题解 | #10进制 VS 2进制#

#include <iostream>
using namespace std;
string divide2(string s) {//大数除二
    string ans;
    int remainer = 0;
    for (int i = 0; i < s.size(); i++) {
        int temp = remainer * 10 + s[i] - '0';
        ans += temp / 2 + '0';
        remainer = temp % 2;
    }

    int pos = 0;
    while (ans[pos] == '0') pos++;
    ans = ans.substr(pos);
    if (ans.size() == 0) ans = "0";
    return ans;
}
string multiply2(string s) {//大数乘二
    string ans;
    int carry = 0;
    for (int i = s.size() - 1; i >= 0; i--) {
        int temp = (s[i] - '0') * 2 + carry;
        ans = to_string(temp % 10) + ans;
        carry = temp / 10;
    }
    if (carry != 0) ans = to_string(carry) + ans;
    return ans;
}
string add(string s,int a) {//大数加0或1
    string ans;
    int carry = a;
    for (int i = s.size() - 1; i >= 0; i--) {
        int temp = s[i] - '0' + carry;
        ans = to_string(temp % 10) + ans;
        carry = temp / 10;
    }
    if (carry != 0) ans = to_string(carry) + ans;
    return ans;
}
string toBinary(string decStr) {//转二进制取余除二
    string ans;
    if (decStr == "0") ans = "0";
    while (decStr != "0") {
        ans = to_string((decStr[decStr.size() - 1] - '0') % 2) + ans;
        decStr = divide2(decStr);
    }
    return ans;
}
string reverseToDec(string binaryStr) {//二进制转10进制从高到低，逆序则从低到高
    string ans = "0";
    for (int i = binaryStr.size()-1; i >=0; i--) {
        ans = multiply2(ans);
        ans = add(ans, binaryStr[i] - '0');
    }
    return ans;
}
int main() {
    string a;
    while (cin >> a ) { // 注意 while 处理多个 case
        cout << reverseToDec(toBinary(a)) << endl;
    }
}
// 64 位输出请用 printf("%lld")