Primitive Roots
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1441 Accepted Submission(s): 367
Problem 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, that is, the positive integers less than or equal to n that are relatively prime to n. Write a program which given any positive integer n( 2 <= n < 1000000) outputs all primitive roots of n in ascending order.
Input
Multi test cases.
Each line of the input contains a positive integer n. Input is terminated by the end-of-file seperator.
Output
For each n, outputs all primitive roots of n in ascending order in a single line, if there is no primitive root for n just print -1 in a single line.
Sample Input
4 25
Sample Output
3 2 3 8 12 13 17 22 23
Source
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题意
求出n的所有原根,若不存在原根,输出-1
分析:
原根的定义:对于n的原根x,则满足(x^y) mod n=1的最小y=phi(n),phi(n)是n的欧拉函数值(小于等于n且与n互质的数的个数,包括1)。
如果g是模m的原根,整数d>=0,则g的d次幂是模m的原根的一个充要条件是d和phi(m)互质。
如果gcd(g,m)=1,g^d=1(mod m),则d为phi(m)的一个因子。换句话说如果g是m的原根,那么对于phi(m)的所有因子d(不包含phi(m)本身),g^d=1(mod m)是不成立的。我们可以通过枚举phi(m)的质因子,以及g^phi(m)=1(mod m)是否成立来判断g是否是模m的原根。
有些不存在原根的数字用另一条性质判断:
模m有原根的充要条件是m=2,4,p^n,2×(p)^n, (p为奇质数,n为任意数)
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 1000000;
bool f[N];//素数表
void getPrime()
{
memset(f,0,sizeof f);
int m=sqrt(N+0.5);
for(int i=2;i<=m;i++)
if(!f[i])
for(int j=i*i;j<N;j+=i)
f[j]=1;
}
//求欧拉函数
int phi(int x){
if(!f[x]) return x-1;//x是素数
int ans = x;
for(int i=2; i*i<=x; i++){
if(x%i==0){
while(x%i==0) x/=i;
ans = ans - ans/i;
}
}
if(x>1) ans = ans - ans/x;
return ans;
}
//最大公约数
int gcd(int a, int b){
return b==0?a:gcd(b,a%b);
}
//快速幂
ll quick_mod(ll a, ll b, ll mod){
long long s = 1;
while(b){
if(b&1) s = (s*a)%mod;
a = a*a%mod;
b>>=1;
}
return s;
}
vector<int> V;
vector<int> G;
void cal(int x){
G.clear();
if(!f[x]) return;
else{
for(int i=2; i*i<=x; i++){
if(x%i==0){
G.push_back(i);
if(i*i!=x) G.push_back(x/i);
}
}
}
}
//判断n是否存在原根
bool exist(int n){
if(n%2==0)//偶数
n/=2;
if(!f[n]) return 1;//奇素数
for(int i=3; i*i<=n; i+=2){
if(n%i==0){
while(n%i==0) n/=i;
return n==1;
}
}
return 0;
}
void solve(int n){
if(n==2){
puts("1");
return;
}
if(n==4){
puts("3");
return;
}
if(!exist(n)){
puts("-1");
return;
}
int p = phi(n);
cal(p);
int x = -1;
for(int i=2; i<n; i++){
bool flag = 1;
if(quick_mod(i, p, n)!=1) continue;
for(int j=0; j<G.size(); j++){
if(quick_mod(i, G[j], n)==1){
flag = 0;
break;
}
}
if(flag){
V.resize(1);
V[0] = x = i;
break;
}
}
if(x==-1){
puts("-1");
return;
}
for(int i=2; i<p; i++){
if(gcd(i, p)==1)
V.push_back(quick_mod(x, i, n));
}
sort(V.begin(), V.end());
vector<int>::iterator it=unique(V.begin(), V.end());
V.erase(it, V.end());
for(int i=0; i<V.size(); i++){
if(i) putchar(' ');
printf("%d", V[i]);
}
puts("");
}
int main(){
getPrime();
int n;
while(~scanf("%d", &n)) solve(n);
return 0;
}