高精度加法

https://www.acwing.com/problem/content/793/

使用vector的高精度

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

vector<int> add(vector<int> A, vector<int> B) {
    //将长度更长的作为第一个参数
    if(A.size() < B.size()) return add(B, A);
    vector<int> ans;
    int t = 0;

    for(int i = 0; i < A.size(); i++) {
        t += A[i];
        if(i < B.size()) t += B[i];
        ans.push_back(t % 10);
        t /= 10;
    }
    //处理一下最后的进位
    while(t) {
        ans.push_back(t % 10);
        t /= 10;
    }
    return ans;
}

int main() {
    ios::sync_with_stdio(0); 
    cin.tie(0);

    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> ans = add(A, B);
    for(int i = ans.size() - 1; i >= 0; i--) cout << ans[i];
    return 0;
}

使用数组模型高精度（更快一点）

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 15;
int ans_size; // 答案的长度
int a[N], b[N], ans[N];

void add(int a[], int b[], int len) {
    int t = 0;
    for(int i = 0; i < len; i++) {
        t += a[i] + b[i];
        ans[ans_size++] = (t % 10);
        t /= 10;
    }
    //处理最后一位进位
    while(t) {
        ans[ans_size++] = (t % 10);
        t /= 10;
    }
    return ;
}

int main() {
    ios::sync_with_stdio(0); 
    cin.tie(0);

    string s1, s2;
    cin >> s1 >> s2;
    //逆序存储两个数
    for(int i = s1.size() - 1; i >= 0; i--) {
        a[s1.size() - 1 - i] = s1[i] - '0';
    }
    for(int i = s2.size() - 1; i >= 0; i--) {
        b[s2.size() - 1 - i] = s2[i] - '0';
    }

    int len = max(s1.size(), s2.size());
    add(a, b, len);

    //逆序输出答案
    for(int i = ans_size - 1; i >= 0; i--) cout << ans[i];
    return 0;
}

高精度减法

https://www.acwing.com/problem/content/description/794/

使用vector的高精度

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;

bool cmp(vector<int> A, vector<int> B) {
    if(A.size() != B.size()) return A.size() > B.size();
    else {
        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> ans;
    int t = 0;

    for(int i = 0; i < A.size(); i++) {
        t = A[i] - t;
        if(i < B.size()) t -= B[i];
        ans.push_back((t + 10) % 10);
        if(t < 0) t = 1;
        else t = 0;
    }

    //去除前导0
    while(ans.size() > 1 && ans.back() == 0) ans.pop_back();
    return ans;
}

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> ans;
    if(cmp(A, B)) ans = sub(A, B);
    else {
        cout << "-";
        ans = sub(B, A);
    }
    for(int i = ans.size() - 1; i >= 0; i--) cout << ans[i];
    return 0;
}

数组模拟高精度（快一点）

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 15;
int ans_size;
int a[N], b[N], ans[N];

bool cmp(string s1, string s2) {
    if(s1.size() != s2.size()) return s1.size() > s2.size();
    else {
        for(int i = 0; i < s1.size(); i++) {
            if(s1[i] != s2[i]) return s1[i] > s2[i];
        }
    }
    return true;
}

void sub(int a[], int b[], int len) {
    int t = 0;
    for(int i = 0; i < len; i++) {
        t = a[i] - b[i] - t;
        ans[ans_size++] = ((t + 10) % 10);
        if(t < 0) t = 1;
        else t = 0;
    }

    //这里因为a > b，所以在循环结束后必然t > 0
    //处理前导0
    while(ans_size > 1 && ans[ans_size - 1] == 0) ans_size--;
}

int main() {
    ios::sync_with_stdio(0); 
    cin.tie(0);

    string s1, s2;
    cin >> s1 >> s2;
    //逆序存储两个数
    for(int i = s1.size() - 1; i >= 0; i--) {
        a[s1.size() - 1 - i] = s1[i] - '0';
    }
    for(int i = s2.size() - 1; i >= 0; i--) {
        b[s2.size() - 1 - i] = s2[i] - '0';
    }
    int len = max(s1.size(), s2.size());
    if(!cmp(s1, s2)) {
        cout << "-" ;
        sub(b, a, len);
    }else sub(a, b, len);

    //逆序输出答案
    for(int i = ans_size - 1; i >= 0; i--) cout << ans[i];
    return 0;
}

高精度乘法

https://www.acwing.com/problem/content/795/

使用vector

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;

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

    for(int i = 0; i < A.size(); i++) {
        t += A[i] * b;
        ans.push_back(t % 10);
        t /= 10;
    }

    //加上最后一位进位
    if(t) ans.push_back(t);
    //去除前导0
    while(ans.size() > 1 && ans.back() == 0) ans.pop_back();
    return ans;
}

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');
    vector<int> ans = mul(A, b);

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

数组模拟

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 15;
int ans_size;
int a[N], ans[N];

void mul(int a[], int b, int len) {
    int t = 0;
    for(int i = 0; i < len; i++) {
        t += a[i] * b;
        ans[ans_size++] = (t % 10);
        t /= 10;
    }

    //处理进位位
    while(t) {
        ans[ans_size++] = (t % 10);
        t /= 10;
    }

    //去除前导0
    while(ans_size > 1 && ans[ans_size - 1] == 0) ans_size--;
}

int main() {
    ios::sync_with_stdio(0); 
    cin.tie(0);

    string s1;
    int b;
    cin >> s1 >> b;
    for(int i = s1.size() - 1; i >= 0; i--) {
        a[s1.size() - 1 - i] = s1[i] - '0';
    }
    int len = s1.size();
    mul(a, b, len);

    for(int i = ans_size - 1; i >= 0; i--) cout << ans[i];
    return 0;
}

高精度除法

https://www.acwing.com/problem/content/796/

使用vector

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;

vector<int> div(vector<int> A, int b, int &r) {
    vector<int> ans;
    for(int i = 0; i < A.size(); i++) {
        r = r * 10 + A[i];
        ans.push_back(r / b);
        r %= b;
    }
    return ans;
}

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

    //不用颠倒次序，从高位算起
    for(int i = 0; i < a.size(); i++) A.push_back(a[i] - '0');
    int r = 0, k = 0;
    vector<int> ans = div(A, b, r);

    //去除前导0
    while(k < ans.size() - 1 && ans[k] == 0) k++;
    for( ; k < ans.size(); k++) cout << ans[k];
    cout << endl << r << endl;
    return 0;
}

数组模拟

ps: 如果b最大可以取到 1 * 10^9，那么r在计算过程中就会爆int。就需要用long long 来存储

下面洛谷这题就是：
https://www.luogu.com.cn/problem/P1480

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 15;
typedef long long LL;
int ans_size, b;
LL r;
int a[N], ans[N];

void div(int a[], int b, int len, LL &r) {
    for(int i = 0; i < len; i++) {
        r = r * 10 + a[i];
        ans[ans_size++] = (r / b);
        r %= b;
    }
}

int main() {
    ios::sync_with_stdio(0); 
    cin.tie(0);

    string s1;
    cin >> s1 >> b;
    //除法可以顺序存储，不需要逆序
    //（你想逆序存储也可以~）
    for(int i = 0; i < s1.size(); i++) {
        a[i] = s1[i] - '0';
    }
    int len = s1.size();

    div(a, b, len, r);
    //去除前导0
    int i = 0;
    while(i < ans_size - 1 && ans[i] == 0) i++;
    for( ; i < ans_size; i++) cout << ans[i];
    // cout << endl << r << endl;
    return 0;
}