P3612 [USACO17JAN]Secret Cow Code S

// Problem: P3612 [USACO17JAN]Secret Cow Code S
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3612
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// User: Pannnn

#include 

using namespace std;

template
void debugVector(const T &a) {
    cout << "[ ";
    for (size_t i = 0; i < a.size(); ++i) {
        cout << a[i] << (i == a.size() - 1 ? " " : ", ");
    }
    cout << "]" << endl;
}

template
void debugMatrix2(const T &a) {
    for (size_t i = 0; i < a.size(); ++i) {
        debugVector(a[i]);
    }
}

template
using matrix2 = vector>;

template
vector> getMatrix2(size_t n, size_t m, T init = T()) {
    return vector>(n, vector(m, init));
}

template
using matrix3 = vector>>;

template
vector>> getMatrix3(size_t x, size_t y, size_t z, T init = T()) {
    return vector>>(x, vector>(y, vector(z, init)));
}

void printBigInteger(vector a) {
    for (size_t i = a.size() - 1; i >= 0; --i) {
        cout << a[i];
    }
}
vector addBigInteger(vector a, vector b) {
    vector res;
    int pre = 0;
    for (size_t i = 0; i < a.size() || i < b.size() || pre; ++i) {
        if (i < a.size()) pre += a[i];
        if (i < b.size()) pre += b[i];
        res.push_back(pre % 10);
        pre /= 10;
    }
    return res;
}

/*
    对于位置n上的字符，
    若n < str.length()，则可立即得到n
    否则
        由于每次扩容，字符串长度翻倍，所以令位置n上的字符由上一半的区间中的字符得来
        由于是循环移位，所以n处的字符，等于前一半区间中的对应位置上的前一位得来 
    
*/
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    
    string str;
    long long n;
    cin >> str >> n;
    
    
    while (n > str.length()) {
        long long end = str.length();
        while (end < n) {
            end *= 2;
        }
        n = end / 2 - (end - n);
        if (n - 1 == 0) {
            n = end / 2;
        } else {
            --n;
        }
    }
    cout << str[n - 1] << endl;
    return 0;
}