You have decided that PGP encryptation is not strong enough for your email. You have decided to supplement it by first converting your clear text letter into Pig Latin before encrypting it with PGP.
Input and Output
You are to write a program that will take in an arbitrary number of lines of text and output it in Pig Latin. Each line of text will contain one or more words. A ``word'' is defined as a consecutive sequence of letters (upper and/or lower case). Words should be converted to Pig Latin according to the following rules (non-words should be output exactly as they appear in the input):
- Words that begin with a vowel (a, e, i, o, or u, and the capital versions of these) should just have the string ``ay'' (not including the quotes) appended to it. For example, ``apple'' becomes ``appleay''.
- Words that begin with a consonant (any letter than is not A, a, E, e, I, i, O, o, U or u) should have the first consonant removed and appended to the end of the word, and then appending ``ay'' as well. For example, ``hello'' becomes ``ellohay''.
- Do not change the case of any letter.
Sample Input
This is the input.
Sample Output
hisTay isay hetay inputay.
雖然題目難度不是很大，但是對(duì)我啟發(fā)挺大，碰到這種字符串處理的題，最好就是統(tǒng)一模式，
a=getchar():
while(flag)
{ while()//判斷是否符合條件
{ ---- ,
if((a=getchar())==EOF)//獲取a=getchar()
flag=0;//判斷文件是否結(jié)束
}
while()//判斷是否符合條件
{ ---- ,
if((a=getchar())==EOF)//獲取a=getchar()
flag=0;//判斷文件是否結(jié)束
}
}
#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
// freopen("s.txt","r",stdin);
// freopen("key.txt","w",stdout);
char a;
int flag=0,e=1,mark;
string temp;
char mar;
a=getchar();
while (e)
{
mark=0;
while((a>='a'&&a<='z')||(a>='A'&&a<='Z'))
{
temp="ay";
if(flag==0)
{
flag++;
if(a=='A'||a=='a'||a=='E'||a=='e'||a=='I'||a=='i'||a=='O'||a=='o'||a=='U'||a=='u')
cout<<a;
else
{
mark=1;
mar=a;
}
}
else
cout<<a;
if((a=getchar())==EOF)
{
e=0;
break;
}
}
if(mark==1)
{
cout<<mar;
}
cout<<temp;
while(a<'A'||a>'z'||(a>'Z'&&a<'a'))
{
cout<<a;
if((a=getchar())==EOF)
{
e=0;
break;
}
}
flag=0;
}
//system("PAUSE");
return 0;
}
posted on 2009-07-03 18:04
luis 閱讀(618)
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