ZOJ Problem Set–1151 Word Reversal

Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

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Source: East Central North America 1999, Practice

这是一道考察输出格式控制的题目；其中reverse函数在VC 10.0中可以直接用，而在早一些的版本中，需要#include<algorithm>, 代码如下：

#include<iostream>
#include<sstream>
#include<algorithm>
using namespace std;
int main()
{
  int blocks;cin>>blocks;
  for(int block = 0; block < blocks; block++)
  {
    string state;
    if(block == 0)//仅在第一次的时候有如下两个空行
    {
      getline(cin,state);//用getline 吃掉输入blocks之后的回车
      getline(cin,state);//题目要求的空行
    }
    int stateCount;cin>>stateCount;
    getline(cin, state);
    while(stateCount--)
    {
      getline(cin, state);
      istringstream is(state);
      string word;
      int i = 0;
      while(is>>word)
      {
        reverse(word.begin(), word.end());
        if(i != 0)
          cout<<" ";
        cout<<word;
        i++;
      }
      cout<<endl;
    }
    if(block != blocks - 1)//最后一次输出没有空行
    {
      cout<<endl;
    }
  }
  
  return 0;
}

 

其中reverse函数是解决这道题的关键，在algorithm都文件中有定义。一下是我自己写的string的反序函数：

void ReverseString(string& str)
{
  size_t strLen = str.length();
  for(size_t i = 0; i < strLen/2;i++)
  {
    str[i] = str[i]^str[strLen - i - 1];
    str[strLen - i - 1] = str[i]^str[strLen - i - 1];
    str[i] = str[i]^str[strLen - i - 1];
  }
}

测试代码如下：

#include<iostream>
#include<string>
using namespace std;
void ReverseString(string& str)
{
  size_t strLen = str.length();
  for(size_t i = 0; i < strLen/2;i++)
  {
    str[i] = str[i]^str[strLen - i - 1];
    str[strLen - i - 1] = str[i]^str[strLen - i - 1];
    str[i] = str[i]^str[strLen - i - 1];
  }
}
int main()
{
  string a = "abc";
  cout<<"Befor reverse , a is "<<a<<endl;
  ReverseString(a);
  cout<<"After reverse, now a is "<<a<<endl;
  return 0;
}

输出结果：