10.高精度除法

 大数除以一个小数

#include <bits/stdc++.h>
using namespace std;
//A/b商是C，余数是r 
vector<int> div(vector<int> &A, int b, int &r) {
    vector<int> C;
    r = 0;
    //注意除法是从最高位开始算的 
    //前面那仨都是从最低位开始算 
    for (int i = A.size() - 1; i >= 0; i--) {
        r = r * 10 + A[i];
        C.push_back(r / b);
        r %= b;
    }
    //C[0]是最高位 
    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, C;
    for (int i = a.length() - 1; i >= 0; i--) {
        A.push_back(a[i] - '0');
    }
    int r;
    C = div(A, b, r);
    for (int i = C.size() - 1; i >= 0; i--) {
        cout << C[i];
    }
    cout << endl << r << endl;
    return 0;
}

 2020年9月29日更新：压8位的做法

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int base = 1000000000;
vector<ll> div(vector<ll> &A, int b, ll &r) {
    vector<ll> C;
    r = 0;
    for (ll i = A.size() - 1; i >= 0; i--) {
        r = r * base + 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<ll> A, C;
    for (ll i = a.size() - 1, s = 0, j = 0, t = 1; i >= 0; i--) {
        s += (a[i] - '0') * t;
        j++;
        t *= 10;
        if (j == 9 || i == 0) {
            A.push_back(s);
            s = 0;
            j = 0;
            t = 1;
        }
    }

    ll r;
    C = div(A, b, r);
    cout << C.back();
    for (ll i = C.size() - 2; i >= 0; i--) {
        cout << setw(9) << setfill('0') << C[i];
    }
    cout << endl << r << endl;
    return 0;
}