题解 | #skew数#

pow头文件为#include<cmath>

#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;

int main() {

    string str = "";

    while (getline(cin, str)) {
        int res = 0;
        int len = str.size();
        for (int i = len - 1, k = 0; i >= 0; i--, k++) {
            res += (str[i] - '0') * (pow(2, k + 1) - 1);
        }
        cout << res << endl;
    }

    return 0;
}