UVA 10361 - Automatic Poetry

“Oh God”, Lara Croft exclaims, “it’s one of these dumb riddles again!”

 

In Tomb Raider XIV, Lara is, as ever, gunning her way through ancient Egyptian pyramids, prehistoric caves and medival hallways. Now she is standing in front of some important Germanic looking doorway and has to solve a linguistic riddle to pass. As usual, the riddle is not very intellectually challenging.

 

This time, the riddle involves poems containing a “Schuttelreim”. An example of a Schuttelreim is the following short poem:

 

Ein Kind halt seinen Schnabel nur,

wenn es hangt an der Nabelschnur.        

 

/*German contestants please forgive me. I had to modify something as they were not appearing correctly in plain text format*/

 

A Schuttelreim seems to be a typical German invention. The funny thing about this strange type of poetry is that if somebody gives you the first line and the beginning of the second one, you can complete the poem yourself. Well, even a computer can do that, and your task is to write a program which completes them automatically. This will help Lara concentrate on the “action” part of Tomb Raider and not on the “intellectual” part.

Input

The input will begin with a line containing a single number n. After this line follow n pairs of lines containing Schuttelreims. The first line of each pair will be of the form

s₁<s₂>s₃<s₄>s₅

 

where the s_i are possibly empty, strings of lowercase characters or blanks. The second line will be a string of lowercase characters or blanks ending with three dots “...”. Lines will we at most 100 characters long.

Output

For each pair of Schuttelreim lines l₁ and l₂ you are to output two lines c₁ and c₂ in the following way: c₁ is the same as l₁ only that the bracket marks “<” and “>” are removed. Line c₂ is the same as l₂ , except that instead of the three dots the string s₄s₃s₂s₅ should appear.

Sample Input

3

ein kind haelt seinen <schn>abel <n>ur

wenn es haengt an der ...

weil wir zu spaet zur <>oma <k>amen

verpassten wir das ...

<d>u <b>ist

...

Sample Output

ein kind haelt seinen schnabel nur

wenn es haengt an der nabel schnur

weil wir zu spaet zur oma kamen

verpassten wir das koma amen

du bist

bu dist

//#define RUN
#ifdef RUN

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <vector>
#include <list>
#include <cctype> 
#include <algorithm>
#include <utility>
#include <math.h>

using namespace std;

#define MAXN 105
//char l1[MAXN];
//char l2[MAXN];
string l1;
string l2;

void play(){
	bool leftB1 = false;
	bool rightB1 = false;
	bool leftB2 = false;
	bool rightB2 = false;

	string s2, s3, s4, s5;
	s2.clear();
	s3.clear();
	s4.clear();
	s5.clear();

	for(int i=0; i<l1.length(); i++){
		//第一次找到<
		if(!leftB1 && !rightB1 && l1[i]=='<' && !leftB2 && !rightB2){
			leftB1 = true;
			continue;
		}

		//第一次找到>
		if(leftB1 && !rightB1 && l1[i]=='>'  && !leftB2 && !rightB2){
			rightB1 = true;
			continue;
		}

		//找到第一个<>中间的字符
		if(leftB1 && !rightB1  && !leftB2 && !rightB2){
			s2 += l1[i];
			continue;
		}

		//第二次找到<
		if(leftB1 && rightB1 && !leftB2 && !rightB2 && l1[i]=='<'){
			leftB2 = true;
			continue;
		}

		//找到第一个<>到第二个<>之间的字符
		if(leftB1 && rightB1 && !leftB2 && !rightB2){
			s3 += l1[i];
			continue;
		}

		//第二次找到>
		if(leftB1 && rightB1 && leftB2 && !rightB2 && l1[i]=='>'){
			rightB2 = true;
			continue;
		}

		//找到第二个<>中间的字符
		if(leftB1 && rightB1 && leftB2 && !rightB2){
			s4 += l1[i];
			continue;
		}

		//找到第二个<>后面的字符
		if(leftB1 && rightB1 && leftB2 && rightB2){
			s5 += l1[i];
			continue;
		}
	}

	int l1start = l1.find('<');
	string s1 = l1.substr(0, l1start);


	//cout << "s2:" << s2 << endl;
	//cout << "s3:" << s3 << endl;
	//cout << "s4:" << s4 << endl;
	//cout << "s5:" << s5 << endl;
	cout << (s1+s2+s3+s4+s5) << endl;
	string combine = s4+s3+s2+s5;
	int l2start = l2.find("...");
	cout << l2.replace(l2start,3,combine) << endl;
}


int main(){

#ifndef ONLINE_JUDGE
	freopen("10361.in", "r", stdin);
	freopen("10361.out", "w", stdout);
#endif

	int n;
	scanf("%d", &n);
	getchar();
	for(int i=0; i<n; i++){
		l1.clear();
		l2.clear();
		//memset(l1, '\0', sizeof(l1));
		//memset(l2, '\0', sizeof(l2));

		//fgets(l1, MAXN, stdin);
		//fgets(l2, MAXN, stdin);

		getline(cin, l1);
		getline(cin, l2);

		//cout << l1 << endl;
		//cout << l2 << endl;
		play();
	}

}


#endif