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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
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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
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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
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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
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電力系統在臺穩定計算式電力系統不正常運行方式的一種計算。它的任務是已知電力系統某一正常運行狀態和受到某種擾動，計算電力系統所有發電機能否同步運行
1運行說明：
請輸入初始功率S0，形如a+bi
請輸入無限大系統母線電壓V0
請輸入系統等值電抗矩陣B
矩陣B有以下元素組成的行矩陣
1正常運行時的系統直軸等值電抗Xd
2故障運行時的系統直軸等值電抗X d
3故障切除后的系統直軸等值電抗
請輸入慣性時間常數Tj
請輸入時段數N
請輸入哪個時段發生故障Ni
請輸入每時段間隔的時間dt
標簽：
電力系統
正
計算
運行
上傳時間：
2015-06-13
上傳用戶：it男一枚
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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
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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
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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
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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
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本代碼為編碼開關代碼，編碼開關也就是數字音響中的
360度旋轉的數字音量以及顯示器上用的（單鍵飛梭開
關）等類似鼠標滾輪的手動計數輸入設備。
我使用的編碼開關為5個引腳的，其中2個引腳為按下
轉輪開關（也就相當于鼠標中鍵）。另外3個引腳用來
檢測旋轉方向以及旋轉步數的檢測端。引腳分別為a,b,c
b接地a，c分別接到P2.0和P2.1口并分別接兩個10K上拉
電阻，并且a，c需要分別對地接一個104的電容，否則
因為編碼開關的觸點抖動會引起輕微誤動作。本程序不
使用定時器，不占用中斷，不使用延時代碼，并對每個
細分步數進行判斷，避免一切誤動作，性能超級穩定。
我使用的編碼器是APLS的EC11B可以參照附件的時序圖
編碼器控制流水燈最能說明問題，下面是以一段流水
燈來演示。
標簽：
代碼
編碼開關
上傳時間：
2017-07-03
上傳用戶：gaojiao1999
