HDU-1062-Text Reverse（细节题）

题目链接：点击打开链接

Text Reverse

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 26663    Accepted Submission(s): 10393

[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.

虽然是道水题，但是有好多需要注意的地方，就贴一下

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
char str[1010];
int main()
{
int t;
scanf("%d",&t);
getchar();
while(t--)
{
gets(str);
int len=strlen(str);
int start=0;
bool flag=0;
for(int i=0;i<len;i++)
{
if(str[i]==' ')
{
if(flag) printf(" ");
for(int k=i-1;k>=start;k--)
printf("%c",str[k]);
start=i+1;
flag=1;
}
if(i==len-1)
{
printf(" ");
for(int k=i;k>=start;k--)
printf("%c",str[k]);
}
}
puts("");
/*
for(int i=0;i<len;i++)
{
if(str[i]==' '||i==len-1) 这样写是 PE的，举个特例：a_b_ 这个程序的答案是：a__b,正确答案应该是：a_b
{
if(flag) printf(" ");
for(int k=(i==len-1?i:i-1);k>=start;k--)
printf("%c",str[k]);
start=i+1;
flag=1;
}
}
puts("");*/
}
return 0;
}