 全文文献 工具书 数字 学术定义 翻译助手 学术趋势 更多   one and only 的翻译结果: 查询用时：0.186秒 在分类学科中查询 所有学科 计算机硬件技术 更多类别查询 历史查询  one and only 一个也只能(4)  一个也只能
 (2) we propose algorithms to modify the preferred list in such a way that each node will be selected as the km preferred node of one and only one other node and proposed a simple mechanism to tolerate node failures. 提出了调整结点的优先列表的算法，以保证每个结点被一个也只能是一个结点选择做自己的第七个优先结点，并相应地提出了一个处理结点失败的算法。 短句来源 If node n has not heard from node i for T(i)out since its receipt of the latest broadcast from node i, it will consider node i failed, then modify the preferred list in such a way that each node will be selected as the kth preferred node of one and only one other node. 如果结点 n自从最后一次收到结点i的广播信息以后T(i)out还没有收到结点i的广播信息，就认为结点i 失败，然后调整各正常的结点的优先列表，以保证每个结点被一个也只能是一个结点选作自己的第k个优先结点，以减少任务的失败率。 短句来源 In this case, it needs to modify the preferential list for all nodes to ensure each node in the preferential list is selected as the k th preferential node by one and only one node else. 这时需要调整各正常结点的优先列表,以保证每个结点被一个也只能是一个结点选择为自已的第k个优先结点,以减少任务的失败率。 短句来源 In a distributed real-time system, if node n has not heard from node i for T~((i))_(out) since its receipt of the latest broadcast from node i, it will be considered that node i failed; then the preferred list is modified in such a way that each node will be selected as the k~(th) preferred node of one and only one other node. 实时分布式系统中,如果结点n自从最后一次收到结点i的广播信息以后超过时间T(i)out还没有收到结点i的广播信息,就认为结点i失败,然后调整各正常的结点的优先列表,以保证每个结点被一个也只能是一个结点选择做自己的第k个优先结点,以减少任务的失败率. 短句来源 “one and only”译为未确定词的双语例句
 For each integer a with 1≤a≤q and (a,q)=1,there exists one and only one b with 1≤b≤q such that ab≡1 mod q. 对任意与q互素且满足1≤a≤q的整数a,存在唯一一个整数b满足1≤b≤q使得ab≡1modq. 短句来源 The dynamic programming equations: Uk(y)= define one, and only one, sequence ψk in D1 with || ψk || D1 ≤L; 动态规划方程在D_1中定义一个唯一的序列ψ_k,且满足||ψk||_(D1)≤L; 短句来源 Let X be a Banach space,and K be a nonempty bounded closed convex subset of X which has a normal structure. Define a mean nonexpanded mapping T on K as:‖Tx-Ty‖≤a‖x-y‖+b‖x-Ty‖, for any x,y in K and any nonnegative real numbers a and b with a+b≤1. Then T has one and only one fixed point in K. X表示Banach空间,K是X中的非空有界闭凸子集且具有正规结构,已知平均非扩张映射T:K→K,满足‖Tx-Ty‖≤a‖x-y‖+b‖x-Ty‖, x,y∈K,a,b≥0,a+b≤1在K中存在唯一的不动点. 短句来源 a map of S into itself,a an element of S. Then there exists one and only one map f from N to S such thatf(1)= a f(a+) = (?) ,恒唯一存在N到S的映射f满足条件 f(1)=a,f(a~+)=(?) 短句来源 Afterwards,it is obtainted that the equation x 3+ax+b=0(a≠0,b≠0) has one and only one root or one simple root and one twofold root or three different roots,and it has no roots in the field F. 然后得到x3+ax +b =0 (a≠ 0 ,b≠ 0 )在域F中有且仅有一根 ,或一个单根与一个二重根 ,或三个互异的根 ,或没有根 . 短句来源 更多 相似匹配句对
 One is R. 研究结果表明,危害贵州省小麦的病原丝核菌至少有两种,一种是R. 短句来源 One of the 5 E. 分离的E . 短句来源 Tao and One “道”与“一” 短句来源 I am Only One 只有我一个(英文) 短句来源 The tapetum is glandular one. 绒毡层为腺质绒毡层. 短句来源 查询“one and only”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  one and only
 It is demonstrated that of all finite strain measures there is one and only one able to realize fully uncoupled separation of the volumetric and the isochoric deformation in a natural, additive manner. Consider a polynomialP (z) of degreen whose zeros are known to lie inn closed disjoint discs, each disc containing one and only one zero. With the algorithm described, it is possible to decompose a directed graph without loops and cycles into subgraphs which have one and only one source, one and only one sink and no other connections to the complement of the subgraph. Most respondents displayed a combination of motives while for only 11% of them altruism ("Achieving something positive for others") was the one and only driving force behind their interest in international volunteering. After reaction, the one and only heteropolyanion is partly reduced [PMo12O40]3- and vanadium is present as vanadyl cations. 