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. 基于visual Studio2013解决面试题之0305广度优先搜索二叉树

     题目

  2. jsp:include怎么设置才能正确显示包含的页面呢

    1.项目的所有jsp都放在WEB-INF文件夹之下,使用的是SpirngMVC进行了过滤,jsp:include只能引入WEB-INF外部的jsp文件,对于改变后缀显示为htm的jsp的WEB-INF ...

  3. Tengine中的proxy_upstream_tries

    upsream xxx { server 192.168.100.100; server 192.168.100.101; server 192.168.100.102; } server { loc ...

  4. objective-c 中数据类型之中的一个 几何数据类型(CGPoint,CGSize,CGRect)

    // CGPoint 结构体数据原型, 用于声明一个点: /* Points. */ struct CGPoint { CGFloat x; CGFloat y; }; typedef struct ...

  5. 手把手教你_android自己主动化实践方案选型

    接到一个android自己主动化的任务,看了看手中的家伙:ranorex,appium,uiautomator 当然先捡商用的试试,简单呀,能够录制回放,只是不是抱特别大的期望,这个爷比較娇气,要是a ...

  6. Python Errors and Exceptions

    1. python中的try{}catch{} 2. raise exception 3. try...except ... else.. 4. finally块 python中的异常处理的keywo ...

  7. Android平均分布的布局图像的下一行

    Android下一行平均分布图片的布局 这是一个非经常见的需求,比方有三个图片button,须要在底部三个平均,比方下个样例: 下面是布局文件 <LinearLayout android:lay ...

  8. xcode5下一个ffmpeg静态库配置

    1.若要安装xcode命令行工具 1).xcode5安装命令行工具方法: 在终端运行命令Using xcode-select --install 2).xcode5之前安装命令行工具方法: 2.xco ...

  9. setStyleSheet来设定窗口部件的样式

    使用setStyleSheet来设置图形界面的外观:QT Style Sheets是一个很有利的工具,允许定制窗口的外观,此外还可以用子类QStyle来完成,他的语法很大比重来源于html的CSS,但 ...

  10. 可编辑的表格:jQuery+PHP实现实时编辑表格字段内容

    在本例中,我们会通过jQuery实现单击将一个文本信息变为可编辑的表单,你可以对文本内容进行编辑,然后点击“确定”按钮,新的内容将发送到后台PHP程序处理,并保存到数据库:当点击“取消”按钮,则页面恢 ...