http://acm.hunnu.edu.cn/online/?action=problem&type=show&id=11354&courseid=0

Problem description

The philosopher Willard Van Orman Quine (1908–2000) described a novel method of constructing a sentence in order to illustrate the contradictions that can arise from self-reference. This operation takes as input a single phrase and produces a sentence from that phrase. (The author Douglas R. Hofstadter refers to this process as to Quine a phrase.) We can define the Quine operation like so:

Quine(A) = "A" A

In other words, if A is a phrase, then Quine(A) is A enclosed in quotes ("), followed by a space, followed by A. For example:

Quine(HELLO WORLD) = "HELLO WORLD" HELLO WORLD

Below are some other examples of sentences that can be created by the Quine operation. Note that Quining allows sentences to be indirectly self-referential, such as the last sentence below.

"IS A SENTENCE FRAGMENT" IS A SENTENCE FRAGMENT"IS THE NAME OF THIS PROBLEM" IS THE NAME OF THIS PROBLEM"YIELDS FALSEHOOD WHEN QUINED" YIELDS FALSEHOOD WHEN QUINED

Your goal for this problem is to take a sentence and decide whether the sentence is the result of a Quine operation.

Input

The input will consist of a sequence of sentences, one sentence per line, ending with a line that has the single word, END. Each sentence will contain only uppercase letters, spaces, and quotation marks. Each sentence will contain between 1 and 80 characters and will not have any leading, trailing, or consecutive spaces.

You must decide whether each sentence is the result of a Quine operation. To be a Quine, a sentence must match the following pattern exactly:

  1. A quotation mark
  2. Any nonempty sequence of letters and spaces (call this phrase A)
  3. A quotation mark
  4. A space
  5. Phrase A—exactly as it appeared in (2)

If it matches this pattern, the sentence is a Quine of the phrase A. Note that phrase A must contain the exact same sequence of characters both times it appears.

Output

There will be one line of output for each sentence in the data set. If the sentence is the result of a Quine operation, your output should be of the form, Quine(A), where A is the phrase to Quine to create the sentence.

If the sentence is not the result of a Quine operation, your output should be the phrase, not a quine.

Sample Input

"HELLO WORLD" HELLO WORLD"IS A SENTENCE FRAGMENT" IS A SENTENCE FRAGMENT"IS THE NAME OF THIS PROBLEM" IS THE NAME OF THIS PROBLEM"YIELDS FALSEHOOD WHEN QUINED" YIELDS FALSEHOOD WHEN QUINED"HELLO" I SAIDWHAT ABOUT "WHAT ABOUT"" NO EXTRA SPACES " NO EXTRA SPACES"NO"QUOTES" NO"QUOTES""END

Sample Output

Quine(HELLO WORLD)Quine(IS A SENTENCE FRAGMENT)Quine(IS THE NAME OF THIS PROBLEM)Quine(YIELDS FALSEHOOD WHEN QUINED)not a quinenot a quinenot a quinenot a quinenot a quine

Judge Tips

A review of quotation marks in strings:  As a reminder, the quotation mark character is a regular character, and can be referred to in C, C++, and Java using the standard single-quote notation, like so:

'"'

However, to place a quotation mark inside a double-quoted string in C, C++, and Java, you must place a backslash (\) in front of it. If you do not it will be interpreted as the end of the string, causing syntax errors. For example:

"This quotation mark \" is inside the string""\"""\"SAID SHE\" SAID SHE"

题意:给出一个字符串,符合"A"A的状况输出Quine(A),否则输出not a quine

思路:首先从左到右找出第一对双引号里的字符串,然后从右到左找出一个双引号后面的字符串,处理好好比较即可

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; char s1[10005],s2[10005];
char str[20005]; int main()
{
int len,i,l1,l2;
while(gets(str))
{
if(!strcmp(str,"END"))
break;
len = strlen(str);
l1 = l2 = 0;
if(str[0] != '"')//第一个必须为“
{
printf("not a quine\n");
continue;
}
for(i = 1; i<len; i++)//找到第一对引号内的字符串
{
if(str[i]!=34)
s1[l1++] = str[i];
else
{
s1[l1] = '"';
break;
}
}
for(i = len-1; i>=0; i--)//从后面开始找到第一个引号出现的位置为止
{
if(str[i] == ' ' && str[i-1] == '"')
{
s2[l2] = '\0';
break;
}
else
s2[l2++] = str[i];
}
char tem;
for(i = 0; i<l2/2; i++)//翻转字符串
{
tem = s2[l2-1-i];
s2[l2-1-i] = s2[i];
s2[i] = tem;
}
if(!strcmp(s1,s2))
printf("Quine(%s)\n",s1);
else
printf("not a quine\n");
} return 0;
}

HUNNU11354:Is the Name of This Problem的更多相关文章

  1. 1199 Problem B: 大小关系

    求有限集传递闭包的 Floyd Warshall 算法(矩阵实现) 其实就三重循环.zzuoj 1199 题 链接 http://acm.zzu.edu.cn:8000/problem.php?id= ...

