // 必须要去前导 0，因为最高位很可能是 0
    return C;
}

int main() {
    string a, b;
    cin >> a >> b; // a = "1222323", b = "2323423423"

    vector<int> A, B;
    for (int i = a.size() - 1; i >= 0; i--)
        A.push_back(a[i] - '0');
    for (int i = b.size() - 1; i >= 0; i--)
        B.push_back(b[i] - '0');

    auto C = mul(A, B);

    for (int i = C.size() - 1; i >= 0; i--)
        cout << C[i];

    return 0;
}

更新：大数相加A+B和大数相乘A*B通用模板

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

vector<int> add(vector<int> A, vector<int> B) {
    // A: 4 3 2 1
    // B: 6 5
    vector<int> C(max(A.size(), B.size()) + 7, 0);  // 数组C开大一点没事，反正可以去前导零的
    for (int i = 0; i < A.size(); i ++) C[i] += A[i];
    for (int i = 0; i < B.size(); i ++) C[i] += B[i];

    // 处理进位
    for (int i = 0; i + 1 < C.size(); i ++) {
        C[i + 1] += C[i] / 10;
        C[i] %= 10;
    }

    // 处理前导零
    while (C.size() > 1 && C.back() == 0) C.pop_back();

    reverse(C.begin(), C.end());
    return C;
}

vector<int> mul(vector<int> A, vector<int> B) {
    // A: 4 3 2 1
    // B: 6 5
    vector<int> C(A.size() + B.size() + 7, 0);  // 数组C开大一点没事，反正可以去前导零的
    for (int i = 0; i < A.size(); i ++) {
        for (int j = 0; j < B.size(); j ++) {
            C[i + j] += A[i] * B[j];
        }
    }

    // 处理进位
    for (int i = 0; i + 1 < C.size(); i ++) {
        C[i + 1] += C[i] / 10;
        C[i] %= 10;
    }

    // 处理前导零 "0000" 去掉前导零
    while (C.size() > 1 && C.back() == 0) C.pop_back();

    reverse(C.begin(), C.end());
    return C;
}

int main() {
    string s1 = "9899", s2 = "100";

    vector<int> A, B;
    for (int i = s1.size() - 1; i >= 0; i --) A.push_back(s1[i] - '0');
    for (int i = s2.size() - 1; i >= 0; i --) B.push_back(s2[i] - '0');

    vector<int> C = add(A, B);
    cout << s1 << "+" << s2 << "=";
    for (int i = 0; i < C.size(); i ++) cout << C[i];
    cout << endl;

    C = mul(A, B);
    cout << s1 << "*" << s2 << "=";
    for (int i = 0; i < C.size(); i ++) cout << C[i];
    cout << endl;

    return 0;
}

附：A*B图解