给你四个字符串

string a = "What are you doing at the end of the world? Are you busy? Will you save us?";
string b = "What are you doing while sending \"";
string c = "\"? Are you busy? Will you send \"";
string d = "\"?";

有一个字符串数组

规律为如下：

f[0] = a;

f[n] = b + f[n-1] + c + f[n-1] + d;

现在给你一个n和k

要你求出f[n] 的第k个字符

如果k > f[n]的长度  就返回 '.'

题解：

开一个num数组，num[i] 表示 f[i] 的长度

由于num数组到后面已经超出了long long的范围了

所以我们只要求到num[55] ， num[56]已经爆了long long

后面的数组赋值给一个很大的数字就行 例如LLONG_MAX long long 范围的最大值

然后通过递归 把在f[n]的第k个位置 推到 在f[0]原来的位置 然后输出答案即可

#include<bits/stdc++.h>
using namespace std;
#define maxn 100005
#define ll long long
string a = "What are you doing at the end of the world? Are you busy? Will you save us?";
string b = "What are you doing while sending \"";
string c = "\"? Are you busy? Will you send \"";
string d = "\"?";
//string f1 = b + a + c + a + d;
int lena = a.size();
int lenb = b.size();
int lenc = c.size(); 
int lend = d.size();
ll num[maxn];
char ans;
ll nn, kk;
char dfs(int n, ll k) {
	if(k >= num[n]){
		return '.';
	}
	if(n == 0) {
		return a[k];
	}
	if(k < lenb) {
		return b[k];
	}
	k -= lenb;
	if(k < num[n-1]) {
		return dfs(n-1, k);	
	}
	k -= num[n-1];
	if(k < lenc) {
		return c[k];	
	}
	k -= lenc;
	if(k < num[n-1]) {
		return dfs(n-1, k);
	}
	k-=num[n-1];
	if(k < lend) {
		return d[k];
	}

}
int main() {
	num[0] = lena;
	for(int i = 1; i <= 55; i++) {
		num[i] = lenb + lenc + lend + 2*num[i-1];
		cout<<num[i]<<endl;
	}
	for(int i = 56; i <= 100000; i++) {
		num[i] = LLONG_MAX;
	}
//	cout<<LLONG_MAX<<endl;
	//9223372036854775807LL
	int q;
	cin>>q;
	while(q--) {
		cin>>nn>>kk;
		ans = dfs(nn, kk-1);
		printf("%c", ans);
	}
	cout<<endl;
	return  0;
}