【开源】int,long long去一边去：高精度大合集！

加法 \(add\)

string add(string s1, string s2) {    //时间复杂度 O(log n)
    string res = "";
    int c = 0, i = 0;
    while (i<s1.size() || i < s2.size() || c>0) {
        int a = (i<s1.size()) ? (s1[s1.size() - i - 1] - '0') : 0;
        int b = (i<s2.size()) ? (s2[s2.size() - i - 1] - '0') : 0;
        res += ((a + b + c) % 10) + '0';
        c = (a + b + c) / 10;
        i++;
    }
    reverse(res.begin(), res.end());
    return res;
}

减法 \(sub\)

string sub(string a, string b) {	//时间复杂度 O(log n)
    string res;
    int c = 0;
    if (a.size() < b.size()) a = string(b.size() - a.size(), '0') + a;
    else if (a.size() > b.size()) b = string(a.size() - b.size(), '0') + b;
    for (int i = a.size() - 1; i >= 0; i--) {
        int d = a[i] - b[i] - c;
        if (d < 0) d += 10, c = 1;
        else c = 0;
        res.push_back(d + '0');
    }
    reverse(res.begin(), res.end());
    while (res.size() > 1 && res[0] == '0') {
        res.erase(res.begin());
    }
    return res;
}

乘法 \(mul\)

string mul(string num1, string num2) {	//时间复杂度 O(n^2)
    string res(int(num1.length()) + int(num2.length()), '0');
    for (int i = int(num1.length()) - 1; i >= 0; i--) {
        for (int j = int(num2.length()) - 1; j >= 0; j--) {
            int pd = (num1[i] - '0') * (num2[j] - '0');
            int p1 = i + j;
            int p2 = i + j + 1;
            int sum = pd + (res[p2] - '0');
            res[p2] = sum % 10 + '0';
            res[p1] += sum / 10;
        }
    }
    int i = 0;
    while (i<int(res.length()) && res[i] == '0') {
        i++;
    }
    return i == int(res.length()) ? "0" : res.substr(i);
}

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#include <iostream>
#include <vector>
using namespace std;
// 将字符串转换为整数数组
vector<int> strToVec(string str) {
	vector<int> vec;
	for (int i = str.length() - 1; i >= 0; i--) {
		vec.push_back(str[i] - '0');
	}
	return vec;
}
// 将整数数组转换为字符串
string vecToStr(vector<int> vec) {
	string str = "";
	for (int i = vec.size() - 1; i >= 0; i--) {
		str += to_string(vec[i]);
	}
	return str;
}
// 高精度加法
vector<int> add(vector<int> num1, vector<int> num2) {
	vector<int> result;
	int carry = 0;
	int len1 = num1.size();
	int len2 = num2.size();
	int maxLen = max(len1, len2);
	for (int i = 0; i < maxLen; i++) {
		int digit1 = i < len1 ? num1[i] : 0;
		int digit2 = i < len2 ? num2[i] : 0;
		int sum = digit1 + digit2 + carry;
		result.push_back(sum % 10);
		carry = sum / 10;
	}
	if (carry != 0) {
		result.push_back(carry);
	}
	return result;
}
// 高精度乘法
vector<int> mul(vector<int> num1, vector<int> num2) {
	int len1 = num1.size();
	int len2 = num2.size();
	// 递归终止条件
	if (len1 == 0 || len2 == 0) {
		return vector<int>();
	}
	// 递归基
	if (len1 == 1 && len2 == 1) {
		vector<int> result;
		int product = num1[0] * num2[0];
		result.push_back(product % 10);
		if (product >= 10) {
			result.push_back(product / 10);
		}
		return result;
	}
	// 将数字分为两部分
	int mid = min(len1, len2) / 2;
	vector<int> num1Low(num1.begin(), num1.begin() + mid);
	vector<int> num1High(num1.begin() + mid, num1.end());
	vector<int> num2Low(num2.begin(), num2.begin() + mid);
	vector<int> num2High(num2.begin() + mid, num2.end());
	// 递归计算
	vector<int> z0 = mul(num1Low, num2Low);
	vector<int> z1 = mul(num1High, num2High);
	vector<int> z2 = mul(add(num1Low, num1High), add(num2Low, num2High));
	z2 = add(z2, z0);
	z2 = add(z2, z1);
	// 合并结果
	vector<int> result;
	result.insert(result.end(), z0.begin(), z0.end());
	result.insert(result.begin() + mid, z2.begin(), z2.end());
	result.insert(result.begin() + 2 * mid, z1.begin(), z1.end());
	return result;
}
int main() {
	string str1, str2;
	cout << "请输入两个整数：" << endl;
	cin >> str1 >> str2;
	vector<int> num1 = strToVec(str1);
	vector<int> num2 = strToVec(str2);
	vector<int> result = mul(num1, num2);
	string strResult = vecToStr(result);
	cout << "两个整数的乘积为：" << endl;
	cout << strResult << endl;
	return 0;
}

