&&&least significant difference
的翻译结果:
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&在分类学科中查询
很抱歉，暂未找到该词条的译词。
The least significant difference (LSD) t test showed that: (1)serum glucose level increased at 30min and 1 80min period (P<0.05 vs. saline control) in Group B and C;
30min一120min时，HOMA一IR显著升高(尸<0.05); 以上各指标在B和C组之间差异无统计学意义(P值皆大于0.05)。
Datas were expressed as means + SEM,analysis of variance was used to test for differences on SPSS 10.0,least significant difference test (LSD-t) for multiple comparison.
统计学处理：所有数据在SPSS10.0软件上行方差分析，各组间两两比较用LSD法进行。
By comparing growth characteristics with least significant difference, different treatments showed a highly significant difference among plant height, root length, root total length of Astragalus sinicus and Medicago sative (p<0.01), but the effect was not significant for subroots.
多重比较说明,不同处理之间对紫云英和紫花苜蓿株高、根系长、根总长的作用差异极显著(P<0.01),对侧根数作用的差异不大。
Cells were transfected by the combination plasimids, then the inhibition rate of proliferation of the transfected cells were tested. The results were analyzed using the methods of statistics including Jin's Q-test, Least significant difference (LSD),cluster analysis.
按照组合用质粒转染细胞,然后对转染细胞的增殖抑制率进行检测,并采用金正均Q值法、单因素方差分析中的LSD法、聚类分析法等统计学方法对结果进行统计分析。
The results of multiple range analysis(Method:95% Least Significant Difference)for feeding rates of small-sized and large-sized
Mesocyclops
revealed a statistically significant difference(
多重统计分析结果表明，广布中剑水蚤对小型和大型裸腹?的摄食率有显著差异（P＝0?0335），捕食者喜食小型猎物。
Finally,the least significant bitplanes of BG are shifted down.
4)向下偏移重要性最差的BG位平面。
The difference was significant.
卡方检验 χ2 =5 .2 0 8,P <0
0 2 5 ,差异显著。
The difference was significant.
有显著性差异 (P <0 .0 1)。
sunasa's was the least.
sunasa最少;
and significant development.
最后指出了我国人寿保险的发展趋势和方向。
查询“least significant difference”译词为用户自定义的双语例句&&&&我想查看译文中含有：的双语例句
为了更好的帮助您理解掌握查询词或其译词在地道英语中的实际用法，我们为您准备了出自英文原文的大量英语例句，供您参考。&&&&&&&&&&&&&&&&&&&& The growth performane of 100 China fir families planted on trial sites inGuangdong Province has been analysed. The trial designed in accordancewith mathematical statistics applies the method of performance levelanalysis to compute the ranks of the increment of every family in theexperimental stands, and adopts the least significant difference test as-wellas contrast method to synthesize and analyse all the levels. The testhas preliminarily selected 20 superior families and determined diffe... &&&&&&&&&&&&本文分析了100个杉木家系在广东省6个试验点造林的生长表现情况,试验按照数理统计的要求设计,采用性状水平分析法(performace Lerel Analysis),对各试验林中各个家系的生长量作出标准得分估算,并应用最小湿著差异检验法和对照对比法,从各个水平进行综合分析。初步选择出20个优良家系,并提出了各个家系在广东省的特别适生区域,为建立优树无性系第1代种子园提供了依据。&&&&&&&& This paper studies how the optimum design of animal threeway cross experiments can be made so that the precision of the experiments is maximized for a given testing expense, or that the testing expense is minimized for a given heterosis(Least significant difference). Two factors are considered: the first is the cost involved in securing an individual with phenotype record, the second is the phenotype variances of the population used in the experment. &&&&&&&&&&&&本文给出了家畜三元杂交试验的优化设计方法和原理。从两个方面分别进行讨论:经费相同时使试验结果精度最高;满足试验可靠性要求而所需费用最少。结果用公式给出,简单易用,易于推广。&&&&&&&& According to the collected data in the northwest part grassland of Jilin Province,the bifactor interaction analysis of variance is conducted to independence between four slopes and three sampling methods of Citellus dauricus population (P>0. 50). And again,the bifactor analysis of variance is obtained to not significant difference among the populations for three sampling methods (P>0. 50), and very significant difference among the four slopes (P<0. 001). Using to the least s... &&&&&&&&&&&&根据吉林省西北部草原4种坡度和3种抽样比例的达乌尔黄鼠密度调查资料,利用双因素有交互作用的方差分析,得到坡度与抽样方法独立(P0.50),坡度间的差异极为显著(P<0.001)。用最小显著差数法检验,得到坡脚与其它3种坡度间的鼠密度差异极为显著(P0.10)。给出了坡度X与鼠密度Y的回归方程Y=0.9X(P<0.05)。 &&&&&&&&相关查询:
