ZOJ 1151 Word Reversal反转单词 （string字符串处理）

链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=151

For each list of words, output a line with each word reversed without changing the order of the words.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Input

You will be given a number of test cases. The first line contains a positive integer indicating the number of cases to follow. Each case is given on a line containing a list of words separated by one space, and each
word contains only uppercase and lowercase letters.

Output

For each test case, print the output on one line.

Sample Input

1

3

I am happy today

To be or not to be

I want to win the practice contest

Sample Output

I ma yppah yadot

oT eb ro ton ot eb

I tnaw ot niw eht ecitcarp tsetnoc

翻译：

对于一串单词，直接把他们输出在一行上，要把每一个单词反转，但每一个单词的位置不要改变；

本程序包括多组測试数据;

输入数据的第一行是一个整数N，然后是一空行，后面跟着N个数据块，每一个数据块的格式在程序描写叙述中说明了，数据块中有一空行。

输出格式由N个输出块组成，每一个输出块之间有一空行；

输入描写叙述：

现给你多组測试数据，第一行是一个正整数，表示接下来測试数据的组数，每组測试占一行，包括一串单词，中间用一个空格隔开，每一个单词仅包括大小写字母；

输出描写叙述：

每组測试数据都打印在一行上；

代码：

#include <iostream>

#include <string>

#include <cstdio>

#include <algorithm>

using namespace std;



string s, t;

int n, cas, flag = 0;

char ss[1000];



int main()

{

    cin >> cas;

    while(cas--) {

        scanf("%d", &n);

        getchar();

        while(n--) {

            cin.getline(ss, 1000);

            s = ss;

            flag = 0;

            t.clear();

            for(int i=0; i<s.size(); i++) {

                if(s[i] != ' ' && i <= s.size()-1) {

                    t += s[i];

                }else {

                    reverse(t.begin(), t.end());

                    if(flag) cout << " ";

                    cout << t;

                    flag = 1;

                    t.clear();

                }

            }

            reverse(t.begin(), t.end());

            cout << " " << t << endl;

        }

        if(cas) cout << endl;

    }

    return 0;

}