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ImgRotate
- Dim DP() As Byte, c As Long, d As Long Dim i As Long, j As Long Dim Color As Long, SRad As Double, CRad As Double Dim p As Double, q As Double, m As Double, n As Double Dim m1 As Double, n1 As Double, m2 As Double, n2 As Double, t As RGB, t1
momi
- 大数的模幂算法(GUI)，用密码学课本中的算法，快速、高效。计算(x的r次方) mod p 的值-Large numbers of modular exponentiation algorithm (GUI), used cryptography algorithm textbooks, fast and efficient. Calculation of (x of the r-th power) mod p value
Euler_fuction
- Euler函数: m = p1^r1 * p2^r2 * …… * pn^rn ai >= 1 , 1 <= i <= n Euler函数： 定义：phi(m) 表示小于等于m并且与m互质的正整数的个数。 phi(m) = p1^(r1-1)*(p1-1) * p2^(r2-1)*(p2-1) * …… * pn^(rn-1)*(pn-1) = m*(1 - 1/p1)*(1 - 1/p2)*……*(1 - 1/pn)
Prime_Numbers
- Threw history,prime numbers has always be a great fascination for mathematicians. Even in modern times,primes numbers continues to fascinate many people.Probably the main reason why prime numbers continues to create such great interest would be beca
lisanduishu
- 设p是素数，a是p的本原根，即a1,a2,a3,…,ap-1在mod p下产生1到p-1的所有值，所以对任何b属于{1,…,p-1},有唯一的i属于{1,…,p-1}使得ai mod p 等于p。称i为模p下以a为底b的离散对数。-P is a prime number based, a is the primitive root p, that is, a1, a2, a3, ..., ap-1 in the mod p under 1 to p-1 for all values, so fo
1-p^2-1mod-p
- 从1-p^2-1中找出与p互素的元素，计算1-p^2-1模p的值，并按模值分类。已验证p=5,7。-Find numbers 1-p^2-1 which is co-prime to p，compute values mod p, and classify them according to values mod p.
ModOfPower
- Write an efficient algorithm to calculate R=B^P mod M