NWERC2016H

题目大意

求两个gray code的间距，gray code生成方式如下

gray_code[n][0...2^{n-1}-1]='0'+gray_code[n-1][0...2^{n-1}]

gray_code[n][2^{n-1}...2^{n}]='1'+gray_code[n-1][2^{n-1}...0]

简要题解

递归算每个gray code是第几个就好

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

using LL = long long;

LL getPos(string t) {
    if (t.length() == 1) {
        return t[0] - '0';
    }
    if (t[0] == '1') {
        return (1LL << t.length()) - getPos(t.substr(1, t.length())) - 1;
    }
    return getPos(t.substr(1, t.length()));
}

int main() {
    int n; cin >> n;
    string str;
    cin >> str;
    LL pos1 = getPos(str);
    cin >> str;
    LL pos2 = getPos(str);
    cout << pos2 - pos1 - 1 << endl;
    return 0;
}