题解 | #10进制 VS 2进制#

//
// Created by Virtus on 2024/2/22.
//
#include <iostream>
#include <string>
#include <vector>

using namespace std;

// 字符串除法
string stringDivide(string str, int x) {
    int remainder = 0;
    for (char & i : str) {
        int current = remainder * 10 + i - '0';
        i = current / x + '0';
        remainder = current % x;
    }
    int pos = 0;
    while (str[pos] == '0') {
        pos++;
    }
    return str.substr(pos);
}

// 字符串乘法
string stringMultiple(string str, int x) {
    int carry = 0;
    for (int i = str.size() - 1; i >= 0; --i) {
        int current = x * (str[i] - '0') + carry;
        str[i] = current % 10 + '0';
        carry = current / 10;
    }
    if (carry != 0) {
        str = '1' + str;
    }
    return str;
}

// 字符串加法
string stringAdd(string str, int x) {
    int carry = x;
    for (int i = str.size() - 1; i >= 0; --i) {
        int current = (str[i] - '0') + carry;
        str[i] = current % 10 + '0';
        carry = current / 10;
    }
    if (carry != 0) {
        str = "1" + str;
    }
    return str;
}

int main() {
    string str;
    while (cin >> str) {
        vector<int> binary;
        while(!str.empty()) {
            int last = str[str.size() - 1] - '0';
            binary.push_back(last % 2);
            str = stringDivide(str, 2);
        }
        string answer = "0";
        for (int i : binary) {
            answer = stringMultiple(answer, 2);
            answer = stringAdd(answer, i);
        }
        cout << answer << endl;
    }
    return 0;
}