HDU——1062Text Reverse（水题string::find系列+reverse）

Text Reverse

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

Total Submission(s): 24157 Accepted Submission(s): 9311



[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.

刚开始发现最后一个单词和前一个连在一块了，改了一个地方就过了,本打算用stringstream重定向，想想好麻烦der，还是用最喜欢的string系列函数吧.还好string支持加号运算-.-|||

#include<iostream>
#include<cmath>
#include<algorithm>
#include<string>
using namespace std;
int main(void)
{
int t,loc;
cin>>t;
getchar();
string s1;
while (t--)
{
getline(cin,s1);
loc=0;
while (s1.find(" ",loc)!=string::npos)//反转前n-1个单词
{
reverse( s1.begin()+loc , s1.begin()+s1.find(" ",loc) );
loc=s1.find(" ",loc)+1;//更新每次反转的begin位置
}
reverse( s1.begin()+s1.find_last_of(" ")+1 , s1.end() );//反转最后一个单词
cout<<s1<<endl;
}
return 0;
}