在学习高精度乘低精度的时候的一点思考
高精度乘低精度是让高精度的每一位都乘低精度的值
那高精度乘高精度可以理解为n个高精度乘低精度相加的值

#include<iostream>
#include<vector>
using namespace std;

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;
}

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

vector<int> Mul(vector<int>& A, vector<int>& B) {
    vector<int> SUM;
    for (int i = 0; i < B.size(); i++) {
        vector<int> C;
        C = mul(A, B[i]);
        for (int j = 0; j < i; j++) {
            C.insert(C.begin(), 0);
        }
        SUM = add(SUM, C);
    }
    return SUM;
}
int main() {
    string a, b;
    vector<int> A, B;
    cin >> 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');
    }
    vector<int> C = Mul(A, B);
    for (int i = C.size() - 1; i >= 0; i--) {
        printf("%d", C[i]);
    }
    return 0;
}