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, if integer B is satisfied with AB mod p = 1, B is the model p multiplication inverse element of. Theorem: the necessary and sufficient condition for a multiplication inverse element with mod P is gcd (A, P) = 1. Proof:First prove adequacyIf gcd (A, P) = 1, according to Euler's theoremObviously, a Phi (P)-1 mod p is the inverse element of the p multiplication mode of.Prove the necessityAssume that the multiplication inverse element of model a p is B.AB 1271 mod pThen AB = KP + 1, so 1 = AB-kPBe

Refer to the following Alpha router configuration method to access and set up a Broadband Router through a browser. After these three steps, you can basically get familiar with this quick installation method. To use this installation method, you must have installed a browser on your workstation and have an ISP account or a fixed valid IP address. If you do not have any of them, you cannot use this installation method. The following describes how to ac

Management Protocol) This TCP/IP protocol allows Internet hosts to participate in multi-point broadcast (multicasting)-a type of broadcast to a computer group Effective means of information IIS Information Server IP (Internet Protocol) Internet Protocol IRC online chat ISP network service provider IPX Internet connection group protocol IPC inter-process communication IRQ interrupt request IP address An IP address is a 32-bit IP address assigned to th

The connect by level syntax is particularly useful when constructing a large amount of data. However, if you do not really understand the meaning of the syntax, misuse will lead to the production of a large amount of data, generally, the incorrect understanding is the Cartesian product of the Base record and the base record. Cartesian product is correct, but not both of them are the base record. One of them is the record of the last level, it is important to know this, otherwise it will produce

521DThere are three strategies for upgrading skills: 1, =; 2, +=;3, *=. The optimal order of upgrades is obviously first using strategy 1, then strategy 2, and then strategy 3.Assignment increment (for each skill only considers the maximum gain assignment operation), plus the multiply (for each skill to give priority to the maximum gain of the addition), in the premise of not exceeding the overall upgrade scheme, the selection of M-type of the largest multiplier of the upgrade scheme. Since the

can be solved at the application layer. Ii. explain the output + ---- + ------------- + ------- + ------------------- + --------- + ------- + ------ + ------- + | Id | select_type | table | type | possible_keys | key | key_len | Ref | rows | Extra | + ---- + ------------- + ------- + ------------------- + --------- + ------- + ------ + ------- + 1. id My understanding is the mark of the smooth execution of SQL statements, from large to small. For example: Mysql> explain select * from (select *

, the basic as much as possible to reduce the join, subqueries and so on. MySQL uses the simplest query , which is the most efficient. As for join, you can put it in the application layer to solve Two. Explain output interpretation +----+-------------+-------+-------+-------------------+---------+--------- +-------+------+-------+ | ID | Select_type | Table | Type | Possible_keys | Key | Key_len | Ref | Rows | Extra | +----+-------------+-------+-------+-------------------+---------+-----

General Idea:MySQL does not have a convenient statistical function like Oracle, can only rely on their own hard computing: through the Time field directly increase the year, the month, and then through the left join associated Time field to calculate the chain, year-over formula can beOriginal table structure:Find sql:[SQL]View PlainCopyPrint? --by year SELECT t5.*, Case when t5.last_energy_year is NULL OR t5.last_energy_year =0 Then 0.00 ELSE FORMAT (((t5.energy_year-t5.last

This article will discuss the similarities and differences between simple nested queries and non-nested queries in SQL, through which you can better understand SQL statements. One day's work was to fix the bug of a project, and then we found that its SQL statements were extremely messy, there were a lot of left join and in operations, and nested queries had only one table nested query ). I don't know where I have seen it. The nested query is slow, so I changed all the nested queries to join. I w

Round 1day1:Today, the examination room played a big mistake, a little summing up a bit.Get the title, a long time did not do 4 questions, the first idea is to do it quickly, rhythm to a little faster.A glance at the T2, ah relatively simple digital DP, a little push will be able to launch it.T3t4 seems to be a divine problem, first put, and then take part of the points.Then go to the battle T1, think of a very naked and violent n^1.5 algorithm, a bit silly dozen a monotonous queue, there are a