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            1062 Text Reverse

            Posted on 2008-10-09 19:32 歲月流逝 閱讀(899) 評論(0)  編輯 收藏 引用

            Text Reverse

            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
            Total Submission(s): 1391    Accepted Submission(s): 577


            Problem Description
            Ignatius likes to write words in reverse way. Given a single line of text which is written by Ignatius, you should reverse all the words and then output them.
             

            Input
            The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
            Each test case contains a single line with several words. There will be at most 1000 characters in a line.
             

            Output
            For each test case, you should output the text which is processed.
             

            Sample Input
            3
            olleh !dlrow
            m'I morf .udh
            I ekil .mca
             

            Sample Output
            hello world!
            I'm from hdu.
            I like acm.
            
            Hint
            Remember to use getchar() to read '\n' after the interger T, then you may use gets() to read a line and process it.
             

            Author
            Ignatius.L
             
            #include <stdio.h>
            #include <string.h>
            int main()
            {
                int i=0,n,length,t,j;
                char ch,p[1001],temp;
                scanf("%d",&n);
                ch=getchar();
                while(n--)
                {
                    gets(p);
                    length=strlen(p);
                    t=0;
                    for(i=0;i<=length;i++)
                    {
                        if(p[i]==' '||p[i]=='\0')
                        {
                           for(j=t;j<(i+t)/2;j++)
                        {
                            temp=p[j];
                            p[j]=p[i-j+t-1];
                            p[i-j+t-1]=temp;
                        }
                        t=i+1;
                        }
               
                    }   
                    printf("%s\n",p);
               
                }
                return 0;
            }
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