杭电ACM 1062

废话不说，先上题：

Text Reverse

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17744 Accepted Submission(s): 6730


[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.

这道题初看感觉不难，但是提交的时候 各种报PE错误 擦 无奈了
经过N次测试后 发现测试数据 各种数据之间有多个空格，数据起始可能有多个空格 基本解决好这两个问题就能提交了
先上未优化前的代码:

#include <stdio.h>
#include <string.h>
char a[1002];
void print(char *q)
{
while((q+1)!=a)
{
printf("%c", *q--);
if (*q==' ') {break;}
}
}
int main()
{
int n,i;
char *p;
while(scanf("%d",&n)!=EOF)
{
getchar();
while(n--){
memset(a,'\0',sizeof(a));
gets(a); p=a;
while(*p==' '){putchar(' ');p++;};
while(1)
{
while(*p!=' ')
{
if(*(p+1)=='\0'){print(p);break;}
if (*(p+1)==' ') {print(p);putchar(' ');break;}
p++;
}p++; p++;
if(*p=='\0')break;
else while(*p==' '){putchar(' ');p++;};
}
putchar('\n');
}
}
return 0;
}

还有用数组的：

#include <stdio.h>
int main()
{
int n,i,j,k;
char array[1001],t;
while(scanf("%d",&n)!=EOF)
{
getchar();
while(n--)
{
i=0;
gets(array);
while(array[i]!='\0')
{
if(array[i]==' ') { i++;continue; }
j=i;
while((array[j+1]!=' ')&&(array[j+1]!='\0')) j++;
k=j;
while(i<j)
{
t=array[i]; array[i]=array[j]; array[j]=t;i++;j--;
}
i=k+1;
}
puts(array);
}
}
return 0;
}