求做一个简单C++题目For most fonts, the lowercase letters b and d are mirror images of each other, as are the letters p and q.Furthermore, letters i, o, v, w, and x are naturally mirror images of themselves. Although othersymmetries exists for

求做一个简单C++题目For most fonts, the lowercase letters b and d are mirror images of each other, as are the letters p and q.Furthermore, letters i, o, v, w, and x are naturally mirror images of themselves. Although othersymmetries exists for
求做一个简单C++题目
For most fonts, the lowercase letters b and d are mirror images of each other, as are the letters p and q.
Furthermore, letters i, o, v, w, and x are naturally mirror images of themselves. Although other
symmetries exists for certain fonts, we consider only those specifically mentioned thus far for the
remainder of this problem.
Because of these symmetries, it is possible to encode certain words based upon how those words would
appear in the mirror. For example the word boxwood would appear as boowxod, and the word ibid as
bidi. Given a particular sequence of letters, you are to determine its mirror image or to note that it is
invalid.
Input: The input contains a series of letter sequences, one per line, followed by a single line with the #
character. Each letter sequence consists entirely of lowercase letters.
Output: For each letter sequence in the input, if its mirror image is a legitimate letter sequence based
upon the given symmetries, then output that mirror image. If the mirror image does not form a legitimate
sequence of characters, then output the word INVALID.
Example Input:
boowxod
bidi
bed
bbb
#
Example Output:
boxwood
ibid
INVALID
ddd
要有具体代码

求做一个简单C++题目For most fonts, the lowercase letters b and d are mirror images of each other, as are the letters p and q.Furthermore, letters i, o, v, w, and x are naturally mirror images of themselves. Although othersymmetries exists for
假设字符串存放在input.txt中,程序从文件中读入数据.
#include
#include
#include
#include
using namespace std;
int main( void )
{
ifstream ifs( "input.txt" );
string strSource( "bdiovwx" );
string str;
while( ifs>> str )
{
if( str == "#" )
{
break;
}

unsigned int n = str.find_first_not_of( strSource, 0 );
if( n == string::npos ) //没找到,说明输入的字符串中的字符都是“mirror"的.
{
string::reverse_iterator rEnd = str.rend(), rIter;
for( rIter = str.rbegin(); rIter != rEnd; ++rIter ) //反向输出
{
if( *rIter == 'b' ) //很抱歉,我实在是想不到高级而且简单的方法了.
{
*rIter = 'd';
}
else if( *rIter == 'd' )
{
*rIter = 'b';
}

cout