大整数除法

这个大整数除法指的是求出 a 除以 b 的整数位结果, 当然余数也可以求, 但不是求精确的小数结果.

举个例子来说思路:

a = 17693 b = 23 求 a / b

len(a) = 5, len(b) = 2

那么
1. 17693 - 23000 < 0, 跳过
2. 17693 - 2300*7 = 1593
3. 1593 - 230*6 = 213
4. 213 - 23*9 = 6
5. 6 - 23 < 0, 跳过

现在很清楚了, 17693 / 23 = 769 余 6

从上面的例子可以很清楚的知道, 其实做除法是一个做减法的过程. 那为什么不直接一次一次减23呢? 因为这样会耗费很多时间, 所以从23的10^len倍减起, 这样时间上会快很多. 当然代码写起来也麻烦不少.

---

整个代码也不过60+行

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

vector<int> num1, num2;

void strToVector(string s1, string s2)
{
    int len1 = s1.size(), len2 = s2.size();
    if (len1 < len2) return;
    for (int i = len1 - 1; i >= 0; --i) {
        num1.push_back(s1[i] - '0');
    }
    for (int i = len2 - 1; i >= 0; --i) {
        num2.push_back(s2[i] - '0');
    }
    while (num1.size() != num2.size()) num2.insert(num2.begin(), 0);
}

bool isBigger(vector<int> v1, vector<int> &v2) // 返回v1是否比v2更大 
{
    int l1 = v1.size(), l2 = v2.size();
    if (l1 != l2) return l1 > l2;
    for (int i = l1 - 1; i >= 0; --i) {
        if (v1[i] == v2[i]) continue;
        if (v1[i] < v2[i]) return 0;
        else return 1;
    }
    return 1;
}

void sub(vector<int> &v1, vector<int> v2) // 记得去掉后缀0, .pop_back() 
{
    while (v2.size() < v1.size()) v2.push_back(0);
    for (int i = 0; i < v1.size() + 0; ++i) {
        if (v1[i] < v2[i]) {
            v1[i] += 10 - v2[i];
            v1[i + 1] -= 1;
        } else v1[i] -= v2[i];
    }
    while (v1.back() == 0) {
        v1.pop_back();
    }
}

int main()
{
    string sn1, sn2;
    cin >> sn1 >> sn2;
    int len1 = sn1.size(), len2 = sn2.size();
    strToVector(sn1, sn2);
    int len = len1 - len2, ans = 0;
    for (int i = 0; i <= len; ++i) {
        int time = 0;
        while (isBigger(num1, num2)) {
            sub(num1, num2); // 做减法, num1是减后的数
            time++; 
        }
        if (isBigger(num1, num2) == isBigger(num2, num1)) time = 1;
        ans = ans * 10 + time;
        num2.erase(num2.begin()); // 去掉一个零前缀0 
    }
    cout << ans;
}
/*
17693 23
*/ 

---

精确到小数位的高精度除法, 不过似乎对能够整除的情况支持的不是很好. 思路和上面的整数除法是一样的.

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

bool isBigger(vector<int> a, vector<int> b)
{
    if (a.size() != b.size()) return a.size() > b.size();
    int len = a.size() - 1;
    while (len >= 0) {
        if (a[len] != b[len]) return a[len] > b[len];
        len--;
    }
    return true; // 当相等时return true 
}

void sub(vector<int> &v1, vector<int> v2) // 记得去掉后缀0, .pop_back() 
{
    while (v2.size() < v1.size()) v2.push_back(0);
    for (int i = 0; i < v1.size() + 0; ++i) {
        if (v1[i] < v2[i]) {
            v1[i] += 10 - v2[i];
            v1[i + 1] -= 1;
        } else v1[i] -= v2[i];
    }
    while (v1.back() == 0) {
        v1.pop_back();
    }
}

int main()
{
    vector<int> a, b;
    char c;
    string ans = "";
    int index = 0, n;

    while (1) {
        c = getchar();
        if (c == ' ' || c == '\n') break;
        a.insert(a.begin(), c - '0');
    }
    while (1) {
        c = getchar();
        if (c == ' ' || c == '\n') break;
        b.insert(b.begin(), c - '0');
    }
    while (isBigger(a, b)) {
        b.insert(b.begin(), 0);
        index++;
    }
    cin >> n;
    for (int i = 0; i <= n; ++i) {
        int time = 0;
        a.insert(a.begin(), 0);
        while (isBigger(a, b)) {
            sub(a, b);
            time++;
        }
        ans += '0' + time;
    }
    for (int i = 0; i < ans.size() - 2; ++i) {
        if (i == index) {
            if (index == 0) cout << "0";
            cout << '.';
        }
        cout << ans[i];
    }
    if (ans[ans.size() - 1] > '4') cout << (char)(ans[ans.size() - 2] + 1);
    else cout << ans[ans.size() - 1];
    cout << endl;
}
/*
213 23
9.2608695652173913043478260869565
*/