模板---高精度（加减乘除）

高精度

高精度加法

• C = A + B, A >= 0, B >= 0

#include <iostream>
#include <vector>

using namespace std;

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;

}

int main()
{
    string a, b;      //按字符串形式输入123456
    vector<int> A, B;
    cin >> a >> b;                      //按照 ‘6’‘5’‘4’‘3’‘2’‘1’压栈
    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 << C[i];
    cout << endl;

    return 0;
}

• 例题：AC 1500 趣味数字

---

高精度减法

• C = A - B, 满足A >= B, A >= 0, B >= 0

#include <iostream>
#include <vector>

using namespace std;

//判断是否有 A >= B
bool cmp(vector<int> &A, vector<int> &B)
{
    if(A.size() != B.size()) return A.size() > B.size();
    
    for (int i = A.size() - 1; i >= 0; i -- ){
        if (A[i] != B[i]) 
            return A[i] > B[i];
    }
    return true;
}

vector<int> sub(vector<int> &A, vector<int> &B)
{
    vector <int> C;
    int t = 0;
    
    for(int i = 0; i < A.size(); i ++ )
    {
        t = A[i] - t;
        if(i < B.size()) t = 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;
}


int main()
{
    string a, b;
    vector<int> A, B;
    cin >> a >> 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 <int> C;
    
    if(cmp(A, B)) C = sub(A, B);
    else{
        C = sub(B, A);
        cout << "-";
    }
    
    for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];
    cout << endl;
     
    return 0;
}

---

高精度乘低精度

• C = A * b, A >= 0, b > 0

#include<iostream>
#include<vector>

using namespace std;

vector<int> mul(vector<int> &A, int b)
{
    vector<int> C;
    int t = 0;

    for(int i = 0; i < A.size() || t; i ++ ) // 要么是i没循环完，要么是t不为0
    {
        if(i < A.size()) t += A[i] * b;
        C.push_back(t % 10);
        t /= 10;
    }

    return C;
}

int main()
{
    vector<int> A, C;
    string a;
    int b;
    cin >> a >> b;

    if( b == 0) cout << "0" << endl;
    else
    {
        for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0'); 

        C = mul (A, b);

        for (int i = C.size() - 1; i >= 0; i -- )  cout << C[i];
    }  
    return 0;
}

---

高精度除以低精度

• A / b = C ... r, A >= 0, b > 0

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

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;
    vector<int> A;
    cin >> a >> b;
    
    for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
    
    int r;
    vector<int> C = div(A, b, r);
    
    for (int i = C.size() - 1; i >=0; i -- ) cout << C[i];
    cout << endl << r << endl;
    
    return 0;
}