题解 | 10进制 VS 2进制

//字符串除法从高位开始考虑 字符串乘法从低位开始考虑,与平时数字计算相似
#include <iostream>
#include <string>
#include <stack>

using namespace std;

string Divide(string s, int x) {//一千位的十进制数用字符串处理
    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) {//十转二
    stack<int> sck;
    while (s.size() != 0) {
        int bit = (s[s.size()-1] - '0') % 2;     //整个数%2 和 最后一位%2的结果一样
        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 用于二转十 实现十进制字符串和十进制整型乘法
    int carry = 0;     //进位
    for (int i = str.size() - 1; i >= 0; i--) {//低位开始考虑
        int cur = x * (str[i] - '0') + carry;
        str[i] = cur % 10 + '0';
        carry = cur / 10;
    }
    if (carry != 0) {
        str = "1" + str; //乘2进位只能是1,如果乘大于2的整数,则应加上进位
    }
    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;   //在两个十进制数正常逐位相加时，每一位产生的进位是 0 或者 1
    }
    return str;
}

string convertFunc2(string s, int x) {//二转十
    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;
}