高精度

高精度

高精度 + 高精度

https://www.luogu.com.cn/problem/P1601

题意：求 A + B 。

#include <bits/stdc++.h>
using namespace std;

const char nl = '\n';
//const int N = ;

// C = A + B
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(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    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) cout << C[i];

    return 0;
}

高精度 - 高精度

https://www.luogu.com.cn/problem/P2142

题意：求 A - B 。（第二个可能比第一个大）

#include <bits/stdc++.h>
using namespace std;

const char nl = '\n';
//const int N = ;

// 判断是否 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 false;
    return true;
}

// C = A - B
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(); // 去掉前导0

    return C;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    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) cout << C[i];
    }else{
        auto C = sub(B, A);
        cout << '-';
        for (int i = C.size() - 1; i >= 0; --i) cout << C[i];
    }

    return 0;
}

高精度 * 低精度

题意：求 C = A * b 。

#include <bits/stdc++.h>
using namespace std;

const char nl = '\n';
//const int N = ;

// C = A * b
vector<int> mul(vector<int> &A, int b){
    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size(); ++i){
        t += A[i] * b;
        C.push_back(t % 10);
        t /= 10;
    }

    while (t){
        C.push_back(t % 10);
        t /= 10;
    }

    return C;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    string a;
    int b;

    cin >> a >> b;

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

    auto C = mul(A, b);

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

    return 0;
}

高精度 / 低精度

https://www.luogu.com.cn/problem/P1480

题意：求 C = A / b 。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

const char nl = '\n';
//const int N = ;

vector<int> mul(vector<int> &A, int b, LL &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(); // 去掉前导0

    return C;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    string a;
    int b;

    cin >> a >> b;

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

    LL r;
    auto C = mul(A, b, r);

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

    return 0;
}