  2. No-args constructor for class X does not exist. Register an InstanceCreator with Gson for this type to fix this problem.

    Gson解析JSON字符串时出现了下面的错误: No-args constructor for class X does not exist. Register an InstanceCreator ...

  3. C - NP-Hard Problem(二分图判定-染色法)

    C - NP-Hard Problem Crawling in process... Crawling failed Time Limit:2000MS     Memory Limit:262144 ...

  4. Time Consume Problem

    I joined the NodeJS online Course three weeks ago, but now I'm late about 2 weeks. I pay the codesch ...

  5. Programming Contest Problem Types

        Programming Contest Problem Types Hal Burch conducted an analysis over spring break of 1999 and ...

  6. hdu1032 Train Problem II (卡特兰数)

    题意: 给你一个数n,表示有n辆火车,编号从1到n,入站,问你有多少种出站的可能.    (题于文末) 知识点: ps:百度百科的卡特兰数讲的不错,注意看其参考的博客. 卡特兰数(Catalan):前 ...

  7. BZOJ2301: [HAOI2011]Problem b[莫比乌斯反演 容斥原理]【学习笔记】

    2301: [HAOI2011]Problem b Time Limit: 50 Sec  Memory Limit: 256 MBSubmit: 4032  Solved: 1817[Submit] ...

  8. [LeetCode] Water and Jug Problem 水罐问题

    You are given two jugs with capacities x and y litres. There is an infinite amount of water supply a ...

  9. [LeetCode] The Skyline Problem 天际线问题

    A city's skyline is the outer contour of the silhouette formed by all the buildings in that city whe ...

随机推荐

  1. spring mvc 接受多对象的处置

    spring mvc 接受多对象的处理 spring mvc感觉非常好用,尤其是对接收对象參数的自己主动绑定非常简便,但对于同一时候传多个对象时有些困扰. 同一时候项目并没有直接使用spring的fo ...

  2. C++ delete 和 delete []

    C++ delete 和 delete [] 简单结论: new delete new [] delete []   文章 : 对 delete [] 的声明 void operator delete ...

  3. OCM读书笔记(2) - PL/SQL 基础

    1. % type 用法,提取% type所在字段的类型 declare     myid dept.deptno % type;    myname dept.dname % type;begin  ...

  4. 超声波模块SRF05

    //////////////////////////////////////////////////////////////////////////////// // //     PIC16F877 ...

  5. Ajax 下拉列表联动显示

    一般处理程序文件 代码 using System;using System.Web;using System.Linq;using System.Data.Linq;using System.Text ...

  6. ASP.NET 应用程序(Application)生命周期概述

    原文:ASP.NET 应用程序(Application)生命周期概述 引用MSDN:ASP.NET 应用程序生命周期概述 本 主题概述应用程序生命周期,列出重要的生命周期事件,并描述如何编写适合应用程 ...

  7. Linux入门基础 #9:管道及重定向

    本文出自   http://blog.csdn.net/shuangde800 ------------------------------------------------------------ ...

  8. [Windows Phone]修改应用程序主题

    问题: Windows Phone模拟器默认情况下是黑色(Dark)主题,开发者往往都是在黑色主题下进行应用开发,加入自定义的颜色,样式等等,而当把手机操作系统主题设为白色(Light)主题之后会发现 ...

  9. 图像编程学习笔记1——bmp文件结构处理与显示

    文本内容转载自<数字图像处理编程入门>,代码为自己实现 1.1图和调色板的概念 如今Windows(3.x以及95,98,NT)系列已经成为绝大多数用户使用的操作系统,它比DOS成功的一个 ...

  10. sort 使用 tab键 作为 分隔符_人生如梦_百度空间

    sort 使用 tab键 作为 分隔符_人生如梦_百度空间 sort 使用 tab键 作为 分隔符 For some reason "\t" doesn't work right, ...