无话可说高精度

高精加

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#include <iostream>
#include <vector>

using namespace std;

const int N = 1e6 + 10;

vector <int> add(vector<int> &A,vector<int> &B)
{
    vector <int> C;

    int t = 0;
    for(int i = 0; i < A.size() || i < B.size(); i ++)
    {
        if(i < A.size()) t += A[i];
        if(i < B.size()) t += B[i];
        C.push_back(t%10);
        t /= 10;
    }
    if(t) C.push_back(1);
    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');


auto C = add(A,B);

for(int i = C.size()-1;i >= 0; i --) printf("%d",C[i]);
    return 0;
}

高精度减

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 #include <iostream>
#include <vector>

using namespace std;

const int N = 1e6 + 10;

bool cmp(vector<int> &A,vector<int> &B)
{
    if(A.size() != B.size()) return A.size() > B.size();

    for(int i = A.size() ; 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 -= 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');

if(cmp(A,B))
{
    auto C = sub(A,B);
    for(int i = C.size()-1; i >= 0; i --) printf("%d",C[i]);
}
else
{
    auto C = sub(A,B);

    printf("-");
    for(int i = C.size()-1;i >= 0; i --) printf("%d",C[i]);
}



    return 0;
}

高精乘单精 

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// 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;
}


高精除单精

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// 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;
}