RSA算法 :首先, 找出三個(gè)數(shù), p, q, r, 其中 p, q 是兩個(gè)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(gè)數(shù)便是 person_key，接著, 找出 m, 使得 r^m == 1 mod (p-1)(q-1)..... 這個(gè) 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(lái), 計(jì)算 n = pq....... m, n 這兩個(gè)數(shù)便是 public_key ，編碼過(guò)程是, 若資料為 a, 將其看成是一個(gè)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(shí)間：
2013-12-14
上傳用戶：zhuyibin
prolog 找路例子程序：
=== === === === === ===
Part 1-Adding connections
Part 2-Simple Path
example
| ?- path1(a,b,P,T).
will produce the response:
T = 15
P = [a,b] ?
Part 3 - Non-repeating path
As an example, the query:
?- path2(a,h,P,T).
will succeed and may produce the bindings:
P = [a,depot,b,d,e,f,h]
T = 155
Part 4 - Generating a path below a cost threshold
As an example, the query:
?- path_below_cost(a,[a,b,c,d,e,f,g,h],RS,300).
returns:
RS = [a,b,depot,c,d,e,g,f,h] ?
RS = [a,c,depot,b,d,e,g,f,h] ?
no
==================================
標(biāo)簽：
Part
connections
example
prolog
上傳時(shí)間：
2015-04-24
上傳用戶：ljt101007
數(shù)字運(yùn)算，判斷一個(gè)數(shù)是否接近素?cái)?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(shí)間：
2015-05-21
上傳用戶：daguda
源代碼\用動(dòng)態(tài)規(guī)劃算法計(jì)算序列關(guān)系個(gè)數(shù)
用關(guān)系"<"和"="將3個(gè)數(shù)a，b，c依次序排列時(shí)，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(gè)數(shù)依序列，設(shè)計(jì)一個(gè)動(dòng)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(shí)O(n*n).
標(biāo)簽：
lt
源代碼
動(dòng)態(tài)規(guī)劃
序列
上傳時(shí)間：
2013-12-26
上傳用戶：siguazgb
