Word Reversal

Description

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 yppah yadot
oT eb ro ton ot eb

I tnaw ot niw eht ecitcarp tsetnoc

直接采用栈的形式保存单词的字符，碰到空格或换行就可以直接打印就行了。

#include 
#include 
const int maxn = 100005;
char str[maxn], stack[maxn];
int is_letter ( char ch )   //判断字母
{
    return ch >= 'A' && ch <= 'Z' || ch >= 'a' && ch <= 'z';
}
int main ( )
{
    int T, n, cas = 0, len;
    //freopen ( "in0.in", "r", stdin );
    scanf ( "%d", &T );
    while ( T -- )
    {
        scanf ( "%d", &n );
        getchar ( );    //吃掉换行
        if ( cas ++ )   //每个测试数据中一个空行
            printf ( "\n" );
        while ( n -- )
        {
            int top = -1;
            fgets ( str, maxn, stdin );
            //使用fgets,需要注意会接收\n
            len = strlen ( str );

            for ( int i = 0; i < len; i ++ )
            {
                if ( is_letter ( str[i] ) )
                    stack[++ top] = str[i];
                else
                {
                    while ( top >= 0 )
                        printf ( "%c", stack[top --] );
                    printf ( "%c", str[i] );
                }
            }
            while ( top >= 0 )  //最后还有可能有没有出栈的数据
                printf ( "%c", stack[top --] );
        }
    }
    return 0;
}