r*y
2 楼
请教。谢谢!
s*q
3 楼
之前在Intel P45主板上,利用主板ICH10自带的Raid功能,组了一个Raid 1
后来,这台机器就送人了,但Raid 1的2块硬盘还在我这
由于是Raid 1,所以硬盘上的数据很容易读出来,但是我是希望:
如果在不擦除这2块硬盘上数据的情况下重新建立Raid 1
如果在不购买P45/P55/X58主板的前提下,有什么方案(比如软raid, raid卡)
可供选择?
后来,这台机器就送人了,但Raid 1的2块硬盘还在我这
由于是Raid 1,所以硬盘上的数据很容易读出来,但是我是希望:
如果在不擦除这2块硬盘上数据的情况下重新建立Raid 1
如果在不购买P45/P55/X58主板的前提下,有什么方案(比如软raid, raid卡)
可供选择?
h*y
4 楼
【 以下文字转载自 ebiz 讨论区 】
发信人: witness (毛毛能), 信区: ebiz
标 题: [zz]如何顺利带iPad通关
发信站: BBS 未名空间站 (Wed Dec 29 23:57:30 2010, 美东)
see picture.
发信人: witness (毛毛能), 信区: ebiz
标 题: [zz]如何顺利带iPad通关
发信站: BBS 未名空间站 (Wed Dec 29 23:57:30 2010, 美东)
see picture.
G*G
5 楼
n个点,进行曲线光滑。
要求,第j个点的光滑,不能用到j+1,j+2,....n以后的点的信息。
请问什么算法最好?
要求,第j个点的光滑,不能用到j+1,j+2,....n以后的点的信息。
请问什么算法最好?
M*s
6 楼
需要在美国报税。
p*i
7 楼
我以前试过用一块ICH(n)下的raid 1盘放到ICH(n+1)的机器上, 启动成功后加第二块硬
盘rebuild raid 1
我migrate之前在windows里force安装了ich(n+1)的ata controller的驱动
如果你还有其他支持intel matrix storage的主板, 可以拿一块盘试试
【在 s***q 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 之前在Intel P45主板上,利用主板ICH10自带的Raid功能,组了一个Raid 1
: 后来,这台机器就送人了,但Raid 1的2块硬盘还在我这
: 由于是Raid 1,所以硬盘上的数据很容易读出来,但是我是希望:
: 如果在不擦除这2块硬盘上数据的情况下重新建立Raid 1
: 如果在不购买P45/P55/X58主板的前提下,有什么方案(比如软raid, raid卡)
: 可供选择?
盘rebuild raid 1
我migrate之前在windows里force安装了ich(n+1)的ata controller的驱动
如果你还有其他支持intel matrix storage的主板, 可以拿一块盘试试
【在 s***q 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 之前在Intel P45主板上,利用主板ICH10自带的Raid功能,组了一个Raid 1
: 后来,这台机器就送人了,但Raid 1的2块硬盘还在我这
: 由于是Raid 1,所以硬盘上的数据很容易读出来,但是我是希望:
: 如果在不擦除这2块硬盘上数据的情况下重新建立Raid 1
: 如果在不购买P45/P55/X58主板的前提下,有什么方案(比如软raid, raid卡)
: 可供选择?
S*A
8 楼
你要几阶的光滑啊?
你的光滑是要通过这些点还是和这些点某些距离函数最小?
单单说光滑太笼统。
你的光滑是要通过这些点还是和这些点某些距离函数最小?
单单说光滑太笼统。
h*c
10 楼
gauss quadrature, 不是太懂
有点得瑟了
有点得瑟了
b*s
13 楼
b spline
a*a
19 楼
then the formula I gave should be it. if you don't require crossing the
points, then it is a different matter.
starting from 0th, you can pick an arbitrary tangent. you have now y0, y0'.
then going to the 1st point, the only thing you can do is a quadratic form.
3 equations for 3 unknowns like I laid out before. y0=a x0^2+b x0+c,y1=a x1^
2+b x1+c,y0'=2 a1 x0+b1)
This will fix the tangent at point 1. then you solve anther equation set of
3. (y1=a x1^2+b x1+c,y2=a x2^2+b x2+c,y1'=2 a1 x1+b1). lather rinse repeat
forward until you reach the end.
This is a reduced form of spline. Typically when one talks about spline, it
mean cubic spline. However cubic spline uses piece wise cubic polynomials, y
=a x^3+b x^2 +c x +d. at least 3 points must be involved for there to be a
solution. That means to determine the piece between point 0 and point 1, you
must use info on point 2, which violates your requirement.
As a result, if I understand your requirement correct, as far as piecewise
spline is concerned, all you can do is a "quadratic" spline so to speak.
the curve is smooth to the 2nd order. if all you can use is (x0,y0),(x1,y1)
for the first segment, then the pieces going up to (x2,y2) can not be more
than 2nd order smooth anyway.
【在 G***G 的大作中提到】![](/moin_static193/solenoid/img/up.png)
: 如果不能用到后面的点,那么它的切向量只能从以前得出。
: 0就假设切向量为0.
points, then it is a different matter.
starting from 0th, you can pick an arbitrary tangent. you have now y0, y0'.
then going to the 1st point, the only thing you can do is a quadratic form.
3 equations for 3 unknowns like I laid out before. y0=a x0^2+b x0+c,y1=a x1^
2+b x1+c,y0'=2 a1 x0+b1)
This will fix the tangent at point 1. then you solve anther equation set of
3. (y1=a x1^2+b x1+c,y2=a x2^2+b x2+c,y1'=2 a1 x1+b1). lather rinse repeat
forward until you reach the end.
This is a reduced form of spline. Typically when one talks about spline, it
mean cubic spline. However cubic spline uses piece wise cubic polynomials, y
=a x^3+b x^2 +c x +d. at least 3 points must be involved for there to be a
solution. That means to determine the piece between point 0 and point 1, you
must use info on point 2, which violates your requirement.
As a result, if I understand your requirement correct, as far as piecewise
spline is concerned, all you can do is a "quadratic" spline so to speak.
the curve is smooth to the 2nd order. if all you can use is (x0,y0),(x1,y1)
for the first segment, then the pieces going up to (x2,y2) can not be more
than 2nd order smooth anyway.
【在 G***G 的大作中提到】
![](/moin_static193/solenoid/img/up.png)
: 如果不能用到后面的点,那么它的切向量只能从以前得出。
: 0就假设切向量为0.
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