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=0;i<l;i++)
        printf("#%c",p[i]);
    printf("
");
}

int main()
{
    int T,r,i,j,k;
    char s[5001],*p=NULL,s2[101];
    scanf("%d",&T);
    while(T--)
    {
        
        scanf("%d",&r);
        getchar();
        while(r--)
        {
            gets(s);
            int l=strlen(s);
            s[l]='';
            //pri(s);
            
            p=s;
            while(!isalpha(*p))
            {
                p++;
            }
            //pri(p);
            
            l=strlen(p);
            for(i=l-1;i>=0;i--)
                if(isalpha(p[i]))
                    break;
            p[i+1]=' ';
            p[i+2]='';
            //pri(p);
            
            l=strlen(p);
            for(i=0;i<l;)
            {
                j=0;
                while(isalpha(p[i]))
                {
                    s2[j++]=p[i++];
                }
                s2[j]='';
                
                for(k=j-1;k>=0;k--)
                    printf("%c",s2[k]);
                
                if(i==l-1)
                printf("
");
                else
                printf(" ");
                i++;
            }
        }
        if(T != 0)
        printf("
");
    }
    return 0;
} 
//易错分析
//gets之前要吃掉换行