-
RSA算法 :首先, 找出三個數(shù), p, q, r, 其中 p, q 是兩個相異的質(zhì)數(shù), r 是與 (p-1)(q-1) 互質(zhì)的數(shù)...... p, q, r 這三個數(shù)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?yàn)?r 與 (p-1)(q-1) 互質(zhì), 用輾轉(zhuǎn)相除法就可以得到了..... 再來, 計(jì)算 n = pq....... m, n 這兩個數(shù)便是 public_key ，編碼過程是, 若資料為 a, 將其看成是一個大整數(shù), 假設(shè) a < n.... 如果 a >= n 的話, 就將 a 表成 s 進(jìn)位 (s
標(biāo)簽：
person_key
RSA
算法
上傳時間：
2013-12-14
上傳用戶：zhuyibin
-
數(shù)字運(yùn)算，判斷一個數(shù)是否接近素數(shù)
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(biāo)簽：
數(shù)字
運(yùn)算
上傳時間：
2015-05-21
上傳用戶：daguda
-
源代碼\用動態(tài)規(guī)劃算法計(jì)算序列關(guān)系個數(shù)
用關(guān)系"<"和"="將3個數(shù)a，b，c依次序排列時，有13種不同的序列關(guān)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(shù)依序列，設(shè)計(jì)一個動態(tài)規(guī)劃算法，計(jì)算出有多少種不同的序列關(guān)系，
要求算法只占用O(n),只耗時O(n*n).
標(biāo)簽：
lt
源代碼
動態(tài)規(guī)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(biāo)簽：
government
streamline
important
alphabet
上傳時間：
2015-06-09
上傳用戶：weixiao99
-
電力系統(tǒng)在臺穩(wěn)定計(jì)算式電力系統(tǒng)不正常運(yùn)行方式的一種計(jì)算。它的任務(wù)是已知電力系統(tǒng)某一正常運(yùn)行狀態(tài)和受到某種擾動，計(jì)算電力系統(tǒng)所有發(fā)電機(jī)能否同步運(yùn)行
1運(yùn)行說明：
請輸入初始功率S0，形如a+bi
請輸入無限大系統(tǒng)母線電壓V0
請輸入系統(tǒng)等值電抗矩陣B
矩陣B有以下元素組成的行矩陣
1正常運(yùn)行時的系統(tǒng)直軸等值電抗Xd
2故障運(yùn)行時的系統(tǒng)直軸等值電抗X d
3故障切除后的系統(tǒng)直軸等值電抗
請輸入慣性時間常數(shù)Tj
請輸入時段數(shù)N
請輸入哪個時段發(fā)生故障Ni
請輸入每時段間隔的時間dt
標(biāo)簽：
電力系統(tǒng)
正
計(jì)算
運(yùn)行
上傳時間：
2015-06-13
上傳用戶：it男一枚
-
上下文無關(guān)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(chǎn)生式規(guī)則集，形如A→α，其中A∈V，且α∈(V∪T)*。V的元素稱為非終結(jié)符，T的元素稱為終結(jié)符，S是一個特殊的非終結(jié)符，稱為文法開始符。
設(shè)G=(V, T, S, P)是一個CFG，則G產(chǎn)生的語言是所有可由G產(chǎn)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(guān)語言（Context-Free Language, CFL），當(dāng)且僅當(dāng)存在一個CFG G，使得L=L(G)。 *⇒
例如，設(shè)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(jié)符都是大寫字母，開始符都是S，終結(jié)符都是小寫字母。
標(biāo)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?yàn)樗恼Z言是由以下兩種形式的規(guī)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(tǒng),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(guī)則和上述規(guī)則形式(2)實(shí)現(xiàn).設(shè)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(shù)據(jù):
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(yīng)關(guān)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(biāo)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(biāo)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(biāo)簽：
converts
Toolbox
complex
logical
上傳時間：
2016-02-12
上傳用戶：a673761058
-
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(biāo)簽：
the
animation
Simulate
movement
上傳時間：
2017-02-11
上傳用戶：waizhang
