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Speculative store buffer
A speculativestorebuffer is speculatively updated in response to speculative storememory operations buffered by aload/storeunit in a microprocessor. I...
分类:其他好文   时间:2014-09-24 01:40:45    阅读次数:200
hdu5009 Paint Pearls (DP+模拟链表)
http://acm.hdu.edu.cn/showproblem.php?pid=50092014网络赛 西安 比较难的题Paint PearlsTime Limit: 4000/2000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Othe...
分类:其他好文   时间:2014-09-23 21:09:25    阅读次数:249
python 报错 wxPyDeprecationWarning: Using deprecated class PySimpleApp.
如题: python报错提示为:wxPyDeprecationWarning:UsingdeprecatedclassPySimpleApp.解决: 将wx.PySimpleApp()替换为wx.App(False)即可。参考: http://stackoverflow.com/questions/15144168/error-using-deprecated-class-pysimpleapp-after-removing-epd
分类:移动开发   时间:2014-09-23 02:33:24    阅读次数:5303
Android学习绘制图形
Android的绘图继承于View组件,重写onDraw(Canvas canvas) 方法时涉及一个绘图API: Canvas 代表了依附于指定View的画布,并且提供了很多绘制方法。Canvas提供的方法还涉及一个API: Paint ,Paint代表了Canvas上的画笔,主要用于设置绘制风格...
分类:移动开发   时间:2014-09-22 15:45:42    阅读次数:188
HDU - 5023 A Corrupt Mayor's Performance Art(区间修改)
Problem Description Corrupt governors always find ways to get dirty money. Paint something, then sell the worthless painting at a high price to someone who wants to bribe him/her on an auction, this ...
分类:其他好文   时间:2014-09-20 22:56:29    阅读次数:258
HDU 5023 A Corrupt Mayor's Performance Art (线段树)
Problem Description Corrupt governors always find ways to get dirty money. Paint something, then sell the worthless painting at a high price to someone who wants to bribe him/her on an auction, this ...
分类:其他好文   时间:2014-09-20 20:23:09    阅读次数:289
Sparse Coding: Autoencoder Interpretation
稀疏编码在稀疏自编码算法中,我们试着学习得到一组权重参数 W(以及相应的截距 b),通过这些参数可以使我们得到稀疏特征向量 σ(Wx + b) ,这些特征向量对于重构输入样本非常有用。稀疏编码可以看作是稀疏自编码方法的一个变形,该方法试图直接学习数据的特征集。利用与此特征集相应的基向量,将学习得到的...
分类:其他好文   时间:2014-09-19 21:00:16    阅读次数:400
Canvas drawText实现中英文居中
@Override protected void onDraw(Canvas canvas) { super.onDraw(canvas); Paint mTextPaint = new Paint(Paint.ANTI_ALIAS_FLAG); ...
分类:其他好文   时间:2014-09-19 17:36:15    阅读次数:315
go channel learning
I spent several hours to figure out some conceptions about channel tonight1. buffered channel and non-buffered channelbuffered channelbufferedchan := ...
分类:其他好文   时间:2014-09-19 01:08:44    阅读次数:256
hdu-5009-Paint Pearls-dp
由题意我们可以知道,花费最多为n。 所以单次最多涂掉sqrt(n)种颜色。 dp[i]:涂到第i个位置,之前的花费最少为多少。 biao[i][j]:在第i个位置,往前涂j-1种颜色,涂到哪个位置。 vis[i]:i颜色最后出现的位置,不存在等于-1。 我们先离散化颜色。 然后很显然转移方程: dp[i]=min(dp[i],dp[biao[i][j]]+(j+1)*(j+1)); ...
分类:其他好文   时间:2014-09-17 16:56:22    阅读次数:201
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