P1303 A*B Problem

// Problem: P1303 A*B Problem
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P1303
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// User: Pannnn

#include 

using namespace std;

// 高精度加法
vector sum(vector a, vector b) {
    vector res;
    
    int p = 0;
    for (int i = 0; i < a.size() || i < b.size(); ++i) {
        if (i < a.size()) {
            p += a[i];
        }
        if (i < b.size()) {
            p += b[i];
        }
        res.push_back(p % 10);
        p /= 10;
    }
    
    if (p) {
        res.push_back(p);
    }
    return res;
}

// vector与单个数字的乘法，由于有前移，所以将res以引用传递
void mul(vector a, int b, vector &res) {
    int p = 0;
    for (int i = 0; i < a.size() || p != 0; ++i) {
        if (i < a.size()) {
            p += a[i] * b;
        }
        res.push_back(p % 10);
        p /= 10;
    }
}

vector mul(vector a, vector b) {
    vector res;
    for (int i = 0; i < b.size(); ++i) {
        // 模拟乘法运算，b的每一位与a相乘，移位后累加到结果中
        // 提前在临时结果vector中添加i个0，以引用传递进去
        vector tRes(i);
        mul(a, b[i], tRes);
        res = sum(tRes, res);
    }
    // 去除结果中不必要的前导0
    while (res.size() > 1 && res.back() == 0) {
        res.pop_back();
    }
    return res;
}

vector reverse(string num) {
    vector res;
    for (int i = num.length() - 1; i >= 0; --i) {
        res.push_back(num[i] - '0');
    }
    return res;
}

int main() {
    string a, b;
    cin >> a >> b;
    vector num1 = reverse(a);
    vector num2 = reverse(b);
    vector res = mul(num1, num2);
    for (int i = res.size() - 1; i >= 0; --i) {
        cout << res[i];
    }
    cout << endl;
    return 0;
}