UVA之10361 - Automatic Poetry

Problem I

Automatic Poetry

Input:standard input

Output:standard output

Time Limit:2 seconds

Memory Limit:32 MB

“Oh God”, Lara Croft exclaims, “it’sone 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 Kindhaltseinen Schnabel nur,

wennes 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

wherethe s_iare 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 willweat most 100 characters long.

Output

For each pair of Schuttelreim lines l₁and l₂you are to output two lines c₁and c₂in the following way: c₁is the same as l₁only that the bracket marks “<” and “>” are removed. Line c₂is the same asl₂,except that instead of the three dots the string s₄s₃s₂s₅should appear.

Sample Input

3

einkind haelt seinen <schn>abel <n>ur

wennes haengt an der ...

weilwir zu spaet zur <>oma <k>amen

verpasstenwir das ...

<d>u <b>ist

...

Sample Output

einkind haelt seinen schnabel nur

wennes haengt an der nabel schnur

weilwir zu spaet zur oma kamen

verpasstenwir das koma amen

dubist

budist

---

TUD Programming Contest

【题意】：

输入：

输入N组测试用例，每组输入两个字符串。

第一个字符串格式：s₁<s₂>s₃<s₄>s₅

s1,s2,s3,s3,s4,s5都可以为空或者不存在或者全是小写字符

第二个字符串格式：s ....

输出：

每组测试用例输出两个字符串。

第一个字符串格式：s1s2s3s4s5

第二个字符串格式：ss4s3s2s5

【代码】：

/*********************************
*日期：2013-4-26
*作者：SJF0115
*题号:题目10361-AutomaticPoetry
*来源：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=15&page=show_problem&problem=1302
*结果：AC
*来源：UVA
*总结：
**********************************/
#include<stdio.h>
#include<string.h>
#defineN110
intmain(){
inti,j,Case,lena,lenb;
chara[N],b[N],s[N],s1[N],s2[N],s3[N],s4[N],s5[N];
//freopen("C:\\Users\\XIAOSI\\Desktop\\acm.txt","r",stdin);
while(scanf("%d\n",&Case)!=EOF){
while(Case--){
gets(a);
gets(b);
lena=strlen(a);
//第一个子串
for(j=0,i=0;a[i]!='<';i++,j++){
s1[j]=a[i];
}
s1[j]='\0';
//printf("%s\n",s1);
//第二个子串
for(i=i+1,j=0;a[i]!='>';i++,j++){
s2[j]=a[i];
}
s2[j]='\0';
//printf("%s\n",s2);
//第三个子串
for(i=i+1,j=0;a[i]!='<';i++,j++){
s3[j]=a[i];
}
s3[j]='\0';
//printf("%s\n",s3);
//第四个子串
for(i=i+1,j=0;a[i]!='>';i++,j++){
s4[j]=a[i];
}
s4[j]='\0';
//printf("%s\n",s4);
//第五个子串
for(i=i+1,j=0;i<lena;i++,j++){
s5[j]=a[i];
}
s5[j]='\0';
//printf("%s\n",s5);
//第二个字符串
for(i=0,j=0;b[i]!='.';i++,j++){
s[j]=b[i];
}
s[j]='\0';
//printf("%s\n",s);
//输出
printf("%s%s%s%s%s\n",s1,s2,s3,s4,s5);
printf("%s%s%s%s%s\n",s,s4,s3,s2,s5);
}
}
return0;
}

#include<iostream>
#include<string>
usingnamespacestd;

#defineMAXN102
stringa1,a2;

voidsolve()
{
intp1=a1.find('<',0);
intp2=a1.find('>',0);
intp3=a1.find('<',p2+1);
intp4=a1.find('>',p2+1);
strings1=a1.substr(0,p1);
strings2=a1.substr(p1+1,p2-p1-1);
strings3=a1.substr(p2+1,p3-p2-1);
strings4=a1.substr(p3+1,p4-p3-1);
strings5=a1.substr(p4+1);
cout<<s1<<s2<<s3<<s4<<s5<<endl;
a2.replace(a2.length()-3,3,s4+s3+s2+s5);
cout<<a2<<endl;
}

intmain()
{
intt;cin>>t;
cin.get();
while(t--)
{
getline(cin,a1);
getline(cin,a2);
solve();
}
}