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&2008中国知网(cnki) 中国学术期刊(光盘版)电子杂志社digit的翻译中文意思-在线英汉词典
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&&英文单词：digit
简明英汉词典digit[5didVit]n.阿拉伯数字, 手指或足趾, 一指宽(约四分之三英寸)美国传统词典[双解]digitdig.itAHD:[d1j“1t] D.J.[6did9it]K.K.[6d!d9!t]n.（名词）(1)A human finger or toe.人的手指或足趾(2)A corresponding part in other vertebrates.手指，足趾：其它脊椎动物相对应的部位(3)A unit of length derived from the breadth of a finger and equal to about ? of an inch (2.0 centimeters).指宽，指幅：从手指宽度派生出的长度单位，等于?英寸（2.0厘米）(4)One of the ten Arabic number symbols, 0 through 9.数字：十个阿拉伯数字符号从0到9中的任意一个(5)Such a symbol used in a system of numeration.位（数）：用于计算系统的符号语源(1)Middle English 中古英语 (2)from Latin digitus [finger, toe] * see deik- 源自 拉丁语 digitus [手指，足趾] *参见 deik- 现代英汉词典digit[5dIdVIt]n.(1)从0到9的任一数字The number 2001 contains four digits.2001是个四位数。(2)手指；脚趾现代英汉综合大辞典digit[5didVit]n.(1)手指或足趾(2)一指宽的长度单位(约 3/4 英寸)(3)阿拉伯数字(即 0, 1, 2...9; 有时0除外)(4)【天】太阳(月亮)直径的 1/12 (用作测定日食、月食的单位)The number 410 contains three digits.数字 410 中包括三个数目字。特殊用法binary digit二进制数字[数位]carry digit进位数(字); 移位数字check digit检验数位coded decimal digit(二进制)编码的十进制数字compensating digits【信】补偿数字data digit不连续数据decimal digit十进制数字(位)decit decimal digit十进数字dialed digit拨号数字equal-order digits等位数(字)equivalent binary digit等效二进位数error per digit每(个)数(位)误差final carry digit终端进位数fixed digit定趾function digit操作数gap digit【自】间隔位guard digit保护位, 保护数位(用来避免溢出时信息丢失)hexadecimal digit十六进制数字high-order digit高位数位independent digit独立数(字)input digit输入数intermediate sum digit中间和数位least significant digit最低(有效)位, 最低有效数字left-hand digit左边的数字, 高数位low-order digit低位数message digit消息数位metering digit数字记录most significant digit最高(有效)位, 最高有效位数字multiplier digit乘数(的)数(字)noise digit【自】噪声数位noisy digit【自】噪声数位nonzero digit非零位normal random digit正规随机数字numerical digit数字octal digit八进制数字operated digits被加数的数位output digit输出数字parity digit奇偶校验位per digit每个字quinary digit五进制数(字)random digits随机数字redundant digit【自】冗余位retained digit保留的数字sandwich digit中间数字, 中间数位second-order digit第二位数sexadecimal digit十六进制数字sign digit符号位significance digit有效位significant digit有效数(字)special indication digit专用指示(数)码sum digit和数位sum-check digit和数校验位ten's digit十进制数top digit最高位, 最高位数字unallowable instruction digit非法字符voltage digit电压-数字(转换器)英文相关词典digitfigure&&&&number&&&&numeral&&&&[七国语言]英汉电子工程大词典digit数字[七国语言]英汉数学大词典digit数字[七国语言]英汉信息大词典digit数字[七国语言]英汉医学大词典digit指, 趾英汉双解计算机词典digit数字;［数位］1. A graphic character that represents an integer, one of the characters 0 to 9.一个表示整数的形象字符(即0到9中的任一个)。 2. A symbol that represents one of the nonnegative integers smaller than the radix.In decimal notation,one of the characters from 0 to 9. | 表示比数基小的非负整数的符号。在十进制记数法中,每位数是0到9的一个数字。 | 3. 同numeric character。美国传统词典digitdig.itAHD:[d1j“1t] D.J.[6did/it]K.K.[6d!d/!t]n.(1)A human finger or toe.(2)A corresponding part in other vertebrates.(3)A unit of length derived from the breadth of a finger and equal to about ? of an inch (2.0 centimeters).(4)One of the ten Arabic number symbols, 0 through 9.(5)Such a symbol used in a system of numeration.语源(1)Middle English (2)from Latin digitus [finger, toe] * see deik- [名词委审定]英汉计算机名词(第二版, 2002)digit数字[名词委审定]英汉昆虫学名词（2000）digit(1)趾(2)趾英汉船舶大词典digit n.