Zoj problem set-1151 word reversal

Time Limit: 2 seconds memory limit: 65536 KB

For each list of words, output a line with each word reversed without changing the order of the words.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. each input block is in the format indicated in the Problem description. there is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Input

You will be given a number of test cases. the first line contains a positive integer indicating the number of cases to follow. each case is given on a line containing a list of words separated by one space, and each word contains only uppercase and lowercase letters.

Output

For each test case, print the output on one line.

Sample Input

1

3
I am happy today
To be or not to be
I want to win the practice contest

Sample output

I ma yppa yadot
Ot EB Ro ton ot EB
I tnaw ot NIW EHT ecitcarp tsetnoc

Source:East central North America 1999, practice

This is a question about output format control. The reverse function can be used directly in VC 10.0. in earlier versions, # include <algorithm> is required,CodeAs follows:

 
# Include <iostream>
 
# Include <sstream>
 
 
# Include <algorithm>
 
 
Using NamespaceSTD;
 
 
 
 
IntMain ()
 
{
 
 
IntBlocks; CIN> blocks;
 
 
For(IntBlock = 0; block <blocks; block ++)
 
 
{
 
StringState;
 
 
If(Block = 0)// There are only the following two blank lines at the first time
 
 
{
 
 
Getline (CIN, State );// Use Getline to eat the carriage return after the input Blocks
 
Getline (CIN, State );// Empty rows required by the question
 
 
}
 
 
IntStatecount; CIN> statecount;
 
 
Getline (CIN, State );
 
While(Statecount --)
 
 
{
 
 
Getline (CIN, State );
 
 
Istringstream is (State );
 
 
StringWord;
 
IntI = 0;
 
 
While(Is> word)
 
 
{
 
 
Reverse (word. Begin (), word. End ());
 
 
If(I! = 0)
 
Cout <" ";
 
 
Cout <word;
 
 
I ++;
 
 
}
 
 
Cout <Endl;
 
}
 
 
If(Block! = Blocks-1)// No blank lines are output at the last time
 
 
{
 
 
Cout <Endl;
 
 
}
 
}
 
 
 
 
Return0;
 
 
}
 
 
 
 
 
 
 
The reverse function is the key to solving this problem, which is defined in the algorithm files. Here is the reverse order function of string written by myself:
 
VoidReversestring (String& Str)
 
 
{
 
 
Size_t strlen = Str. Length ();
 
 
For(Size_t I = 0; I <strlen/2; I ++)
 
 
{
 
STR [I] = STR [I] ^ STR [strlen-I-1];
 
 
STR [strlen-I-1] = STR [I] ^ STR [strlen-I-1];
 
 
STR [I] = STR [I] ^ STR [strlen-I-1];
 
 
}
 
 
}
 
 
The test code is as follows:
 
# Include <iostream>
 
 
# Include <String>
 
 
Using NamespaceSTD;
 
 
 
VoidReversestring (String& Str)
 
 
{
 
 
Size_t strlen = Str. Length ();
 
 
For(Size_t I = 0; I <strlen/2; I ++)
 
 
{
 
STR [I] = STR [I] ^ STR [strlen-I-1];
 
 
STR [strlen-I-1] = STR [I] ^ STR [strlen-I-1];
 
 
STR [I] = STR [I] ^ STR [strlen-I-1];
 
 
}
 
 
}
 
IntMain ()
 
 
{
 
 
StringA ="ABC";
 
 
Cout <"Befor reverse, A is"<A <Endl;
 
Reversestring ();
 
 
Cout <"After reverse, now a is"<A <Endl;
 
 
Return0;
 
 
}
 
 
Output result: