【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";
}
