UVA10361 -自动作诗机

UVA10361 - Automatic Poetry（自动作诗机）

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

第一段实现代码，从别人那转的：

 1 #include <string.h>
 2 #define MAXN 110
 3 void getss(char s[]);
 4 int main()
 5 {
 6     freopen("data.in","r",stdin);
 7     int n;
 8     char s1[MAXN],s2[MAXN],s3[MAXN],s4[MAXN],s5[MAXN],c,line[MAXN];
 9     scanf("%d",&n);
10     c = getchar();
11     while(n--)
12     {
13         getss(s1);
14         getss(s2);
15         getss(s3);
16         getss(s4);
17         getss(s5);
18         gets(line);
19         line[strlen(line) - 3] = ‘\0‘;
20         printf("%s%s%s%s%s\n",s1,s2,s3,s4,s5);
21         printf("%s%s%s%s%s\n",line,s4,s3,s2,s5);
22     }
23     return 0;
24 }
25
26 void getss(char s[])
27 {
28     int i;
29     for(i=0; i<MAXN; i++)
30     {
31         if((s[i] = getchar()) == ‘<‘ || s[i] == ‘>‘ || s[i] == ‘\n‘){
32             s[i] = ‘\0‘;
33             break;
34         }
35     }
36 }

我自己的解法，同大家分享交流：

 1 #include "stdio.h"
 2 #include "string.h"
 3 char s[200],m[200],p[5][200];
 4 int main(){
 5     int n,i,j,q,k;
 6     scanf("%d ",&n);
 7     for(i=0;i<n;i++)
 8     {
 9         q=k=0;
10         fgets(s,sizeof(s),stdin);
11         fgets(m,sizeof(m),stdin);
12         for(j=0;j<strlen(s);j++)
13         {
14             p[q][k]=s[j];
15             k++;
16             if(s[j]==‘<‘||s[j]==‘>‘){p[q][k-1]=‘\0‘;k=0;q++;continue;}
17             printf("%c",s[j]);
18         }
19         p[q][k]=‘\0‘;
20         m[strlen(m) - 4]=‘\0‘;
21         printf("%s%s%s%s%s",m,p[3],p[2],p[1],p[4]);
22     }
23     return 0;
24 }

UVA10361 -自动作诗机