HUNAN NORMAL UNIVERSITY ACM 11354

Is the Name of This Problem

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:32768KB

Total submit users: 23, Accepted users: 21

Problem 11354 : No special judgement

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 SAID
WHAT 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 quine
not a quine
not a quine
not a quine
not a quine

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    char str[1000],a[1000],b[1000];
    int i,len,j,count,flag,t,tag;
    while(gets(str)&&(strcmp(str,"END")!=0))
    {
        if(str[0]!='"')
        {
            printf("not a quine
");
            continue;
        }
        len=strlen(str);
        memset(a,'',sizeof(a));
        memset(b,'',sizeof(b));
        count=0;
        for(i=0; i<len; i++)
        {
            if(str[i]=='"')
                count++;
        }
        if(count>2)
        {
            printf("not a quine
");
            continue;
        }
        flag=1;
        tag=0;
        for(i=0; i<=len; i++)
        {
            if(str[i]=='"')
            {
                flag=!flag;
                t=0;
                if(flag!=0)
                {
                    if(str[i+1]==' ')
                        i+=2;
                    else
                    {
                        tag=0;
                        break;
                    }
                }
                else
                    i+=1;
            }
            if(flag==0)
            {
                a[t++]=str[i];
                continue;
            }
            if(flag!=0)
            {
                b[t++]=str[i];
            }
            if(str[i]=='')
            {
                if(strcmp(a,b)==0)
                    tag=1;
            }
        }
        if(tag)
            printf("Quine(%s)
",a);
        else
            printf("not a quine
");
    }
    return 0;
}