高精度

加法

不圧位代码

#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;
    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];
    cout << endl;

    return 0;
}
/*

	思想转化为字符串模拟加法从左向右加，同时注意进位的情况

*/

压9位的代码

#include <iostream>
#include <vector>

using namespace std;

const int base = 1000000000;

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 % base);
        t /= base;
    }

    if (t) C.push_back(t);
    return C;
}

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

    for (int i = a.size() - 1, s = 0, j = 0, t = 1; i >= 0; i -- )
    {
        s += (a[i] - '0') * t;
        j ++, t *= 10;
        if (j == 9 || i == 0)
        {
            A.push_back(s);
            s = j = 0;
            t = 1;
        }
    }
    for (int i = b.size() - 1, s = 0, j = 0, t = 1; i >= 0; i -- )
    {
        s += (b[i] - '0') * t;
        j ++, t *= 10;
        if (j == 9 || i == 0)
        {
            B.push_back(s);
            s = j = 0;
            t = 1;
        }
    }

    auto C = add(A, B);

    cout << C.back();
    for (int i = C.size() - 2; i >= 0; i -- ) printf("%09d", C[i]);
    cout << endl;

    return 0;
}

/*
	思想利用int的最大范围尽可能的减少空间和时间的消耗，但要注意最后的输出除了头以外都要保证长度为9用零来填充

*/

减法

#include <iostream>
#include <vector>

using namespace std;

//此时是一个倒叙的数据流
bool cmp(vector<int> &A, vector<int> &B)
{
    if (A.size() != B.size()) return A.size() > B.size();
	// 此时在数据流中越往后的元素在元数据中反而在前面
    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;
    for (int i = 0, t = 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;
    }

	//减法可能产生很多前置零需要删除
    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;
}
/*
	感悟：注意所有边界值的判定和给定数据流的定义 千万不要混淆，不要忽略一些奇特的地方

*/

乘法

#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 ++ )
    {
        if (i < A.size()) t += A[i] * b;
        C.push_back(t % 10);
        t /= 10;
    }
    // 模拟数学的乘法 将我们得到的数保存每一次加到底部

    while (C.size() > 1 && C.back() == 0) C.pop_back();

    return C;
}


int main()
{
    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 -- ) printf("%d", C[i]);

    return 0;
}

除法

#include<bits/stdc++.h>
using namespace std;
int main() {
	string a;
	int b;
	cin>>a>>b;
	vector<int>num;
//	除数
	vector<int>temp;
//	余数
	int cur = 0;
	for(int i=0;i<a.size();i++){
		cur*=10;
		cur+=a[i] - '0';
		if (cur >= b) {
			temp.push_back(cur/b) ;
			cur = cur%b;
		} else {
			temp.push_back(0);
		}
		
		/*
			可以改成 cur = cur*10 + a[i];
					temp.push_back(cur/b);
					cur = cur%b;
					整体上会更精简
		*/
		
		
	}
	
	/*
		因为vector的底层是数组在erase掉一个元素的时间复杂度为n但是pop_back是o(1)可以改成
			reverse(temp.begin(),temp.end());
            while(temp.back() == 0 && temp.size() > 1){
                temp.pop_back();
            }
	*/
	while(temp.front() == 0 && temp.size() > 1){
		temp.erase(temp.begin());
	}
	for(int i=0;i<temp.size();i++){
		cout<<temp[i];
	}
	cout<<endl<<cur<<endl;
	
	
}
/*

12313123123123123123
12313
1000009999441494
7501
12313123123123115622

*/

#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;
    vector<int> 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 << C[i];

    cout << endl << r << endl;

    return 0;
}