[算法总结] 模板---高精度!

说在前面

本篇代码全部来自Acwing y总

高精度加法 使用方法

输入 A和B 即可 输出即是A+B

code:

#include 
#include 

using namespace std;

vector add(vector &A, vector &B)
{
    if (A.size() > 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');

    auto C = add(A, B);

    for (int i = C.size() - 1; i >= 0; i -- ) cout > 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');

    vector C;

    if (cmp(A, B)) C = sub(A, B);
    else C = sub(B, A), cout = 0; i -- ) printf("%d", C[i]);

    return 0;
}

高精度除法 使用方法:

输入A和B 输出A/B的商和余数

code:

#include 
#include 
#include 

using namespace std;

vector div(vector &A, int b, int &r)
{
    vector 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;
    vector A;

    int B;
    cin >> a >> B;
    for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');

    int r;
    auto C = div(A, B, r);

    for (int i = C.size() - 1; i >= 0; i -- ) cout