数,数字,数字;n.位 朗文英汉综合电脑词典digit 数字，位数，位(十进制数的) 英汉电信大词典digit n.数字,长度单位 英汉地质大词典digit n.数字,数位,数字符号 英汉纺织大词典digit n.数,数字 英汉广播大词典digit n.数字,数位 英汉航海大词典digit n.数字,数 英汉航空大词典digit n.数,数位 英汉化学大词典digit n.数字,手指,位数 英汉海运大词典digit 手指数字计数单位3/4英寸的长度单位号位数食分(太阳，月直径的1/12)数位数字符号单值数 基本词义digit 手指数字计数单位3/4英寸的长度单位号位数食分(太阳，月直径的1/12) 基本词义digit 数(字)位 基本词义digit 数字，数位数字符号单值数 英汉经贸大词典digit n.数字 英汉计算机大词典digit n.数字(手指,位数 英汉机械大词典digit n.数字,数字 英汉建筑大词典digit n.位,号,数字 最新会计师英汉大词典digit数字英汉农牧林大词典digit n.趾,指,数字 英汉能源大词典digit n.数字,手指,位数,数字,数位,数字符号 英汉水利大词典digit n.位数,数字,十进数的位 英汉贸易大词典digit数字英汉消防大词典digit 数(字) 现代商务英汉大词典digit数字英汉冶金大词典digit n.数字,数位 英汉医学大词典digit n.数字 英汉中医大词典digit n.指,趾
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least significant digit怎么解释
石川沙都子
least significant digit最小有效位数双语对照词典结果：least significant digit[英][li:st siɡˈnifikənt ˈdidʒit][美][list sɪɡˈnɪfɪkənt ˈdɪdʒɪt]最低（有效）位，最低有效数字;
扫描下载二维码Least Significant Non-Zero Digit of n!
Least Significant Non-Zero Digit of n!
Let p(k) be the least significant non-zero decimal digit of k!
first several values of this sequence are
2,6,4,2,2,4,2,8,8,8,6,8,2,8,8,6,8,2,4,4,8,4,6,4,4,8,4,6,8,...
Can we directly determine the kth term for any given k?
Also, what
is the asymptotic distribution of the digits?
To answer these
questions, let L(k) denote the least significant non-zero decimal
digit of the integer k.
Writing n! in the form
n! = (2^a2)(5^a5)(3^a3)(7^a7)...
we can let (n!)' denote this same number divided by its highest
power of 10, i.e.,
(n!)' = (2^(a2-a5))(3^a3)(7^a7)...
Since we've divided out all powers of 10, the least significant
digit of this number is non-zero, as are the least significant digits
of the factors.
Thus we have
L[ L(2^a2 - a5) L(3^a3) L(7^a7) ...]
For any given integer n we can compute the exponent of any prime p
in n! simply by summing the nearest integers to [n/p^j] for j=1,2,..
For example, the exponent of 3 in 89! is given by
[89/3] + [89/9] + [89/27] + [89/81]
29 + 9 + 3 + 1
Furthermore, the least significant decimal digit of 3^k is cyclical
with the four values {1,3,9,7}, so it's easy to see that L(3^42) = 9.
Likewise, the least significant digits of the sequence p^k, k=1,2,..
for every odd prime ending with the digit 3 or 7 has a period of four,
while those ending with 9 have a period of two, and those ending with
1 have a period of one.
Thus, all these periods are divisors of four.
Of course, the least significant digits of 2^k also have a period of
four, i.e., {2,4,8,6}.
Also, as we multiply successive integers to generate n!, we always
have more powers of 2 than of 5, so the value of L(n!) is easily
computed recursively as L(L(n)L((n-1)!)) unless n is a multiple of
5, in which case we need more information.
As a result, the values
of L(n!) come in fixed strings of five, as shown below
Thus, if we know the value of L((5n)!) we automatically know the values
of L((5n+j)!) for j=0,1,2,3,4.
However, the pattern of the values
L((5n)!) is not immediately apparent.
If we tabulate these values we
find that they too come in fixed strings of five, so we only need to
know L((25n)!) to automatically know L((25n+5j)!) for j=0,1,2,3,4.
Continuing in this way, we can tabulate the values of L((5^t n)!) as
shown below
Notice that the pattern of digits for L(625n!) is the same as for
In general it appears that the pattern for L((5^j n)!) is the
same as for L((5^(j+4) n)!).
In addition, on each level there are
precisely four distinct blocks of 5 sequential digits, one block
beginning with each of the digits 0,2,4,6,8.
