zoukankan      html  css  js  c++  java
  • google在线测试练习2

    Problem

    Given a list of space separated words, reverse the order of the words. Each line of text contains L letters and W words. A line will only consist of letters and space characters. There will be exactly one space character between each pair of consecutive words.

    Input

    The first line of input gives the number of cases, N.
    N test cases follow. For each test case there will a line of letters and space characters indicating a list of space separated words. Spaces will not appear at the start or end of a line.

    Output

    For each test case, output one line containing "Case #x: " followed by the list of words in reverse order.

    Limits

    Small dataset

    N = 5
    1 ≤ L ≤ 25

    Large dataset

    N = 100
    1 ≤ L ≤ 1000

    思路:先一个一个单词反转,再将整个字符串反转。

    #include<iostream>
    #include<fstream>
    #include<string>
    using namespace std;
    int main()
    {
    	int n_case;
    	ifstream ifile("B-large-practice.in");
    	ofstream ofile("resultb2.txt");
    	ifile >> n_case;
    	for(int i = 0; i <= n_case; i++)
    	{
    		char line[1002];
    		ifile.getline(line, 1002);
    		string words(line);
    		int begin = 0;
    		int end = 0;
    		if(i == 0)
    			continue;
    		for(int i = 1; i < words.length(); i++)
    		{
    			if(words[i] ==  ' ')
    			{
    				int sum = begin + i - 1;
    				for(int j = begin; j <= sum / 2; j++)
    				{
    					char tmp = words[j];
    					words[j] = words[sum - j];
    					words[sum - j] = tmp;
    				}
    				begin = i + 1;
    			}
    		}
    		int sum = begin + words.length() - 1;
    		for(int j = begin; j <= sum / 2; j++)
    		{
    			char tmp = words[j];
    			words[j] = words[sum - j];
    			words[sum - j] = tmp;
    		}
    		sum = words.length() - 1;
    		for(int j = 0; j <= sum / 2; j++)
    		{
    			char tmp = words[j];
    			words[j] = words[sum - j];
    			words[sum - j] = tmp;
    		}
    		ofile << "Case #" << i << ": " << words << endl;
    	}
    	return 0;
    }
    


    版权声明:本文博主原创文章,博客,未经同意不得转载。

  • 相关阅读:
    Linux stress 命令
    接口安全设计
    2019年MTP管理技能提升培训笔记
    Docker运行图形化程序
    CentOS7搭建本地YUM仓库,并定期同步阿里云源
    搭建私服-docker registry
    linux剪贴板
    如何搭建Docker私有仓库
    彻底解决 LINK : fatal error LNK1123: 转换到 COFF 期间失败: 文件无效或损坏
    为人父母始知天下事---“宝宝哭了”的问题来说说什么是分析,什么是设计
  • 原文地址:https://www.cnblogs.com/lcchuguo/p/4887991.html
Copyright © 2011-2022 走看看