Word Reversal （简单字符串处理）

题目描述：

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

 //题意描述：输入一个字符串，不改变单词的顺序，将每个单词反转输出
 //解题思路: 用gets读入，去头去尾后，判断出一个单词，将其复制到另一个字符串中逆序输出，直至字符串尾即可
 #include<stdio.h>
 #include<ctype.h>
 #include<string.h>

 void pri(char p[])
 {
     int i,l=strlen(p);
     for(i=;i<l;i++)
         printf("#%c",p[i]);
     printf("\n");
 }

 int main()
 {
     int T,r,i,j,k;
     char s[],*p=NULL,s2[];
     scanf("%d",&T);
     while(T--)
     {

         scanf("%d",&r);
         getchar();
         while(r--)
         {
             gets(s);
             int l=strlen(s);
             s[l]='\0';
             //pri(s);

             p=s;
             while(!isalpha(*p))
             {
                 p++;
             }
             //pri(p);

             l=strlen(p);
             for(i=l-;i>=;i--)
                 if(isalpha(p[i]))
                     break;
             p[i+]=' ';
             p[i+]='\0';
             //pri(p);

             l=strlen(p);
             for(i=;i<l;)
             {
                 j=;
                 while(isalpha(p[i]))
                 {
                     s2[j++]=p[i++];
                 }
                 s2[j]='\0';

                 for(k=j-;k>=;k--)
                     printf("%c",s2[k]);

                 if(i==l-)
                 printf("\n");
                 else
                 printf(" ");
                 i++;
             }
         }
         if(T != )
         printf("\n");
     }
     return ;
 }
 //易错分析
 //gets之前要吃掉换行