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…
模板题,可用于求一个数的所有原根. #include<bits/stdc++.h> using namespace std; typedef long long ll; ,inf=0x3f3f3f3f; int n,fac[N],nf; vector<int> ans; int Pow(int x,int p,int mod) { ; ,x=(ll)x*x%mod))ret=(ll)ret*x%mod; return ret; } int phi(int x) { int ret=…
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…
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…
