uva 10361（字元串）

題目：“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.

Note: German users, 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 <

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
	int i=0;
	int j=0;
	int k=0;
	int n=0;
	int p;	
	scanf("%d",&p);
	getchar();
	for(i=0;i<p;i++) {
		char temp[5][5000]={{"\0"},{"\0"},{"\0"},{"\0"},{"\0"}};
		char str[5000]={"\0"};
		gets(str);
		for(j=0;str[j];j++) {
			if(str[j]=='<'||str[j]=='>') {
				k++;
				n=0;
			}
			else
			{
				temp[k][n++]=str[j];
				printf("%c",str[j]);

			}

		}
		printf("\n");
//		getchar();
//		scanf("%s",str);
		gets(str);
		for(j=0;str[j];j++) {
			if(str[j]!='.')
				printf("%c",str[j]);
		}
		printf("%s%s%s%s\n",temp[3],temp[2],temp[1],temp[4]);
		k=0;
		n=0;
	}
	return 0;

}
           

n>ur

wenn es haengt an der ...

weil wir zu spaet zur <>oma <k>amen

verpassten wir das ...

<d>u <b>ist

...Universidad de Valladolid OJ: 10361 – Automatic Poetry 2/2

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

題意：一次循環兩句話，第一句帶有<>，每一分句有一個标志，标志如s1<s2>s3<s4>s5，輸出時要去掉<>；第二句話帶有...輸出時去掉...補上s4s3s2s5。

思路：設一個二維數組分别裝下S1,S2,S3,S4,S5，在判斷<>時輸出第一句話并且存儲S1，S2，S3，S4，S5。然後輸入第二句話，先輸出非‘.’，再輸出s4s3s2s5.