There are two typical use cases for
super:
In a class hierarchy withsingle inheritance, super can be used to refer to parent classes withoutnaming them explicitly, thus making the code more mainta...
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编程语言 时间:
2015-06-05 21:17:28
阅读次数:
218
因为有了Node.js,JavaScript可以被用于服务端编程。通过各种扩展,Node.js可以变得非常强大。今天分享下怎样用C++创建Node.js扩展。 参考原文:Making Dynamsoft Barcode SDK an Addon for Node.js 搭建Nodejs开...
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编程语言 时间:
2015-06-02 11:30:59
阅读次数:
2022
CD MakingTime Limit: 20 SecMemory Limit: 256 MB题目连接http://acm.uestc.edu.cn/#/problem/show/65DescriptionTom has N songs and he would like to record the...
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其他好文 时间:
2015-06-02 00:17:06
阅读次数:
195
girlfriend.h
class girlfriend
{
public:
string name;
int age;
face look;
string education_background;
protected:
void making_food(boyfriend &me);
void go_shopping(boyfriend &m...
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其他好文 时间:
2015-05-31 18:26:41
阅读次数:
94
iOS Programming Localization 本地化 Internationalization is making sure your native cultural information is not hard-coded into your application. 国际化确保你的...
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移动开发 时间:
2015-05-27 12:13:48
阅读次数:
226
题目如下:
A long-distance telephone company charges its customers by the following rules:
Making a long-distance call costs a certain amount per minute, depending on the time of day when the call is...
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编程语言 时间:
2015-05-26 21:35:30
阅读次数:
1019
Spring中配置和读取Properties文件 public class PropertiesFactoryBeanextends PropertiesLoaderSupportimplements FactoryBean, InitializingBean Allows for making a properties file from a classpath locat...
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编程语言 时间:
2015-05-25 20:47:03
阅读次数:
214
SyncTrayzor is a little tray utility forSyncthingon Windows.It hosts and wraps Syncthing, making it behave more like a native Windows applicationand l...
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Windows程序 时间:
2015-05-21 17:11:02
阅读次数:
1583
Problem Description
ZLGG found a magic theory that the bigger banana the bigger banana peel .This important theory can help him make a portal in our universal. Unfortunately, making a pair of porta...
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其他好文 时间:
2015-05-13 22:06:01
阅读次数:
153
首先可以明确一个方面,那就是如果将X改成Y,那么Y肯定是这N个数中的某一个(为什么仔细想想)
之后可以得到一个状态转移那就是dp[i][j]代表已经考虑了i位的情况下,结尾为j的最小更改数。
状态转移为dp[i][j] = min(dp[i-1][k] + abs(a[i] - b[j])) 这样的话可以写出一个初步的代码:
#include
#include
#include
using...
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其他好文 时间:
2015-05-12 23:03:49
阅读次数:
146