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

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

巴特西

Word Reversal （简单字符串处理）

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