![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
/*
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.
*/
#include<iostream>
using namespace std;
int main()
{
int n;
cin>> n;
getchar();
while(n--)
{
char a[1001];
memset(a,0,sizeof(a));
gets(a);
char*p,*q,*k;
k= q= p= a;
while(*p!=0)
{
int c=0;
while(*p==' ')
{
p++;
cout<<' ';
}
k= p; //k标记字母开始的第一个位子
while(*p!=' '&& *p!=0) //把p移到最后字母的后一个
p++;
q= p; //q标记最后一个字符的后一个位子
p--; //倒退一个,p就是标记最后一个字母了,输出全都字母
while(p>=k)
{
cout<< *p;
p--;
}
p= q;
}
cout<< endl;
}
return 0;
}
对于字符串的输入还不是很了解的同学可以看看http://hi.baidu.com/__afterward__/blog/item/de8ff4b22ce3fa8ccbefd0e5.html
Text Reverse
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2704 Accepted Submission(s): 805
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.