英语人>网络例句>anti isomorphic lattice 相关的网络例句
anti isomorphic lattice相关的网络例句

查询词典 anti isomorphic lattice

与 anti isomorphic lattice 相关的网络例句 [注:此内容来源于网络,仅供参考]

Explicitly, they are consist of three classes: If H is semisimple, then H k* for some finite group; If H is not semisimple and the characteristic of k is zero, then H is isomorphic the dual of the cross product between one so called Andruskiewitsch-Schneider algebra and a group algebra; If H is not semisimple and the characteristic of k is not zero, then H is isomorphic the dual of the cross product between one special algebra and a group algebra.

具体地讲,它们共分三类:①如果H是半单的,则H同构与一个群代数的对偶;②如果H是非半单的并且基础域的特征是0的话,则H同构一个所谓Andruskiewitsch-Schneider代数与一个群代数交差积的对偶;③如果H是非半单的并且基础域的特征不是0的话,则H同构于某个特定代数与一个群代数交差积的对偶。

More precisely,we shall say that two graphs are isomorphic if there is a one-one correspondence between their vertices which has the property that two vertices are joined by an edge in one graph if and only if the corresponding vertices are joined by an edge in the other [3],Another graph isomorphic to the graphs in Figs.

更准确地说,如果两个图的顶点间存在一一对应,使得在一个图中有一条边连接两个顶点在另一个图中相应也有且仅有两个对应顶点由一条边相连,则说这两个图是同构的。

In this paper a new necessary and sufficient condition of graphic isomorphism is proved which is stated as following: two graphs are isomorphic if and only if their subgraphs are isomorphic and the new vertices as well as their adjancy edges are corresponding with the property of isomorphism.

提出了一个新的判定图同构的充分必要条件,即在子图同构的前提下,根据新增顶点及相应关联边的关系,判断母图同构的充分必要条件。

Explicitly, they are consist of three classes: If H is semisimple, then H k* for some finite group; If H is not semisimple and the characteristic of k is zero, then H is isomorphic the dual of the cross product between one so called Andruskiewitsch-Schneider algebra and a group algebra; If H is not semisimple and the characteristic of k is not zero, then H is isomorphic the dual of the cross product between one special algebra and a group algebra.

具体地讲,它们共(来源:5fbfA02BC论文网www.abclunwen.com)分三类:①如果H是半单的,则H同构与一个群代数的对偶;②如果H是非半单的并且基础域的特征是0的话,则H同构一个所谓Andruskiewitsch-Schneider代数与一个群代数交差积的对偶;③如果H是非半单的并且基础域的特征不是0的话,则H同构于某个特定代数与一个群代数交差积的对偶。

Kaspersky Security Suite CBE Win7 is essential similar to Kaspersky Internet Security 2010, with both has the same build version number of 9.0.0.736, and includes complete protection features such as anti-virus, anti-malware, anti-rootkit-, anti-Trojan, anti-worm, anti-spam, parental control, two-way firewall, Urgent Detection System, Security Analyzer, Safe Run, Auto-Run Disable, iSwift and iChecker Scanning, virtual keyboard, URL Advisor, whitelisting and application control.

卡巴斯基安全套装CBE Win7是必不可少的相似,卡巴斯基互联网安全2010年,都具有相同的建设版本9.0.0.736数量,包括诸如反完整的保护功能的病毒,反恶意软件,反rootkit -,反木马,反病毒,反垃圾邮件,父母控制,双向防火墙,紧急检测系统,安全分析,安全运行,自动运行禁用,iSwift和iChecker扫描,虚拟键盘,网址顾问,白名单和应用控制。

In 2000, F. LUCA proved that Fermat number are anti-sociable numbers, and in 2005, M. H. LE proved all powers of 2 are anti-sociable numbers. We have used the method of M. H. LE to obtain some new results of the anti-sociable numbers. For every integer n containing prime divisors that are 1 mod 4, let p mod 4 be an arbitrary prime divisor of n. There is at least one anti-sociable number in n^2, p^2n^2, p^4n^2, and p^6n^2. Therefore we can prove that anti-sociable numbers have positive density in perfect square numbers. We also give a method to find the exact anti-sociable numbers.

