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.
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.
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
...
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
每次输入2个字符串,第一次输出第一个串去掉括号剩下的,第二次输出去掉三个点之后的第2个字符串,外加第一串括号里的字符交换之后的字串。
#include <stdio.h>
#include <string.h>
void sort(char s[],char s1[])
{int i,l=strlen(s),start[2],end[2],sum=0;
for (i=0;i<l;i++)
{if (s[i]=='<') start[sum]=i;
if (s[i]=='>') {end[sum]=i;++sum;}
}
for (i=0;i<l;i++)
if ((s[i]!='<')&&(s[i]!='>')) putchar(s[i]);
printf("\n");
sum=0;
l=strlen(s1)-1;
while (s1[l]=='.') --l;
for (i=0;i<=l;i++) putchar(s1[i]);
for (i=start[1]+1;i<=end[1]-1;i++)
putchar(s[i]);
for (i=end[0]+1;i<=start[1]-1;i++)
putchar(s[i]);
for (i=start[0]+1;i<=end[0]-1;i++)
putchar(s[i]);
l=strlen(s);
for (i=end[1]+1;i<l;i++)
putchar(s[i]);
printf("\n");
};
void main()
{char s[1000],s1[1000];
int t;
scanf("%d\n",&t);
while (t--)
{
gets(s);
gets(s1);
sort(s,s1);
}
}