C++

$\color{#cc33ff}{— > 算法基础课题解}$

思路：

高精度除法：高精度 / 低精度

$\color{blue}{易懂版}$

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

//C = A / b   商是C，余数是r
vector<int> div(vector<int> &A, int b, int &r){ // 余数r通过引用传回去
    vector<int> C; //商
    r = 0; //余数
    for (int i = A.size() - 1; i >= 0; i --){ //从最高位开始处理
        r = r * 10 + A[i]; //个位留出来加上A的个位
        C.push_back(r / b); //商就是余数整除b
        r %= b; //商完之后的余数就是 余数%b
    }

    // 此时C从前往后存的是由高位到低（个）位
    reverse(C.begin(), C.end()); //为与前面的加法，减法，乘法相同，reverse一下(主要目的是方便去除前导0)
    while (C.size() > 1 && C.back() == 0) C.pop_back();  //去除前导0
    return C;
}

int main(){
    string a;
    int b;
    cin >> a >> b;// a = "123456"

    vector<int> A; 
    for (int i = a.size() - 1; i >= 0; i --) A.push_back(a[i] - '0');//A = [6, 5, 4, 3, 2, 1]

    int r;
    vector<int> C;
    C = div(A, b, r);

    for (int i = C.size() - 1; i >= 0; i --) printf("%d", C[i]);
    cout <<endl << r;
    return 0;
}

$\color{red}{易记版}$

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

//C = A / b   商是C，余数是r
vector<int> div(vector<int> &A, int b, int &r){ 
    vector<int> C; 
    r = 0; 
    for (int i = A.size() - 1; i >= 0; i --){ 
        r = r * 10 + A[i]; 
        C.push_back(r / b); 
        r %= b; 
    }

    reverse(C.begin(), C.end()); 
    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');

    int r;
    vector<int> C;
    C = div(A, b, r);

    for (int i = C.size() - 1; i >= 0; i --) printf("%d", C[i]);
    cout <<endl << r;
    return 0;
}