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原创 eclipse emf 读写xml文件
import java.io.IOException;import java.util.Collections;import java.util.Map;import org.eclipse.emf.ecore.resource.Resource;import org.eclipse.emf.ecore.resource.ResourceSet;import org.eclipse.e
2016-03-26 09:23:56
2125
原创 树莓派驱动ov7670
卡片电脑树莓派确实强大,但是官方推荐的摄像头有点贵,正好手头有ov7670摄像头 就试试能不能驱动主要用到树莓派的GPIO接口 和 i2c接口 i2c这里要看看树莓派的官方文档 插上i2c线后先执行 i2cdetect -y 1 来确定地址 我是树莓派B版 所以是1 如果是A版的话执行 i2cdetect -y 0初始化之后就是各种时序好读取图像了 直接上代码 接口定义都写的比
2014-05-21 21:16:46
8773
原创 C# 读取 文件正由另一进程使用,因此该进程无法访问该文件。
System.IO.FileStream fs = new System.IO.FileStream(fileName, System.IO.FileMode.Open,System.IO.FileAccess.Read,FileShare.ReadWrite);这样就可以读了
2012-02-13 18:46:34
879
原创 CutSameLine for text
#include #include#includeusing namespace std;string str;char chars[1000];set strs;int old,ne;int main (){ while(scanf("%s",chars)!=EOF) { string str(chars); old=strs
2011-12-22 13:07:32
469
原创 android 输入对话框 确认对话框
protected Dialog onCreateDialog(int id) { if (id == 0) {// 注销 AlertDialog.Builder builder = new AlertDialog.Builder(this); builder.setMessage("你确认要注销吗?") .setCancelable(false) .setPositiveButton("确定", new DialogInterface.OnC
2011-05-21 20:14:00
6151
原创 转发:android 调用系统的接口
显示网页:1. Uri uri = Uri.parse("");2. Intent it = new Intent(Intent.ACTION_VIEWuri);3. startActivity(it);显示地图:1. Uri uri = Uri.parse("geo:38.-77.");2. Intent it = new Intent(Intent.Action_VIEWuri);3. startActivity(it);路径规划:1. Uri uri = Uri.parse("");2. Intent
2011-05-10 16:43:00
1637
原创 c# 全局快捷键
c# 全局快捷键 热键 alt+z 好像是(UInt32)Keys决定的 using System.Runtime.InteropServices;using Microsoft.Win32; public Form1() { InitializeComponent(); //注册热键
2011-04-28 16:41:00
4102
原创 java实现文件加密解密
<br />import java.io.File;import java.io.FileInputStream;import java.io.FileOutputStream;import java.security.MessageDigest;import java.security.NoSuchAlgorithmException;import java.util.Scanner;import javax.crypto.Cipher;import javax.crypt
2011-04-23 16:39:00
22216
原创 hdu Priest John's Busiest Day
Problem DescriptionJohn is the only priest in his town. October 26th is the John's busiest day in a year because there is an old legend in the town that the couple who get married on that day will be forever blessed by the God of Love. This year N couples
2011-03-15 22:28:00
896
原创 Problem 2581 Bomb the Bridge
<br /><br />Problem 2581 Bomb the Bridge<br /> <br /> <br />Problem Description<br />You want to destroy a bridge with bombs .The lower-left corner of the bridge is at (0,0) and upper-right corner is at ( w , l ). There are already b bombs exploded , the i
2011-03-01 22:37:00
762
原创 Problem - 3777 Page Count
<br /><br />Problem - 3777 Page Count<br /> <br />Problem Description<br />When you execute a word processor's print command, you are normally prompted to specify the pages you want printed. You might, for example, enter:<br /> <br />10-15,25-28,8-4,13-20
2011-03-01 22:12:00
567
原创 hdu Problem - 1597 find the nth digit
find the nth digitProblem Description假设:S1 = 1S2 = 12S3 = 123S4 = 1234.........S9 = 123456789S10 = 1234567891S11 = 12345678912............S18 = 123456789123456789..................现在我们把所有的串连接起来S = 1121231234.......123456789123456789112345678912.........那么你
2011-03-01 21:43:00
1169
原创 指定时间内结束指定程序名的多个进程 (c++)
<br />#include <iostream>#include <fstream>#include <windows.h>#include <cstdlib>#include <string>#include <cstring>using namespace st
2011-01-07 17:56:00
985
转载 java 执行dos命令并回显
package main;import java.io.BufferedReader;import java.io.InputStream;import java.io.InputStreamReader;public class Dos { /** * @param args */ public static void main(String[] args) { try { String cmd = "cmd /c ping baidu.com"; Runtime rt = Runtime.
2010-10-30 10:59:00
4094
原创 poj 1050
貌似二维最大字串和先看下一维数组的最大字段和已知n个整数(可能为负整数)组成的序列a1,a2,…,an,求该序列的子段和的最大值。当所有整数均为负整数时定义其最大子段和为0。只介绍dp算法:b[j]是a1,a2,…,aj中以aj结尾的最大子段和。1、证明b[j]满足最有子结构如果b[j]是a1,a2,…,aj中以aj结尾的最大子段和,那么b[j-1]一定是a1,a2,…,a(j-1)以a(j-1)结尾的最大子段和。否则,设b’是a1,a2,…,a(j-1)以a(j-1)结尾的最大子段和,那么,b’+a[j]
2010-10-28 21:35:00
752
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原创 POJ 1179
哎 悲剧啊 竟然搞了4个小时! 枚举断边 再dp 将存数组和符号的数组扩大一倍能使思维更简单特别注意负负得正的情况 所以max min都要记下来 #includeusing namespace std;#define MAXN 101struct Node { long mmax; long mmin;} f[MAXN][MAXN];int main() { long n, num[101], l, i, k, j, len, maxx[101], wa
2010-10-27 22:49:00
743
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