词语吧>英语词典>exponentiation翻译和用法

exponentiation

n.  取幂,求幂,乘方

计算机

英英释义

noun

  • the process of raising a quantity to some assigned power
      Synonym:involution

    双语例句

    • Implementation of SPA Resistant RSA Exponentiation Algorithm
      抵御简单功耗分析的RSA模幂算法实现
    • First countermeasures for the exponentiation computation of RSA cryptographic algorithm were summarized.
      综述了RSA密码算法中模幂运算的主要攻击方法及其防御措施。
    • The modular exponentiation part of the design is based on the M-ary arithmetic, which can save times of calculation.
      模幂部分基于M-ary算法,减少了所需模乘运算的次数。
    • Modular exponentiation of larger-number has universal application in cryptography, and it is the base operation in most public-key cryptography algorithms.
      大数模幂在密码学领域有广泛的应用,它是公钥密码的基础。
    • This paper presents a new algorithm to realize modular exponentiation multiplication by converting multiplication and modular operation into the simple shift and addition operation, thus avoiding modular operation on large number.
      提出一种宏观累加模的快速模幂乘的算法,将乘法运算和求模运算转换成简单的移位运算和加法运算,从而避免了求模运算和减少大数相乘次数。
    • We designed a highly efficient proxy signature scheme. Compared with other scheme, it needs no modular exponentiation and pairing in the signing algorithm. Thus, it is suitable for the low end devices.
      设计了一个高效的代理签名方案,和其他已提出的代理签名方案相比,它的签名算法没有计算量较重的模指数运算和配对运算,比较适合计算能力较弱的低端计算设备。
    • The paper discusses how to improve the algorithms of the exponentiation calculation and the modular calculation to increase the calculation speed of RSA.
      文章就如何改进大数乘幂算法和取余算法以提高RSA算法的运算速度进行了探讨。
    • For example, let's add a new operator to the language, the^ operator, which will perform exponentiation; in other words, 2^ 2 is2 squared or4.
      例如,我们向这种语言添加一个新的运算符,即^运算符,它将执行求幂运算;也就是说,2^2等于2的平方或4。
    • In this case, the exponentiation operator is another form of binary operator so the existing BinaryOp case class serves.
      在本例中,求幂运算符是另一种形式的二进制运算符,所以使用现有BinaryOpcase类就可以。
    • Running this code verifies that exponentiation works ( modulo the bug I mentioned earlier), so half of the battle is now complete.
      运行这段代码确保可以求幂(忽略我之前提到的bug),这样就完成了一半的工作。