joj1146

 1146: Word Reversal

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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

 

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This problem is used for contest: 36 

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用C写这种字符串处理的还是有点#$%^&*的。缓冲区的问题，比较容易写错(更可能是我太水)。感觉像在回顾《The C programming language》的导言。

#include <stdio.h>

int main(void)
{
    //freopen("in.txt","r",stdin);

    int m, n, i, j, k;
    int flag=1;

    char c[100];
    char ch;

    while(scanf("%d", &n) == 1)
    {
        while (n > 0)
        {
            n--;
            scanf("%d", &m);
            getchar(); 

            if(flag==1)
            {
                flag = 0;
            }
            else
            {
                printf("\n");
            }

            i = 0;
            while(m > 0)
            {

                ch = getchar();

                if(ch!='\n' && ch!=' ')
                {
                    c[i++]=ch;
                }
                else if(ch==' ')  //一个单词结束
                {
                    if(i != 0)
                    {
                        for(j=0; j<i; j++)
                        {
                            printf("%c", c[i-j-1]);
                        }

                        i = 0;  //清0
                        printf(" ");
                    }
                    else    //处理一行的第一个字母是空格
                    {
                        printf(" ");
                    }
                }
                else    //一行结束
                {
                    for(j=0; j<i; j++)
                    {
                        printf("%c", c[i-j-1]);
                    }

                    i=0;
                    printf("\n");

                    m--;
                }
            }
        }
    }

    return 0;
}