文件名称:Primitive-root
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给定大素数P,如果求其原根?给出算法描述。并编程实现具体功能。
利用随机数生成大素数q,P是形如2*q+1的大素数,P也为素数,若P不为素数,则重新生成大素数q,重复上述步骤。生成随机数g,1<g<p-1,要满足:g^2 mod p!=1, g^q mod p!=1, 则g为P的本原根。-Given a large prime number P, if you find its original roots? Algorithm descr iption is given. And specific programming functions.
Use of random number generation of large prime numbers q, P is of the form 2* q 1 large prime number, P is a prime number, if P is not prime, then regenerate the
NCKU prime q, repeat the above steps. Generate random numbers g, 1 <g <p-1, to meet: g ^ 2 mod p! = 1, g ^ q mod p! = 1, then
P g is the primitive root.
利用随机数生成大素数q,P是形如2*q+1的大素数,P也为素数,若P不为素数,则重新生成大素数q,重复上述步骤。生成随机数g,1<g<p-1,要满足:g^2 mod p!=1, g^q mod p!=1, 则g为P的本原根。-Given a large prime number P, if you find its original roots? Algorithm descr iption is given. And specific programming functions.
Use of random number generation of large prime numbers q, P is of the form 2* q 1 large prime number, P is a prime number, if P is not prime, then regenerate the
NCKU prime q, repeat the above steps. Generate random numbers g, 1 <g <p-1, to meet: g ^ 2 mod p! = 1, g ^ q mod p! = 1, then
P g is the primitive root.
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下载文件列表
求原根/readme.txt
求原根/root.py
求原根
求原根/root.py
求原根
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