time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

What are you doing at the end of the world? Are you busy? Will you save us?

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?

Input

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).

Output

One line containing q characters. The i-th character in it should be the answer for the i-th query.

Examples
input
3
1 1
1 2
1 111111111111
output
Wh.
input
5
0 69
1 194
1 139
0 47
1 66
output
abdef
input
10
4 1825
3 75
3 530
4 1829
4 1651
3 187
4 584
4 255
4 774
2 474
output
Areyoubusy
Note

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。

Codeforces 897 C.Nephren gives a riddle-递归的更多相关文章

  1. Codeforces Round #449 (Div. 2)-897A.Scarborough Fair(字符替换水题) 897B.Chtholly's request(处理前一半) 897C.Nephren gives a riddle(递归)

    A. Scarborough Fair time limit per test 2 seconds memory limit per test 256 megabytes input standard ...

  2. Codeforces Round #449 [ C/A. Nephren gives a riddle ] [ D/B. Ithea Plays With Chtholly ]

    PROBLEM C/A. Nephren gives a riddle 题 http://codeforces.com/contest/896/problem/A codeforces 896a 89 ...

  3. CodeForces - 896A Nephren gives a riddle

    A. Nephren gives a riddle time limit per test 2 seconds memory limit per test 256 megabytes input st ...

  4. CF&&CC百套计划1 Codeforces Round #449 A. Nephren gives a riddle

    http://codeforces.com/contest/896/problem/A 第i个字符串嵌套第i-1个字符串 求第n个字符串的第k个字母 dfs #include<map> # ...

  5. Codeforces 897C Nephren gives a riddle:模拟【珂学】

    题目链接:http://codeforces.com/contest/897/problem/C 题意: 给你一些字符串: A: [What are you doing at the end of t ...

  6. 寒假特训——搜索——H - Nephren gives a riddle

    What are you doing at the end of the world? Are you busy? Will you save us? Nephren is playing a gam ...

  7. A. Nephren gives a riddle

    What are you doing at the end of the world? Are you busy? Will you save us? Nephren is playing a gam ...

  8. CodeForces - 896D :Nephren Runs a Cinema(卡特兰数&组合数学---比较综合的一道题)

    Lakhesh loves to make movies, so Nephren helps her run a cinema. We may call it No. 68 Cinema. Howev ...

  9. Codeforces 897 B.Chtholly's request-思维题(处理前一半)

      B. Chtholly's request   time limit per test 2 seconds memory limit per test 256 megabytes input st ...

随机推荐

  1. The 2016 ACM-ICPC Asia Shenyang Regional Contest

    A. Thickest Burger 大数 × 2 + 小数 #include <cstdio> #include <algorithm> using namespace st ...

  2. input框中的必填项之取消当前input框为必填项

    html5新增了一个required属性,可以使用这个属性对文本框设置必填项,直接在input文本框上添加required即可 . 效果如图:   

  3. 图文详解安装PHP运行环境

    一.什么是PHP运行环境 能够理解人与计算机交流时语言软件,通常指解释PHP编程语言的软件. 例如: PHP(代码) 需要PHP超文本预编译器(软件). Java需要JVM虚拟机 二.安装PHP运行环 ...

  4. leetcode 【 Set Matrix Zeroes 】python 实现

    题目: Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place. cl ...

  5. C 语言 习题 1-9

    练习1-9 编写一个将输入复制到输出的程序,并将其中连续的多个空格用一个空格代替. #include <stdio.h> int main(int argc, char const *ar ...

  6. JS 关于 URL 的编码或解码方法

    URL的合法字符 URL的合法字符表示再浏览器的地址栏中不会被转义的字符,有两种: URL元字符:分号(;),逗号(’,’),斜杠(/),问号(?),冒号(:),at(@),&,等号(=),加 ...

  7. CSU-2220 Godsend

    题目链接 http://acm.csu.edu.cn:20080/csuoj/problemset/problem?pid=2220 题目 Description Leha somehow found ...

  8. 实用拜占庭容错算法PBFT

    实用拜占庭容错算法PBFT 实用拜占庭容错算法PBFT 96 乔延宏 2017.06.19 22:58* 字数 1699 阅读 4972评论 0喜欢 11 分布式架构遭遇的问题 分布式架构会遭遇到以下 ...

  9. algorithm 头文件

    非修改性序列操作(12个) 循环 对序列中的每个元素执行某操作 for_each() 查找 在序列中找出某个值的第一次出现的位置 find() 在序列中找出符合某谓词的第一个元素 find_if() ...

  10. 【转】UGUI(小地图的实现)与游戏关卡选择的简单实现

    http://www.jianshu.com/p/68637029e9df 游戏中小地图的实现(场景用简单Cube组成先搭建如下图场景,真实场景实现方法也是一样) 图1-1小地图效果图 1.创建好场景 ...