Codeforces Round #449 (Div. 2) C. Nephren gives a riddle 递归

http://codeforces.com/contest/897/problem/C

又是正视了自己的递归水平，55555。

题意：递归又递归的字符串   下面q个询问，第fn个字符串的第k个字符是啥，如果fn没k这么长就是.

做法：递归长度，n很大，先处理n，如果n k一直很大，可以先判断他是在哪个范围，至少得突破最左边的边缘区，就是34那一块，所以

先不断减小n，k达到可以处理的阶段。

另外就是不断递归了



#include<bits/stdc++.h>
using namespace std;
string p = "What are you doing at the end of the world? Are you busy? Will you save us?", q = "What are you doing while sending \"", r = "\"? Are you busy? Will you send \"", s = "\"?";
long long x[100001], y, z, t;
char f(long long a, long long b)
{
if(b>=x[a]) return '.';
if(a == 0) return p[b];
if(b<q.size()) return q[b];
b-=q.size();
if(b<x[a-1]) return f(a-1, b);
b-=x[a-1];
if(b<r.size()) return r[b];
b-=r.size();
if(b<x[a-1]) return f(a-1, b);
b-=x[a-1];
if(b<s.size()) return s[b];
return '.';
}
int main(void)
{
x[0] = p.size();
for(int i=1;i<=55;i++) x[i] = 2*x[i-1]+q.size()+r.size()+s.size();
for(int i=56;i<=100000;i++) x[i] = LLONG_MAX;
for(cin>>t;t--;) cin>>y>>z, cout<<f(y, z-1);
}

f0 = "What are you doing at the end of the world? Are you busy? Will you save us?"
p1 = "What are you doing while sending \""
p2 = "\"? Are you busy? Will you send \""
p3 = "\"?"
f = []
def dfs(n, k):
if k > f
: return '.'
if n == 0: return f0[k-1]
if k <= 34: return p1[k-1]
if k <= f[n-1] + 34: return dfs(n - 1, k - 34)
if k <= f[n-1] + 66: return p2[k - f[n-1] - 34 - 1]
if k <= 2 * f[n-1] + 66: return dfs(n - 1, k - f[n-1] - 66)
return p3[k - f[n-1] * 2 - 66 - 1]

f.append(75)
cnt = 0
for i in range(0, 100):
f.append(f[cnt] * 2 + 68)
cnt += 1

def solve():
n, k =map(int, input().split(' '))
while n > 56 and k > 34:
k -= 34
n -= 1
if k <= 34:
print(p1[k - 1], end='')
else:
print(dfs(n, k), end='')
def main():
case = 1
case = int(input())
for case in range(case):
solve()

main()