LUCA证明了Fermat数都是孤立数;2005年,乐茂华教授证明了2的方幂都是孤立数,用乐茂华教授的方法给出孤立数的一些新的结果:对于任意含有4w+1型素因子的正整数n,设p为n的任意一个4w+1型素因子,则在n^2,p^2n^2,p^4n^2,p^6n^2里至少有一个是孤立数,因此可以证明孤立数在完全平方数里有正密度,另外也给出求解确定孤立数的方法。

For the main-feeding, anti-inflammatory, anti-mutation, anti-corrosion, the anti-cancer, anti-ulcer, anti-virus, catharsis, inhibiting disease-free, pesticide, purge, spasmolytic, hemostat .

主要用于拒食素,消炎,抗突变,防腐,抗癌的,抗溃疡,抗病毒,导泻,免抑制疫力,杀虫剂,通便,解痉,止血剂。

The company persists in its tenet of "all-around green" in its researches and developments of anti-aging and beautifying projects; complies totally with the requirements on green anti-aging beautifications of World Health Organization of "no skin breaking, no injection, no drugs, no dependence, no wound and no pollution"; designs products and brings forward technical standards based on the guideline of "simple, safe, swift and effective". Since its establishment, the company has successfully structured five anti-aging and beautifying technological conceptions and formed four series products, including "basic beautifying series, involucra anti-aging series, youth planting series and health managing series". It has constituted the standards of healthy anti-aging and beautifications for the professional anti-aging and beautifying field of China.

公司在研发抗衰老美容项目中坚持全面绿色的宗旨,完全按照联合国卫生组织对绿色抗衰老美容的要求,不破皮、不打针、不吃药、不依赖、不创伤、不污染,以&简单、安全、快捷、有效&的方针来设计产品和提出技术标准,公司自成立以来,先后成功推出了五大抗衰老美容科技理念,完整地形成了&美容基础系列产品、皮膜抗衰老系列产品、青春种植个性化基因产品、健康管理系列产品&体系,在中国专业抗衰老美容领域制定了健康抗衰老美容标准。

In addition, CRT has anti-glare surface, anti-static the AGAS (Anti-GlareAnti-Static) coating, anti-reflective, anti-static ARAS (Anti-ReflectionAnti-Static) coating, hand touch, but also in the top left under the hand, do not believe you see the display from the side, you can see a handprint on your screen, do you want to help the Public Security Bureau's busy uncle in advance display extracted harm "murderer" of the fingerprint it?

另外,CRT的表面有防强光、防静电的AGAS(Anti-GlareAnti-Static)涂层,防反射、防静电的ARAS(Anti-ReflectionAnti-Static)涂层,用手触摸,还会在上面留下手印,不信你从侧面看显示器,就能看到一个个手印在你的屏幕上,难道你想帮公安局叔叔们的忙,提前提取出伤害显示器&凶手&的指纹吗?

In addition, CRT's surface anti-glare, anti-static the AGAS (Anti-GlareAnti-Static) coating, anti-reflective, anti-static of the ARAS (Anti-ReflectionAnti-Static) coating, hand-touch, but also to stay above The fingerprint, do not believe you can display from the side, you can see one in your handprint on the screen, do you want to help the public security bureau uncle's busy, in advance extract hurt Monitors "murderer" fingerprint?

另外,CRT的表面有防强光、防静电的AGAS(Anti-GlareAnti-Static)涂层,防反射、防静电的ARAS(Anti-ReflectionAnti-Static)涂层,用手触摸,还会在上面留下手印,不信你从侧面看显示器,就能看到一个个手印在你的屏幕上,难道你想帮公安局叔叔们的忙,提前提取出伤害显示器&凶手&的指纹吗?

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你认为你是一个大人物?

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所有的州都是平等的,任何州都不能获得联邦政府的特别待遇。

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