除法&取余 \(divi\)

两个正数相除，商为\(quotient\),余数为\(residue\)

int compare(string str1, string str2) {
	if (str1.length() > str2.length()) return 1;
	else if (str1.length() < str2.length())  return -1;
	else return str1.compare(str2);
}
string sub(string a, string b) {	//时间复杂度 O(log n)
	string res;
	int c = 0;
	if (a.size() < b.size()) a = string(b.size() - a.size(), '0') + a;
	else if (a.size() > b.size()) b = string(a.size() - b.size(), '0') + b;
	for (int i = a.size() - 1; i >= 0; i--) {
		int d = a[i] - b[i] - c;
		if (d < 0) d += 10, c = 1;
		else c = 0;
		res.push_back(d + '0');
	}
	reverse(res.begin(), res.end());
	while (res.size() > 1 && res[0] == '0') {
		res.erase(res.begin());
	}
	return res;
}
string mul(string num1, string num2) {   //时间复杂度 O(n^2)
	string res(int(num1.length()) + int(num2.length()), '0');
	for (int i = int(num1.length()) - 1; i >= 0; i--) {
		for (int j = int(num2.length()) - 1; j >= 0; j--) {
			int pd = (num1[i] - '0') * (num2[j] - '0');
			int p1 = i + j;
			int p2 = i + j + 1;
			int sum = pd + (res[p2] - '0');
			res[p2] = sum % 10 + '0';
			res[p1] += sum / 10;
		}
	}
	int i = 0;
	while (i<int(res.length()) && res[i] == '0') {
		i++;
	}
	return i == int(res.length()) ? "0" : res.substr(i);
}
void divi(string str1, string str2, string& quotient, string& residue) {
	quotient = residue = "";
	if (str2 == "0") {
		quotient = residue = "ERROR";
		return;
	}
	if (str1 == "0") {
		quotient = residue = "0";
		return;
	}
	int res = compare(str1, str2);
	if (res < 0) {
		quotient = "0";
		residue = str1;
		return;
	}
	else if (res == 0) {
		quotient = "1";
		residue = "0";
		return;
	}
	else {
		int len1 = str1.length();
		int len2 = str2.length();
		string tempstr;
		tempstr.append(str1, 0, len2 - 1);
		for (int i = len2 - 1; i < len1; i++) {
			tempstr = tempstr + str1[i];
			tempstr.erase(0, tempstr.find_first_not_of('0'));
			if (tempstr.empty()) tempstr = "0";
			for (char ch = '9'; ch >= '0'; ch--) {
				string str, tmp;
				str = str + ch;
				tmp = mul(str2, str);
				if (compare(tmp, tempstr) <= 0) {
					quotient = quotient + ch;
					tempstr = sub(tempstr, tmp);
					break;
				}
			}
		}
		residue = tempstr;
	}
	quotient.erase(0, quotient.find_first_not_of('0'));
	if (quotient.empty()) quotient = "0";
}