CF897C Nephren gives a riddle

Posted 2020-10-16 王宜鸣

思路：

递归。

比赛的时候脑抽了len[]没算够，wa了几次。

实现：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 using ll = long long;
 4 const string s = "What are you doing at the end of the world? Are you busy? Will you save us?";
 5 const string t = "What are you doing while sending \"";
 6 const string v = "\"? Are you busy? Will you send \"";
 7 const ll len_s = s.length();
 8 const ll len_t = t.length();
 9 const ll len_v = v.length();
10 ll len[100005];
11 char dfs(ll n, ll k)
12 {
13     if (n == 0) 
14     {
15         if (k >= 1 && k <= len_s) return s[k - 1];
16         else return ‘.‘;
17     }
18     else if (n >= 54)
19     {
20         if (k <= len_t) return t[k - 1];
21         return dfs(n - 1, k - len_t);
22     }
23     else 
24     {
25         if (k <= len_t) return t[k - 1];
26         else if (k > len_t && k <= len_t + len[n - 1])
27             return dfs(n - 1, k - len_t);
28         else if (k > len_t + len[n - 1] && k <= len_t + len[n - 1] + len_v)
29             return v[k - len_t - len[n - 1] - 1];
30         else if (k > len_t + len[n - 1] + len_v && k <= len_t + 2 * len[n - 1] + len_v)
31             return dfs(n - 1, k - len_t - len[n - 1] - len_v);
32         else if (k == len_t + 2 * len[n - 1] + len_v + 1)
33             return ‘\"‘;
34         else if (k == len_t + 2 * len[n - 1] + len_v + 2)
35             return ‘?‘;
36         else return ‘.‘;
37     }
38 }
39 int main()
40 {
41     len[0] = (ll)len_s;
42     for (int i = 1; i <= 53; i++)
43     {
44         len[i] = len[i - 1] * 2 + len_t + len_v + 2;
45     }
46     ll q, n, k;
47     cin >> q;
48     while (q--)
49     {
50         cin >> n >> k;
51         cout << dfs(n, k);
52     }
53     return 0;
54 }