-
RSA算法 :首先, 找出三個數, p, q, r, 其中 p, q 是兩個相異的質數, r 是與 (p-1)(q-1) 互質的數...... p, q, r 這三個數便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 這個 m 一定存在, 因為 r 與 (p-1)(q-1) 互質, 用輾轉相除法就可以得到了..... 再來, 計算 n = pq....... m, n 這兩個數便是 public_key ，編碼過程是, 若資料為 a, 將其看成是一個大整數, 假設 a < n.... 如果 a >= n 的話, 就將 a 表成 s 進位 (s
標簽：
person_key
RSA
算法
上傳時間：
2013-12-14
上傳用戶：zhuyibin
-
數字運算，判斷一個數是否接近素數
A Niven number is a number such that the sum of its digits divides itself. For example, 111 is a Niven number because the sum of its digits is 3, which divides 111. We can also specify a number in another base b, and a number in base b is a Niven number if the sum of its digits divides its value.
Given b (2 <= b <= 10) and a number in base b, determine whether it is a Niven number or not.
Input
Each line of input contains the base b, followed by a string of digits representing a positive integer in that base. There are no leading zeroes. The input is terminated by a line consisting of 0 alone.
Output
For each case, print "yes" on a line if the given number is a Niven number, and "no" otherwise.
Sample Input
10 111
2 110
10 123
6 1000
8 2314
0
Sample Output
yes
yes
no
yes
no
標簽：
數字
運算
上傳時間：
2015-05-21
上傳用戶：daguda
-
源代碼\用動態規劃算法計算序列關系個數
用關系"<"和"="將3個數a，b，c依次序排列時，有13種不同的序列關系：
a=b=c,a=b<c,a<b=v,a<b<c,a<c<b
a=c<b,b<a=c,b<a<c,b<c<a,b=c<a
c<a=b,c<a<b,c<b<a
若要將n個數依序列，設計一個動態規劃算法，計算出有多少種不同的序列關系，
要求算法只占用O(n),只耗時O(n*n).
標簽：
lt
源代碼
動態規劃
序列
上傳時間：
2013-12-26
上傳用戶：siguazgb
-
The government of a small but important country has decided that the alphabet needs to be streamlined and reordered. Uppercase letters will be eliminated. They will issue a royal decree in the form of a String of B and A characters. The first character in the decree specifies whether a must come ( B )Before b in the new alphabet or ( A )After b . The second character determines the relative placement of b and c , etc. So, for example, "BAA" means that a must come Before b , b must come After c , and c must come After d .
Any letters beyond these requirements are to be excluded, so if the decree specifies k comparisons then the new alphabet will contain the first k+1 lowercase letters of the current alphabet.
Create a class Alphabet that contains the method choices that takes the decree as input and returns the number of possible new alphabets that conform to the decree. If more than 1,000,000,000 are possible, return -1.
Definition
標簽：
government
streamline
important
alphabet
上傳時間：
2015-06-09
上傳用戶：weixiao99
-
上下文無關文法（Context-Free Grammar, CFG）是一個4元組G=(V, T, S, P)，其中，V和T是不相交的有限集，S∈V，P是一組有限的產生式規則集，形如A→α，其中A∈V，且α∈(V∪T)*。V的元素稱為非終結符，T的元素稱為終結符，S是一個特殊的非終結符，稱為文法開始符。
設G=(V, T, S, P)是一個CFG，則G產生的語言是所有可由G產生的字符串組成的集合，即L(G)={x∈T* | Sx}。一個語言L是上下文無關語言（Context-Free Language, CFL），當且僅當存在一個CFG G，使得L=L(G)。 *⇒
例如，設文法G：S→AB
A→aA|a
B→bB|b
則L(G)={a^nb^m | n,m>=1}
其中非終結符都是大寫字母，開始符都是S，終結符都是小寫字母。
標簽：
Context-Free
Grammar
CFG
上傳時間：
2013-12-10
上傳用戶：gaojiao1999
-
一:需求分析
1. 問題描述
魔王總是使用自己的一種非常精練而抽象的語言講話,沒人能聽懂,但他的語言是可逐步解釋成人能聽懂的語言,因為他的語言是由以下兩種形式的規則由人的語言逐步抽象上去的:
-----------------------------------------------------------
(1) a---> (B1)(B2)....(Bm)
(2)[(op1)(p2)...(pn)]---->[o(pn)][o(p(n-1))].....[o(p1)o]
-----------------------------------------------------------
在這兩種形式中,從左到右均表示解釋.試寫一個魔王語言的解釋系統,把
他的話解釋成人能聽得懂的話.
2. 基本要求:
用下述兩條具體規則和上述規則形式(2)實現.設大寫字母表示魔王語言的詞匯 小寫字母表示人的語言的詞匯 希臘字母表示可以用大寫字母或小寫字母代換的變量.魔王語言可含人的詞匯.
(1) B --> tAdA
(2) A --> sae
3. 測試數據:
B(ehnxgz)B 解釋成 tsaedsaeezegexenehetsaedsae若將小寫字母與漢字建立下表所示的對應關系,則魔王說的話是:"天上一只鵝地上一只鵝鵝追鵝趕鵝下鵝蛋鵝恨鵝天上一只鵝地上一只鵝".
| t | d | s | a | e | z | g | x | n | h |
| 天 | 地 | 上 | 一只| 鵝 | 追 | 趕 | 下 | 蛋 | 恨 |
標簽：
語言
抽象
分
上傳時間：
2014-12-02
上傳用戶：jkhjkh1982
-
We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.
標簽：
represented
integers
group
items
上傳時間：
2016-01-17
上傳用戶：jeffery
-
The XML Toolbox converts MATLAB data types (such as double, char, struct, complex, sparse, logical) of any level of nesting to XML format and vice versa.
For example,
>> project.name = MyProject
>> project.id = 1234
>> project.param.a = 3.1415
>> project.param.b = 42
becomes with str=xml_format(project, off )
"<project>
<name>MyProject</name>
<id>1234</id>
<param>
<a>3.1415</a>
<b>42</b>
</param>
</project>"
On the other hand, if an XML string XStr is given, this can be converted easily to a MATLAB data type or structure V with the command V=xml_parse(XStr).
標簽：
converts
Toolbox
complex
logical
上傳時間：
2016-02-12
上傳用戶：a673761058
-
：介紹了一種基于數字信號處理器（DSP）的移相調頻（Phase-Shifted and Frequency-Varied，PSFV）PWM控制
逆變電源，給出了主電路拓撲結構，分析了其控制原理并設計了其控制程序流程圖。新穎的PSFV 控制能夠實現輸出
電壓90％的調整率，輸出電流波動小于單純移相調功PWM方式，并在輕載時保持連續。功率開關器件零電壓零電流
通斷（Zero-Voltage-Zero-Current Switching，ZVZCS）軟開關的實現，有利于進一步提高開關頻率和降低開關損耗。采用
高性能的專用DSP 芯片TMS320F2812 實現了系統的數字控制，滿足了系統控制的靈活性和實時性，減小了系統的
體積和生產成本。仿真分析和實驗結果證明了此控制模式的可行性與合理性。
標簽：
Frequency-Varied
Phase-Shifted
DSP
and
上傳時間：
2013-12-04
上傳用戶：kristycreasy
-
漢諾塔！！！
Simulate the movement of the Towers of Hanoi puzzle Bonus is possible for using animation
eg. if n = 2 A→B A→C B→C
if n = 3 A→C A→B C→B A→C B→A B→C A→C
標簽：
the
animation
Simulate
movement
上傳時間：
2017-02-11
上傳用戶：waizhang
