38 lines
884 B
C++
38 lines
884 B
C++
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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//原始字符串
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string s;
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//预求的位置值
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LL x;
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//在n个长度的字符串中，查找位置x的字符
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char dfs(LL n, LL x) {
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//如果回到原始串，那么直接返回指定位置的字符
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if (n == s.size()) return s[x - 1];//之所以x-1，是因为字符串下标是从0开始的
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//字符串的中间点
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LL mid = n >> 1;
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//如果在后半部分(转化为前半部分的位置)
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if (x > mid)
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return dfs(mid, (x == mid + 1 ? mid : x - mid - 1));
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//如果在前半部分
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else
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return dfs(mid, x);
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}
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int main() {
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cin >> s >> x;
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//原始长度
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LL n = s.size();
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//不断乘2，翻倍，找出能够容纳x这个数字的最短长度
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while (n < x) n <<= 1;
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//递归
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cout << dfs(n, x);
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}
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