我在RSA学习总结的第三部分关于Mille-Rabin素数测试的正确性证明里需要用到此定理,由于证明太长,故另开一章于此.(为啥我说话突然文绉绉了Orz,可能是这周辩论打多了) 结论是对素数p,modulo p的原根存在,个数为与ø(p-1),modulo p2的原根个数为(p-1)ø(p-1)个 对奇素数p,modulo p^n的原根存在,个数为pn-2(p-1)ø(p-1) (n>=3) 首先证明对任意素数p,modulo p的原根存在 以下是证明思路(符号的意思在第二张图,完整证明里有)…
我在RSA学习总结的第三部分关于Mille-Rabin素数测试的正确性证明里需要用到此定理,由于证明太长,故另开一章于此.(为啥我说话突然文绉绉了Orz,可能是这周辩论打多了) 结论是对素数p,modulo p的原根存在,个数为与ø(p-1),modulo p2的原根个数为(p-1)ø(p-1)个 对奇素数p,modulo p^n的原根存在,个数为pn-2(p-1)ø(p-1) (n>=3) 首先证明对任意素数p,modulo p的原根存在 以下是证明思路(符号的意思在第二张图,完整证明里有)…
终于有一个SPOJ题目是我自己独立做出来的,ORZ,太感动了. 题目意思是给你一个素数,问你一个数r是否满足,r,r^2,r^3,……,r^p-1,全不相同. 以前做过这种类型的题目额.是这样的. 根据欧拉定理我们知道,对于任意一个x<p,我们都有 x^(p-1)==1(mod p),这样我们只要判断x是否对于p-1的所有因数y是否都不满足 x^y!=1,如果存在等于1的情况,那说明就是NO咯. 上代码: #include <cstdio> #define ll long long us…
Primitive Roots Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 5709 Accepted: 3261 Description We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set (ximodp)∣1≤i≤p−1{ (x_i mod p) | 1 \leq i \leq…
Primitive Roots   Description We say that integer x, 0 < x < n, is a primitive root modulo n if and only if the minimum positive integer y which makes x y = 1 (mod n) true is φ(n) .Here φ(n) is an arithmetic function that counts the totatives of n,…
Primitive Roots http://poj.org/problem?id=1284 Time Limit: 1000MS   Memory Limit: 10000K       Description We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (xi mod p) | 1 <= i <= p-1 } is equal…
题目传送门 Primitive Roots Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5434   Accepted: 3072 Description We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (xi mod p) | 1 <= i <= p-1 }…
Primitive Roots Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 3381   Accepted: 1980 Description We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (xi mod p) | 1 <= i <= p-1 } is eq…
Primitive Roots Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2479   Accepted: 1385 Description We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (xi mod p) | 1 <= i <= p-1 } is eq…
一.题目 We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (x i mod p) | 1 <= i <= p-1 } is equal to { 1, ..., p-1 }. For example, the consecutive powers of 3 modulo 7 are 3, 2, 6, 4, 5, 1, and thus…