ZOJ1151――Word Reversal

원제:
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
분석:
개인 적 으로 창고 지식 을 쓰 는 게 좋 을 것 같 아서...
오늘 은 예전 의 지식 을 복습 하고 예전 의 지식 을 많이 잊 어 버 린 것 같 지만 지금 은 복습 하고 기억 하 는 것 이 빠 르 고 활용 할 수 있 습 니 다.
원본 코드:

#include<stdio.h> #include<algorithm> #include<iostream> #include<stack> #include<string.h> using namespace std; stack<char>z; int main() {     int k;     int n;     char a[999];     scanf("%d",&k);     while(k--)     {         scanf("%d",&n);         getchar();         while(n--)         {             gets(a);             int l=strlen(a);             int i=0;             while(1)             {                 while(a[i]!=' '&&i<l)                 {                     z.push(a[i]);                     i++;                 }                 while(!z.empty())                 {                     printf("%c",z.top());                     z.pop();                 }                 if(i!=l)                 {                     printf(" ");                     i++;                 }                 else break;             }             printf("
");         }         if(k)             printf("
");     }     return 0; }