uva10361(字符串)

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=15&problem=1302&mosmsg=Submission+received+with+ID+10379415

10361 - Automatic Poetry

Time limit: 3.000 seconds

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 riddlesagain!”

In Tomb Raider XIV, Lara is, as ever, gunning her way throughancient Egyptian pyramids, prehistoric caves and medival hallways.Now she is standing in front of some important Germanic lookingdoorway and has to solve a linguistic riddle to pass. As usual, theriddle 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.

A Schuttelreim seems to be a typical German invention. The funnything about this strange type of poetry is that if somebody givesyou the first line and the beginning of the second one, you cancomplete the poem yourself. Well, even a computer can do that, andyour task is to write a program which completes them automatically.This will help Lara concentrate on the “action” part of Tomb Raiderand 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 lowercasecharacters or blanks. The second line will be a string of lowercasecharacters or blanks ending with three dots“...”. Lines will we atmost 100 characters long.

Output

For each pair of Schuttelreim lines l₁andl₂ you are to output two lines c₁ andc₂in the following way: c₁ is the same asl₁ only that the bracket marks “<” and“>” are removed. Line c₂is the same asl₂ , except that instead of the three dots the strings₄s₃s₂s₅shouldappear.

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：读取字符串，每个测试例子有两行，第一行中有5个子串，中间用"<",">"分开。第二行有一个串，以“...”结尾。

最后每个实例打印两行，第一行简单的把输入的第一行打印出来，只去掉<和>符号，第二行按照原来第二行的打印，并把三个点去掉，换成s4,s3,s2,s5。即输出buf,s4,s3,s2,s5.

教训：题意首先要读懂，读懂题意再做会事半功倍的。。。

以下两个代码都是A的，但是自己感觉第二个代码目前更适合自己。。

代码1
#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
char str[1000];
int main()
{
 int t;
 scanf("%d",&t);
 getchar();
 while(t--)
 {
  string s[6];
  memset(str,0,sizeof(str));
  gets(str);
  int len=strlen(str),d=0;
  for(inti=0;i<len;i++)
  {
   if(str[i]!='<'&&str[i]!='>')
   {
    putchar(str[i]);
    s[d]+=str[i];
   }
   elseif(str[i]=='>'||str[i]=='<')d++;
  }
  puts("");
  gets(str);
  len=strlen(str);
  for(inti=0;i<len;i++)
  {
   if(str[i]!='.')
    putchar(str[i]);
   elsebreak;
  }
  printf("%s%s%s%s\n",s[3].c_str(),s[2].c_str(),s[1].c_str(),s[4].c_str()); 
 }
 return 0;
}

代码2
#include<stdio.h>
#include<string.h>
#define MAXN 110
char s1[MAXN],s2[MAXN],s3[MAXN],s4[MAXN],s5[MAXN];
char buf[MAXN];
int main()
{
 //freopen("uva10361.txt","r",stdin);
 int cases;
 char c;
 int i;
 scanf("%d",&cases);
 c=getchar();// 消除数字后面的回车
 while(cases--)
 {
  memset(s1,0,sizeof(s1));//数据初始化
  memset(s2,0,sizeof(s2));
  memset(s3,0,sizeof(s3));
  memset(s4,0,sizeof(s4));
  memset(s5,0,sizeof(s5));
  memset(buf,0,sizeof(buf));
  c=getchar();// 数据输入, 输入s1
  for(i=0;c!='<';i++)
  {
   s1[i]=c;
   c=getchar();
  }
  c=getchar();// 输入s2
  for(i=0;c!='>';i++)
  {
   s2[i]=c;
   c=getchar();
  }
  c=getchar();// 输入s3
  for(i=0;c!='<';i++)
  {
   s3[i]=c;
   c=getchar();
  }
  c=getchar();// 输入s4
  for(i=0;c!='>';i++)
  {
   s4[i]=c;
   c=getchar();
  }
  c=getchar();// 输入s5
  for(i=0;c!='\n';i++)
  {
   s5[i]=c;
   c=getchar();
  }
  c=getchar();// 输入buf
  for(i=0;c!='.';i++)
  {
   buf[i]=c;
   c=getchar();
  }
  c=getchar();// 消除两个点和一个换行
  c=getchar();
  c=getchar();
  printf("%s%s%s%s%s\n",s1,s2,s3,s4,s5);//数据输出
  printf("%s%s%s%s%s\n",buf,s4,s3,s2,s5);
 }
 return 0;
}