高精度运算

Posted 2022-02-10 MangataTS

高精度加法

两个大数的高精度加法：

模板题

题目连接：https://www.acwing.com/problem/content/793/

代码

vector<int> add(vector<int> &A, vector<int> &B)

    if (A.size() < B.size()) return add(B, A);

    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size(); i ++ )
    
        t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % 10);
        t /= 10;
    

    if (t) C.push_back(t);
    return C;

高精度减法

两个大数的减法

模板题

题目连接：https://www.acwing.com/problem/content/794/

代码

// C = A - B, 满足A >= B, A >= 0, B >= 0
vector<int> sub(vector<int> &A, vector<int> &B)

    vector<int> C;
    for (int i = 0, t = 0; i < A.size(); i ++ )
    
        t = A[i] - t;
        if (i < B.size()) t -= B[i];
        C.push_back((t + 10) % 10);
        if (t < 0) t = 1;
        else t = 0;
    

    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;

高精度乘法

a是高精度、b是低精度

模板题

题目连接：https://www.acwing.com/problem/content/795/

代码

// C = A * b, A >= 0, b >= 0
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;

高精度除法

模板题

题目连接：https://www.acwing.com/problem/content/796/

代码

// A / b = C ... r, A >= 0, b > 0
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;

Python

因为python的int本身就是一个大数，所以如果有大数类的题目，可以直接使用python写