栈|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
【栈|Word Reversal】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;
}
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