2031. Overturned Numbers

Time limit: 1.0 second

Memory limit: 64 MB
Little Pierre was surfing the Internet and came across an interesting puzzle:
What is the number under the car?
It took some time before Pierre solved the puzzle, but eventually he understood that there were overturned numbers 86, 88, 89, 90, and 91 in the picture and the answer was the number 87.
Now Pierre wants to entertain his friends with similar puzzles. He wants to construct a sequence of
n numbers such that its overturning produces a consecutive segment of the positive integers. Pierre intends to use one-digit integers supplemented with a leading zero and two-digit integers only.To avoid ambiguity, note that when the digits 0, 1, and
8 are overturned, they remain the same, the digits 6 and 9 are converted into each other, and the remaining digits become unreadable symbols.

Input

The only line contains the number n of integers in a sequence (1 ≤
n ≤ 99).

Output

If there is no sequence of length n with the above property, output “Glupenky Pierre” (“Silly Pierre” in Russian).Otherwise, output any of such sequences. The numbers in the sequence should be separated with a space.

Samples

input output
2
11 01
99
Glupenky Pierre

Problem Author: Nikita Sivukhin

Problem Source: Ural Regional School Programming Contest 2014

解析:题目要求翻转后为连续序列的序列,直接枚举就可以。

AC代码:

#include <bits/stdc++.h>
using namespace std; int main(){
int n;
while(scanf("%d", &n) != EOF){
if(n == 1) puts("01");
else if(n == 2) puts("11 01");
else if(n == 3) puts("06 68 88");
else if(n == 4) puts("16 06 68 88");
else puts("Glupenky Pierre");
}
return 0;
}

URAL 2031. Overturned Numbers (枚举)的更多相关文章

  1. 递推DP URAL 1586 Threeprime Numbers

    题目传送门 /* 题意:n位数字,任意连续的三位数字组成的数字是素数,这样的n位数有多少个 最优子结构:考虑3位数的数字,可以枚举出来,第4位是和第3位,第2位组成的数字判断是否是素数 所以,dp[i ...

  2. 递推DP URAL 1009 K-based Numbers

    题目传送门 题意:n位数,k进制,求个数分析:dp[i][j] 表示i位数,当前数字为j的个数:若j==0,不加dp[i-1][0]; 代码1: #include <cstdio> #in ...

  3. ural 2070. Interesting Numbers

    2070. Interesting Numbers Time limit: 2.0 secondMemory limit: 64 MB Nikolay and Asya investigate int ...

  4. ural 1150. Page Numbers

    1150. Page Numbers Time limit: 1.0 secondMemory limit: 64 MB John Smith has decided to number the pa ...

  5. URAL 1792. Hamming Code (枚举)

    1792. Hamming Code Time limit: 1.0 second Memory limit: 64 MB Let us consider four disks intersectin ...

  6. URAL 1002 Phone Numbers(KMP+最短路orDP)

    In the present world you frequently meet a lot of call numbers and they are going to be longer and l ...

  7. URAL 1012 K-based Numbers. Version 2(DP+高精度)

    题目链接 题意 :与1009一样,不过这个题的数据范围变大. 思路:因为数据范围变大,所以要用大数模拟,用java也行,大数模拟也没什么不过变成二维再做就行了呗.当然也可以先把所有的都进行打表,不过要 ...

  8. ural 1118. Nontrivial Numbers

    1118. Nontrivial Numbers Time limit: 2.0 secondMemory limit: 64 MB Specialists of SKB Kontur have de ...

  9. ural 1013. K-based Numbers. Version 3(动态规划)

    1013. K-based Numbers. Version 3 Let’s consider K-based numbers, containing exactly N digits. We def ...

随机推荐

  1. 1682. [HAOI2014]贴海报

    1682. [HAOI2014]贴海报 ★★☆   输入文件:ha14d.in   输出文件:ha14d.out   简单对比 时间限制:1 s   内存限制:256 MB [题目描述] Byteto ...

  2. TortoiseSVN客户端不能记住用户名和密码

    TortoiseSVN客户端重新设置用户名和密码 在第一次使用TortoiseSVN从服务器CheckOut的时候,会要求输入用户名和密码,这时输入框下面有个选项是保存认证信息,如果选了这个选项,那么 ...

  3. jboss之启动加载过程详解

    今天看了看jboss的boot.log和server.log日志,结合自己的理解和其他的资料,现对jboss的启动和加载过程做出如下总结: boot.xml是服务器的启动过程的日志,不涉及后续的操作过 ...

  4. eclipse中添加maven

    收藏一下,一篇很好的例子 maven相关插件:链接:http://pan.baidu.com/s/1i3Ks95j 密码:7pgh eclipse:链接:http://pan.baidu.com/s/ ...

  5. windows如何统计端口的连接数

    习惯了linux的系统管理员,对linux的命令行工具总是印象极深,几乎所有的管理都可以在命令行下完成.命令行工具是linux系统管理的主流. 而使用windows是,因为图形化的界面,大家习惯了图形 ...

  6. 梦想CAD控件安卓文字样式

    增加文字样式 用户可以增加文字样式到数据库,并设置其字体等属性,具体实现代码如下: // 增加文字样式 //getCurrentDatabase()返回当前数据库对象 //getTextstyle() ...

  7. 虚拟机找不到本机vmnet0,vmnet8,无法连接xshell,解决方案

    首先出现这个问题肯定是第一次下载虚拟机把之前的注册表覆盖了,网卡找不到,首先卸载VMware 并且将C\ProgramData下的VMware文件夹删除掉 ,下载cceaner,点击注册表清除干净,再 ...

  8. Rest 参数(...)

    javascript 之Rest 参数(...) ES6 Rest参数 Rest就是为解决传入的参数数量不一定, rest parameter(Rest 参数) 本身就是数组,数组的相关的方法都可以用 ...

  9. java求两个集合的交集和并集,比较器

    求连个集合的交集: import java.util.ArrayList; import java.util.List; public class TestCollection { public st ...

  10. 代码分析工具splint安装介绍

    官网 http://www.splint.org/ splint能干什么? splint是一个静态检查C语言代码安全弱点和编写错误的开源程序.(不支持C++) splint会进行多种常规检查,包括 空 ...