Codeforces 897 C.Nephren gives a riddle-递归
Nephren is playing a game with little leprechauns.
She gives them an infinite array of strings, f0... ∞.
f0 is "What are you doing at the end of the world? Are you busy? Will you save us?".
She wants to let more people know about it, so she defines fi = "What are you doing while sending "fi - 1"? Are you busy? Will you send "fi - 1"?" for all i ≥ 1.
For example, f1 is
"What are you doing while sending "What are you doing at the end of the world? Are you busy? Will you save us?"? Are you busy? Will you send "What are you doing at the end of the world? Are you busy? Will you save us?"?". Note that the quotes in the very beginning and in the very end are for clarity and are not a part of f1.
It can be seen that the characters in fi are letters, question marks, (possibly) quotation marks and spaces.
Nephren will ask the little leprechauns q times. Each time she will let them find the k-th character of fn. The characters are indexed starting from 1. If fn consists of less than k characters, output '.' (without quotes).
Can you answer her queries?
The first line contains one integer q (1 ≤ q ≤ 10) — the number of Nephren's questions.
Each of the next q lines describes Nephren's question and contains two integers n and k (0 ≤ n ≤ 105, 1 ≤ k ≤ 1018).
One line containing q characters. The i-th character in it should be the answer for the i-th query.
3
1 1
1 2
1 111111111111
Wh.
5
0 69
1 194
1 139
0 47
1 66
abdef
10
4 1825
3 75
3 530
4 1829
4 1651
3 187
4 584
4 255
4 774
2 474
Areyoubusy
For the first two examples, refer to f0 and f1 given in the legend.
题意看了好久，才明白什么意思。
除了f0 ，其他的都是在固定的那两句话和问号中插入上一个f的内容。就是假设a b c ，在a和b中间和b和c中间插入上一个f的内容。
这个题先处理一下1e5的数据的长度。首先先初始化，初始化的数组存的长度只要大于1e5就可以，因为题目说f的长度大于1e5就输出.(输出点.)
为什么一定要初始化呢，假设是第53个f的长度大于1e5,53之后的f的长度都应该是大于1e5的，但是处理的时候53以后的已经跳出了。
所以在一开始赋初值的的时候就先把长度赋值大于1e5就可以。
我写的1e5+1，哈哈哈。
然后就是递归处理，依次判断然后减去就可以。
对于问号的处理可以先递归模拟一下前几个f就可以。
其他的没什么坑了。
直接代码:
1 //C.Sorting Railway Cars 递归
2 #include<iostream>
3 #include<cstring>
4 #include<cstdio>
5 #include<cmath>
6 #include<algorithm>
7 using namespace std;
8 typedef long long ll;
9 const int N=1e5+10;
10 char a0[]={"What are you doing at the end of the world? Are you busy? Will you save us?"};
11 char a1[]={"What are you doing while sending \""};
12 char a2[]={"\"? Are you busy? Will you send \""};
13 char a3[]={"\"?"};
14 ll len[N];
15 ll l0,l1,l2,l3;
16 char digui(ll n,ll k){
17 if(n==0)return a0[k-1];
18 if(k<=l1)return a1[k-1];k-=l1;
19 if(k<=len[n-1])return digui(n-1,k);k-=len[n-1];
20 if(k<=l2)return a2[k-1];k-=l2;
21 if(k<=len[n-1])return digui(n-1,k);k-=len[n-1];
22 return a3[k-1];
23 }
24 int main(){
25 l0=strlen(a0);
26 l1=strlen(a1);
27 l2=strlen(a2);
28 l3=strlen(a3);
29 len[0]=l0;
30 for(ll i=1;i<=1e5;i++)
31 len[i]=1e18+1;
32 for(int i=1;i<=1e5;i++){
33 len[i]=len[i-1]*2+l1+l2+l3;
34 if(len[i]>1e18)break;
35 }
36 char ans[N];
37 ll q,n,k;
38 while(~scanf("%lld",&q)){
39 memset(ans,0,sizeof(ans));
40 ll h=0;
41 while(q--){
42 scanf("%lld%lld",&n,&k);
43 if(k>len[n])ans[h++]='.';
44 else ans[h++]=digui(n,k);
45 }
46 for(int i=0;i<h;i++)
47 cout<<ans[i];
48 }
49 return 0;
50 }
cf的测评姬小姐姐最近可能心情不好，打cf的时候测题好慢。
加油啊，简直要菜哭了。
咸鱼加油，好好对待我的id，ZERO。
向学长学习ZEROm。