From the above tabulations we can extract the essential patterns
for L((5^k n)!)
Look-Up Table for L() Patterns
------------------------------
This table represents all we need to determine the value of L(n!) for
any integer n.
First we convert n to the base 5, so we have
n = d_0 + d_1*5 + d_2*5^2 + ... + d_h*5^h
Now we enter the above table at the row h (mod 4) in the block whose
first digit is 0 (because the coefficient of 5^(h+1) is zero), and
determine the digit in the (d_h)th position of this block.
digit be denoted by s_h.
Then we enter the table at row h-1 (mod 4)
in the block that begins with s_h, and determine the digit in the
(d_(h-1))th position of this block.
Let this digit be denote by
We continue in this way down to s_0, which is the least
significant non-zero digit of n!.
To illustrate, consider the case of the decimal number n=1592.
the base 5 this is n=22332.
Now we enter the above table at row
k=4=0 (mod 4) in the block beginning with 0, which is 06264.
leading digit of n (in the base 5) is 2, so we check the digit in
position 2 of this block to give L((2*5^4)!) = 2.
Then we enter
the table at row k=3 (mod 4) in the block beginning with 2, which
is 26648, to find L((2*5^4 + 2*5^3)!) = 6.
Then in the row k=2 (mod 4), the block beginning with 6 is 64244,
and we find L((2*5^4 + 2*5^3 + 3*5^2)!) = 4.
From this we know we're
in the block 48226 in row k=1 (mod 4), so we have L((2*5^4 + 2*5^3 +
3*5^2 + 3*5)!) = 2.
Finally, we enter the row k=0 (mod 4) in block
22428 to find the result
L((2*5^4 + 2*5^3 + 3*5^2 + 3*5 + 2)!)
To streamline this process, let's define an array A(4,5,5) where the
first index signifies the row (0,1,2,3), the second is the block
selector (0,2,4,6,8) in that row, and the third is the digit number
(0,1,2,3,4) in that block.
If it's understood that the first index
is to be taken modulo 4, and if we let dj denote the jth digit of the
base-5 representation of n, then the above evaluation be written in
0, d4) = s4
A(3, s4, d3) = s3
A(2, s3, d2) = s2
A(1, s2, d1) = s1
A(0, s1, d0) = s0
L((d4*5^4 + d3*5^3 + d2*5^2 + d1*5 + d0)!)
This shows how we can easily determine the value of L(n!) by means of
k look-ups (in a simple fixed 4x5x5 table) where k is the number of
base-5 digits of n.
From this we can also rigorously determine the
distribution of digits, which the table's symmetry seems to imply
must be uniform.
Just checking empirically, we find the following
distribution of the values of L(n!) for n from 2 to 10^t with t=4,5,6.
(This excludes n=1 for which L(n!)=1.)
Naturally a similar analysis can be performed with respect to the
least significant digit of n! in any other base.
For example, in
the base 3, we find that the blocks on the even levels are 112 and
221, and the blocks on the odd levels are 122 and 211.
information we can construct the function table shown below.
where "p" denotes the parity of the exponent of 3 for the current
To illustrate, suppose we wish to determine the least
significant non-zero base-3 digit of (139!).
The number 139 written
in the base 3 is 12011, so the exponent of 3 for the leading digit
is 4, which has parity 0.
Thus the parity string of the exponents
Beginning with the most significant digit, 1, and a
"previous output" of 1 (which is always the initial "previous output")
we enter the table in the row 1 1 to find that the output is p+1,
which equals 1 (because the current exponent parity is p=0).
we take this output and the next input digit, 2, and enter the table
at the row 1 2 to get the output 2.
Then we take this output and
the next input digit, 0, and enter the table in the row 2 0 to find
the output 2.
Next we enter at 2 1 to find he output 2-p, and on
this level we have p=1, so the output is 1.
Finally we enter the
table at row 1 1 to find the output p+1, and on this level we have
p=0, so the final output is 1.
This process essentially acts as a kind of "filter", taking consecutive
digits from the set {0,1,2} and outputing digits from the set {1,2},
just as the decimal algorithm takes a stream of digits from the set
{0,1,...,9} and outputs a stream of digits from the set {2,4,6,8}.
For the base-3 filter, a continuous stream of "0" input digits will
leave the output unchanged, i.e., it will retain the previous output
On the other hand, a continuous stream of "2" input digits
will cause the output to oscillate between 1 and 2 on each step.
continuous stream of "1" input digits will act like "0" when the
parity is even, and will act like "1" when parity is odd, with the
result being that the output will change state on the odd steps.
