P1303 A * B Problem

题目传送门
//P1303.cpp

#include <bits/stdc++.h>

using namespace std;

/**
 * 功能：高精度乘低精度模板
 * @param A
 * @param b
 * @return
 */
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 /= 10;
    }
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

int main() {
    string a;
    int b;
    vector<int> A, C;
    cin >> a >> b;
    //倒着放的噢~
    for (int i = a.size() - 1; i >= 0; i--) A.push_back(a[i] - '0');
    C = mul(A, b);
    //倒着输出噢~
    for (int i = C.size() - 1; i >= 0; i--) printf("%d", C[i]);
    return 0;
}

//P1303_2.cpp

#include <bits/stdc++.h>

using namespace std;

/**
 * 功能：高精度乘高精度模板
 * @param A
 * @param b
 * @return
 */
vector<int> mul(vector<int> &A, vector<int> &B) {
    //初始化大小
    vector<int> C(A.size() + B.size());
    //先放里再说
    for (int i = 0; i < A.size(); i++)
        for (int j = 0; j < B.size(); j++)
            C[i + j] += A[i] * B[j];

    //处理余数
    for (int i = 0, t = 0; i < C.size(); i++) {
        t += C[i];
        if (i >= C.size()) C.push_back(t % 10);
        else C[i] = t % 10;
        t /= 10;
    }
    //去掉前导0
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

int main() {
    string a, b;
    cin >> a >> b;
    //准备动作
    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');
    //计算
    vector<int> C = mul(A, B);
    //倒序输出
    for (int i = C.size() - 1; i >= 0; i--) cout << C[i];
    return 0;
}