更多 This paper gives a general discussion on the fundamental postulates of the theory of special relativity and emphasizes that the theory of special relativity has one and only one postulate, i.e. Einstein's principle of relativity. Finally, from Maxwell equations a rigorous derivations of the mathematical expressions for the fundamental postulate are presented. 本文对狭义相对论的基本假设作一番探讨并强调狭义相对论有一条,且仅有一条基本假设,即爱因斯坦的相对性原理.最后从麦克斯韦方程组,严格推导狭义相对论基本假设的数学表达式. According to the concept of implication of the multiple-output cubes, a method of forming a simplified covering table used in multiple-output function is presented. It can be applied to solve the cover-matrix complementation of a single-output function as well as of a multiple-output function. By this method, one can obtain all the meaningful irredundant covers of a multiple-output function, including minimum cover. For a large covering table, the application o f the modified operation of sharp-product can acquire... According to the concept of implication of the multiple-output cubes, a method of forming a simplified covering table used in multiple-output function is presented. It can be applied to solve the cover-matrix complementation of a single-output function as well as of a multiple-output function. By this method, one can obtain all the meaningful irredundant covers of a multiple-output function, including minimum cover. For a large covering table, the application o f the modified operation of sharp-product can acquire one and only one group of nonredundant cover. This method can also be used for elimination of redundant output connections. The generation and solution of the covering table can be simultaneously carried out row by row so that only a part or even none of the covering table need to be stored. The result of computation shows that this method is suitable for solving covering problems of a logic function with multiple input/output variables and other large-scale covering problems. 本文根据多输出多维体蕴涵的概念,提出构成多输出函数简化覆盖表的方法,将求解单输出函数的覆盖矩阵取补法推广到多输出函数的求解。该方法可求得函数全部有意义的无冗余覆盖(包括最小覆盖)。对于大覆盖表运用修改的锐积运算可只求得唯一一组无冗余覆盖。这个方法同时可用于消除输出冗余连接。覆盖表的产生和求解,只需存储它的某一部分或全部不存。因此本算法很简单,程序实现方便,解覆盖几乎不需要增加内存,并且计算量较小。现已编成程序,试算结果表明适于求解输入、输出变量较多的逻辑函数覆盖问题和其它规模较大的覆盖问题。 In this short paper we shall prove she following theorem.Theorem Let N be a set which contains i, a→a a map of N into itself,which satisfies recursion theorem. R. Let S be a set, (?) a map of S into itself,a an element of S. Then there exists one and only one map f from N to S such thatf(1)= a f(a+) = (?)(f(a)),a∈N.. Then we haveP I . 1≠a+. for any a∈N; PⅡ . a+ = b(?)a=b, for any a,b∈N; PⅢ. (Axiom of complete induction. ) If M is a subset of N and M satisfies the following condition:1∈M,"a∈M(?)a∈M",then... In this short paper we shall prove she following theorem.Theorem Let N be a set which contains i, a→a a map of N into itself,which satisfies recursion theorem. R. Let S be a set, (?) a map of S into itself,a an element of S. Then there exists one and only one map f from N to S such thatf(1)= a f(a+) = (?)(f(a)),a∈N.. Then we haveP I . 1≠a+. for any a∈N; PⅡ . a+ = b(?)a=b, for any a,b∈N; PⅢ. (Axiom of complete induction. ) If M is a subset of N and M satisfies the following condition:1∈M,"a∈M(?)a∈M",then M=N. 这篇短文证明了如下定理. 定理 设集N包含1,a(?)a~+是N到自身的一个映射且满足递归定理: R.对于任意的非空集S,S内任意给定的元a及S到自身的映射(?),恒唯一存在N到S的映射f满足条件 f(1)=a,f(a~+)=(?)(f(a)),a∈N.则N中必成立 PⅠ.1≠a~+,对任何a∈N. PⅡ.a~+=b~+(?)a=b,对任何a,b∈N. PⅢ.完全归纳法原理:若M是N的满足条件 1∈M,"a∈M(?)a~+∈M" 的子集,则M=N. << 更多相关文摘 相关查询

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