joj1146

  1146 : Word Reversal

Result
TIME Limit
MEMORY Limit
Run Times
AC Times
JUDGE
3s
8192K
2658
690
Standard
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

 
This problem is used for contest:  36  
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이런 문자열을 C로 쓰면 약간 #$%^&*로 처리됩니다.버퍼 구역의 문제는 비교적 틀리기 쉽다.'The C programming language'의 안내문을 되돌아보는 것 같다.

 1 #include <stdio.h>
 2 
 3 int main(void)
 4 {
 5     //freopen("in.txt","r",stdin);
 6 
 7     int m, n, i, j, k;
 8     int flag=1;
 9 
10     char c[100];
11     char ch;
12 
13     while(scanf("%d", &n) == 1)
14     {
15         while (n > 0)
16         {
17             n--;
18             scanf("%d", &m);
19             getchar(); 
20 
21             if(flag==1)
22             {
23                 flag = 0;
24             }
25             else
26             {
27                 printf("
");
28             }
29 
30             i = 0;
31             while(m > 0)
32             {
33 
34                 ch = getchar();
35 
36                 if(ch!='
' && ch!=' ')
37                 {
38                     c[i++]=ch;
39                 }
40                 else if(ch==' ')  // 
41                 {
42                     if(i != 0)
43                     {
44                         for(j=0; j<i; j++)
45                         {
46                             printf("%c", c[i-j-1]);
47                         }
48 
49                         i = 0;  // 0
50                         printf(" ");
51                     }
52                     else    // 
53                     {
54                         printf(" ");
55                     }
56                 }
57                 else    // 
58                 {
59                     for(j=0; j<i; j++)
60                     {
61                         printf("%c", c[i-j-1]);
62                     }
63 
64                     i=0;
65                     printf("
");
66 
67                     m--;
68                 }
69             }
70         }
71     }
72 
73     return 0;
74 }