查询词典 signed numbers

Memorize numbers module — In this task, subjects must recall numbers that are given at baseline, then carry out an arithmetic "distractor" task, then reproduce the numbers given at the outset."If you have 2 numbers, it's easy, but if you have 10 numbers and then a distractor task, your working memory is working very hard to get this information," Dr. Penner notes.

记数模组—在这个任务中，受试者必须记住一开始给予的数字，之后进行算数诱答任务，接著再现一开始给予的数目；Penner博士指出，如果你有两个数字，那是简单的，但是如果有10个数字，然后进行诱答任务，你的操作记忆会难以得到此资讯。

By introducing the methods for transforming exact numbers and triangular fuzzy numbers trapezoidal fuzzy numbers, the ranking method is also practicable for ranking reciprocal judgment matrix with exact numbers and triangular fuzzy numbers.

同时给出精确值、三角模糊数的互补判断矩阵转化为梯形模糊数互补判断矩阵的方法，因此本文方法同样适合于处理精确值、三角模糊数的互补判断矩阵的排序问题。

Traditionally, Forth has been implemented on two's-complement machines where there is a one-to-one mapping of signed numbers to unsigned numbers - any single cell item can be viewed either as a signed or unsigned number.

传统地， Forth 已经被实作在二补数的机器上，它有一个有号数对於无号数的一对一对应-任何的单一单元元素可以看做是有号或无号数。

An AutoNumber field can generate three kinds of numbers: sequential numbers that increment by one, random numbers, and Replication ID numbers.

"自动编号"字段可以生成三种类型的编号：每次增加 1 的顺序编号、随机编号及"同步复制 ID"。

An AutoNumber field can generate three kinds of numbers: sequential numbers that increment by one, random numbers, and Replication ID (also referred to as GUIDs— globally unique identifiers) numbers.

"自动编号"字段可以生成三种编号：每次加一的顺序编号、随机编号和同步复制 ID（也称 GUID——全局唯一标志符）编号。

This invention discloses a multi-queue sequence buffer management circuit and a method based on a pipeline applying a pipeline structure including: an arbitration circuit selecting one for process from read, write and distribution buffer requests, a buffer slot state module designing state of the slot requiring operation and queue numbers and assigning idle slots, a buffer slot filter module filtering the slot, a buffer slot filter module filtering the slot states not belonging to the current operation queues nor idle aligned in terms of the head pointer, a queue slot selection module computing continuous idle slot numbers from the slot pointed by the head pointer and refreshing the head pointer and selecting preparing slots, a queue slot prior queuing module refreshing the read pointer and result numbers of the current operation queues with the pointer of the first prepare slots and their numbers which can support multi-queue to share one buffer space, queues can access the buffer in overlap.

本发明公开一种基于流水线的多队列顺序化缓冲管理电路及方法，本发明电路采用流水线结构，包括：仲裁电路，从读、写、分配缓冲请求中选取一路进行处理；缓冲槽口状态设置模块，设置请求操作的槽口状态和队列号，分配空闲槽口；缓冲槽口滤除模块，滤除不属于当前操作队列且非空闲态的槽口状态，按头指针对齐；队列槽口选择模块，计算头指针指向槽口起的连续空闲槽口数并更新头指针，选出预备态的槽口；队列槽口优先排队模块，用第一个预备态槽口的指针和预备态槽口数分别更新当前操作队列的读指针和结果数；本发明可以支持多个队列共享一个缓冲空间，各类指令队列能对缓冲进行交叉访问，并对指令结果的写入读出进行顺序化管理。

Well, it could be that you have a huge collection of ISBN numbers, and that there are so many of them that for weird reasons some have the wrong check digit perhaps the numbers were scanned from a printed list and the OCR got the numbers wrong, perhaps the numbers were manually captured...

因为你可能有数量巨大的 ISBN 的编码，而其中又有大量的编码不知道什么原因，校验码是错的（可能是编码是从一个打印出来的列表里头扫描并 OCR 出来的，编码就错了，也可能是编码被手工捕获的。。。

Superreal and surreal numbers extend the real numbers by adding infinitesimally small numbers and infinitely large numbers, but still form fields .

这一原则允许真正的第一道命令声明R以将重新成为真正的第一道命令声明*河

Op1 + op2 The addition operator will add two numbers. op1 - op2 The subtraction operator will subtract two numbers. op1 * op2 The multiplication operator will multiply two numbers. op1 / op2 The division operator will divide two numbers. op1 % op2 The modulus operator will return the remainder of the division of two integer operands. op1 xx op2 The exponentiation operator will raise op1 to the power of op2.++op1 The pre-increpment operator will increase the value of op1 first, then assign it. op1++ The post-increment operator will increase the value of op1 after it is assigned.--op1 The pre-decrement operator will decrease the value of op1 before it is assigned. op1-- The post-decrement operator will decrease the value of op1 after it is assigned.

op1 + op2 对两个数值做加法操作 op1 - op2 对两个数值做减法操作 op1 * op2 对两个数值做乘法操作 op1 / op2 对两个数值做除法操作 op1 % op2 求两个整型数值的余数 op1 xx op2 求幂操作：求 op1 的 op2 次幂++op1 前加操作： op1 的值先增加，然后将值赋给自身 op1++后加操作： op1 的值先赋给自身，再增加值--op1 前减操作： op1 的值先减少，然后将值赋给自身 op1—后减操作： op1 的值先赋给自身，再减少值

If we rechange these two numbers:one other to four times as many as its original value,the average of the five numbers will become 13.Which two numbers have been changed\?I only know that the total sum of the original numbers is 45(5×9),thd total sum of the 5numbers after the first change is 40(5×8),and the total sum of the 5numbers after the second change is 65(13×5).What can Ido next\?Please help me with the problem.Thank you!Mathematics Maniac,Zhongwanting

这就要求琬婷一方面要用王牌外语准确地把她想要问的问题清晰地表达出来，只有这样网上"数字博士"才能明白你要问的问题；另一方面琬婷还必须能够正确理解网上"数学博士"对她所提问题的英语答复，这两者是相互依赖的，否则就不能正确地帮助你理解和解决问题，这就要求琬婷必须用王牌外语进行正确地表达和理解。

The dissecting of samples in group2 were difficult. The root of pulmonary artery and ascending aorta failed to be unfolded because fibrous tissue was tough, right and left fibrous trigone were too firm to be solved by hand. Cardiac muscle fibers couldn't be stripped along myofibrillar trajectory since they were prone to break because of their friability.

组2的心脏解剖困难，表现为纤维组织坚韧，游离肺动脉非常困难；徒手无法松解左、右纤维三角，肺动脉和主动脉根部的游离非常困难；心肌纤维坚硬、质脆，解剖时容易断离成碎块，无法沿纤维走行方向剥离。

We have battled against the odds in a province that has become increasingly violent.

我们对在一个争夺日益激烈省的可能性。