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.

程序思路：

通过空格区分单词，每检测到一个空格更新指针，一个指向头的位置，一个指向尾的位置。

程序：

#include "iostream"

#include "string"

#include "stdio.h"

using namespace std;

int main()

{

    int T;

    string str;

    int len;

    int cnt = 0;

    int start, end;

    cin >> T;

    getchar();

    for (int i = 0;i < T;i++)

    {

        getline(cin, str);

        len = str.size();

        str[len] = ' ';

        start = 0;

        for (int j = 0;j < len + 1;j++)

        {

            if (str[j] == ' ')

            {

                end = j - 1;

                for (int n = end;n >= start;n--)

                    cout << str[n];

                start = j + 1;

                if(j!=len)

                    cout << ' ';

                else

                    cout << endl;

            }

        }

    }

    return 0;

}

补充：

由于每个空格检测一次，所以还需要在接收到的字符串的最后补上一个空格。程序思路不难，很容易搞懂。