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UVA之10361 - Automatic Poetry

2024-07-02 13:20:32   226人阅读

Problem I

Automatic Poetry

Input: standard input

Output: standard output

Time Limit: 2 seconds

Memory Limit: 32 MB

 

“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 si 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

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】：

[cpp] view plaincopy
  1. /********************************* 
  2. *   日期：2013-4-26 
  3. *   作者：SJF0115 
  4. *   题号: 题目10361 - Automatic Poetry 
  5. *   来源：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=15&page=show_problem&problem=1302 
  6. *   结果：AC 
  7. *   来源：UVA 
  8. *   总结： 
  9. **********************************/  
  10. #include<stdio.h>  
  11. #include<string.h>  
  12. #define N 110  
  13. int main (){  
  14.     int i,j,Case,lena,lenb;  
  15.     char a[N],b[N],s[N],s1[N],s2[N],s3[N],s4[N],s5[N];  
  16.     //freopen("C:\\Users\\XIAOSI\\Desktop\\acm.txt","r",stdin);    
  17.     while(scanf("%d\n",&Case) != EOF){  
  18.         while(Case--){  
  19.             gets(a);  
  20.             gets(b);  
  21.             lena = strlen(a);  
  22.             //第一个子串  
  23.             for(j = 0,i = 0;a[i] != ‘<‘;i++,j++){  
  24.                 s1[j] = a[i];  
  25.             }  
  26.             s1[j] = ‘\0‘;  
  27.             //printf("%s\n",s1);  
  28.             //第二个子串  
  29.             for(i = i+1,j = 0;a[i] != ‘>‘;i++,j++){  
  30.                 s2[j] = a[i];  
  31.             }  
  32.             s2[j] = ‘\0‘;  
  33.             //printf("%s\n",s2);  
  34.             //第三个子串  
  35.             for(i = i+1,j = 0;a[i] != ‘<‘;i++,j++){  
  36.                 s3[j] = a[i];  
  37.             }  
  38.             s3[j] = ‘\0‘;  
  39.             //printf("%s\n",s3);  
  40.             //第四个子串  
  41.             for(i = i+1,j = 0;a[i] != ‘>‘;i++,j++){  
  42.                 s4[j] = a[i];  
  43.             }  
  44.             s4[j] = ‘\0‘;  
  45.             //printf("%s\n",s4);  
  46.             //第五个子串  
  47.             for(i = i+1,j = 0;i < lena;i++,j++){  
  48.                 s5[j] = a[i];  
  49.             }  
  50.             s5[j] = ‘\0‘;  
  51.             //printf("%s\n",s5);  
  52.             //第二个字符串  
  53.             for(i = 0,j = 0;b[i] != ‘.‘;i++,j++){  
  54.                 s[j] = b[i];  
  55.             }  
  56.             s[j] = ‘\0‘;  
  57.             //printf("%s\n",s);  
  58.             //输出  
  59.             printf("%s%s%s%s%s\n",s1,s2,s3,s4,s5);  
  60.             printf("%s%s%s%s%s\n",s,s4,s3,s2,s5);  
  61.         }  
  62.     }  
  63.     return 0;  
  64. }  
  65.   
  66.       


[cpp] view plaincopy
  1. #include<iostream>  
  2. #include<string>  
  3. using namespace std;  
  4.   
  5. #define MAXN 102  
  6. string a1, a2;  
  7.   
  8. void solve()  
  9. {  
  10.     int p1 = a1.find(‘<‘, 0);  
  11.     int p2 = a1.find(‘>‘, 0);  
  12.     int p3 = a1.find(‘<‘, p2+1);  
  13.     int p4 = a1.find(‘>‘, p2+1);  
  14.     string s1 = a1.substr(0, p1);  
  15.     string s2 = a1.substr(p1+1, p2-p1-1);  
  16.     string s3 = a1.substr(p2+1, p3-p2-1);  
  17.     string s4 = a1.substr(p3+1, p4-p3-1);  
  18.     string s5 = a1.substr(p4+1);  
  19.     cout<<s1<<s2<<s3<<s4<<s5<<endl;  
  20.     a2.replace(a2.length()-3, 3, s4+s3+s2+s5);  
  21.     cout<<a2<<endl;  
  22. }  
  23.   
  24. int main()  
  25. {  
  26.     int t;  cin>>t;  
  27.     cin.get();  
  28.     while(t--)  
  29.     {  
  30.         getline(cin, a1);  
  31.         getline(cin, a2);  
  32.         solve();  
  33.     }  
  34. }  


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