杭电 1062 Text Reverse

Text Reverse

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11832 Accepted Submission(s): 4487


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each test case, you should output the text which is processed.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

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.

[align=left]Author[/align]
Ignatius.L

　　这个题调试了一段时间才AC，提交了三四次，都是PE！
　　这个题应该注意什么呢？每个单词之间可以有一个或多个空格，如果用scanf去获得每一行的一个单词，在输出的时候，就可能在应该输出多个空格的地方，只输出了一个空格！导致输出格式的错误！另外，我提交的代码AC了，说明oj提供给这个题测试样例的每一行的单词之间应该是没有制表符的！
以下是代码：

View Code

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char *argv[])
{
int n, Len, i, flag, j;
char s[1005];
scanf( "%d", &n );
getchar();
//printf( "%d\n", n );
while( n-- )
{
gets(s);
//printf( "%s\n",s );
//getchar();
Len = strlen(s);
//printf( "%d\n", Len );
flag = 0;
for( i = 0; i < Len; i++ )
if( s[i] == ' ' )
{
for( j = i-1; j >= flag; j-- )
printf( "%c", s[j] );
flag = i+1;
printf( "%c",s[i] );
}
else if( i == Len-1 )
{
for( j = i; j >= flag; j-- )
printf( "%c", s[j] );
flag = i+1;
}
printf( "\n" );
}

s//ystem("PAUSE");
return 0;
}