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. boost::asio网络传输错误码的一些实验结果(recv error_code)

    错误码很重要,可以由此判断网络连接到底发生了神马事情,从而驱动高层逻辑的行为.只有笼统的错误码判断的网络层是不够规范的,鄙人觉得有些错误码还是需要在网络层就区分开的,特此记录一些当前实验的错误码以及发 ...

  2. 经常使用MD5算法代码

    经常使用的MD5算法代码日期: 2014年8月4日作者: 铁锚 MD5,全称为 Message Digest Algorithm 5(消息摘要算法第五版).详情请參考 维基百科:MD5  MD5加密后 ...

  3. BAPI总的数据库提交和回滚

    BAPI事物中的数据提交和回滚必须通过调用SAP标准业务对象BAPI SERVICE(对象类型SAP0001)的BAPI方法bapiservic.transactioncommit和bapiservi ...

  4. js中静态函数与变量

    一 私有变量和函数 js中没有概念上的私有,公有也没有静态和非静态相关概念,有的只能是通过作用于来模仿 函数的块级作用域使得函数内部成员可以不被外部所访问,比如我们使用块级作用于定义一个类 //定义一 ...

  5. SDL 简介

    SDL 简介 什么是SDL? 即 Simple DirectMedia Layer,使用 LGPL 许可证. 免费的跨平台多媒体应用编程接口 用于游戏.游戏开发工具.模拟器.样本演示.多媒体应用等 它 ...

  6. MySQL 採用Xtrabackup对数据库进行全库备份

    1,xtrabackup简单介绍 关于数据库备份以及备份工具.參考:http://blog.itpub.net/26230597/viewspace-1460065/,这里来介绍xtrabackup已 ...

  7. 将Ojective-C代码移植转换为Swift代码

    相比于Objective-C,Swift语言更加简练.有时我们需要把原来写的一些Objective-C代码转换成Swift,下面总结了各种常见的情况. 1,构造函数的迁移 Objective-C为: ...

  8. Swift - 炫酷放射弹出按钮菜单(改造自AwesomeMenu)

    这个是一个第三方按钮菜单组件,原版是使用Objective-C编写的名为AwesomeMenu的组件,地址是:https://github.com/levey/AwesomeMenu 这里改造成了Sw ...

  9. JSP/Servlet-----charset 、pageEncoding差别

    一.JSP/Servlet中的几个编码的作用         在JSP/Servlet中有4个地方可设置编码(例如以下).当中前两个仅仅能用于JSP中,后两个可用于JSP和Servlet 中.    ...

  10. hdu4055 Number String

    Number String Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Tota ...