题解 | #skew数#

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main()
{
    string skew;
    while (cin >> skew)
    {
        reverse(skew.begin(), skew.end());
        int res = 0;
        for (int i = 0; i < skew.size(); ++i)
            res += (skew[i]-'0') * ((2 << i) - 1); //并不需要去用pow，2的次方直接用位运算即可
        cout << res << endl;
    }
    return 0;
}