题解 | #10进制 VS 2进制#

#include <iostream>
#include <string>
#include <stack>

using namespace std;

string Divide(string s, int x) {
    int remainder = 0;
    for (int i = 0; i < s.size(); i++) {
        int cur = remainder * 10 + s[i] - '0';
        s[i] = cur / x + '0';
        remainder = cur % x;
    }
    int pos = 0;
    while (s[pos] == '0') {
        pos++;
    }
    return s.substr(pos);
}

string convertFunc1(string s, int x) {
    stack<int> sck;
    while (s.size() != 0) {
        int bit = (s[s.size() - 1] - '0') % 2;
        sck.push(bit);
        s = Divide(s, x);
    }
    string res;
    while (!sck.empty()) {
        res.push_back(sck.top() + '0');
        sck.pop();
    }
    return res;
}

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

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

string convertFunc2(string s, int x) {
    string ans = "0";
    for (int i = s.size() - 1; i >= 0; i--) {   //从最低二进制位开始
        ans = Multiple(ans, x);
        ans = Add(ans, s[i] - '0');
    }
    return ans;
}


int main() {
    string s;
    while (cin >> s) {
        cout << convertFunc2(convertFunc1(s, 2), 2);
    }
    return 0;
}

不太懂，留着学习