HDUOJ--------Text Reverse

Text Reverse Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 13449    Accepted Submission(s): 5140 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.

 基础题：

代码：

 1 #include<iostream>

 2 #define maxn 1005

 3 using namespace std;

 4 char str[maxn][maxn];

 5 int main()

 6 {

 7     int n,i,cnt,j;

 8     char c;

 9     bool ifhave;

10     //freopen("test.in","r",stdin);

11     //freopen("teset.out","w",stdout);

12     scanf("%d",&n);

13     getchar();

14     while(n--)

15     {

16         j=i=0;

17         ifhave=false;

18      while(c=getchar(),c!='\n')

19      {

20          if(c==' ')

21              j++,i=0;

22          else

23             str[j][i++]=c;

24          

25      }

26      if(c==' ')

27          ifhave=true,j--;

28      cnt=j;

29      for(j=0;j<=cnt;j++)

30      {

31          for(i=strlen(str[j])-1; i>=0 ; i--)

32              printf("%c",str[j][i]);

33             if(j!=cnt) printf(" ");

34      }

35      if(ifhave)printf(" ");

36      puts("");

37      memset(str,'\0',sizeof str);

38     }

39     return 0;

40 }

比较高级的代码：

 1 #include<stdio.h>

 2 #include<string.h>

 3 int main()

 4 {

 5   int t;

 6   scanf("%d\n",&t);

 7   while(t--)

 8   {

 9     char s[1005]={};

10     while(scanf("%[^ \n]",s)!=EOF)

11     {

12       printf("%s",strrev(s));

13       s[0]=0;

14       putchar(getchar());

15     }

16   }

17   return 0;

18 }