exponentiation

英

美

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

• 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。
• It is shown by time complexity that the new algorithm obtains the speed improvement of encryption and decryption compared with the modular exponentiation by the repeated squaring method.
  文章给出一种新的RSA的快速算法，结合模n和底数a对指数m动态地取最优的幂后进行模幂乘运算，时间复杂性分析表明新算法可以减少加密和解密的计算量。
• On estimation of optimal window size in m_ary algorithm in modular exponentiation and point multiplication;
  定义了大窗口和小窗口，指出经典蚁群算法实质上是大窗口蚁窗算法。
• During the period of transforming, RSA should perform the modular exponentiation multiplication algorithm of large number.
  在变换过程中，RSA必需经历大数的模幂乘运算。
• First, computers have circuits for performing arithmetic operations, such as: addition, subtraction, division, multiplication and exponentiation.
  第一，计算机具有进行加、、、及取幂等各种算术运算的电路。
• The hardware architecture is made up of modular controller, modular exponentiation controller, data register, and modular multiplication operation units.
  算法的硬件结构由模乘控制器、模幂控制器、数据寄存器和模乘运算单元构成。
• 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。
• This paper introduced a method called "sliding-window" used in multiple-precision integer exponentiation arithmetic, and studied its application combined with Montgomery reduction, and how to calculate the "window size" related to the bits of the multiple-precision integer.
  介绍了多精度整数求幂运算中的“滑动窗口”算法，并结合Montgomery约简算法，对“滑动窗口”算法进行了应用研究，分析了根据多精度整数的位数来确定相应的窗口大小。
• A concrete instance is presented in which the S-Box iteration is much less time-consuming than modular exponentiation.
  给出一个实例，在该实例中S盒的迭代远远快于模指数计算。
• 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。