Primitive Roots（原根个数）

最新推荐文章于 2019-06-12 13:00:07 发布

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本文介绍了一种计算模素数p下原根数量的方法，并给出了一段C++代码实现。利用欧拉函数φ(n)，可以快速计算出特定范围内所有奇素数p的原根个数。

Primitive Roots

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 3 is a primitive root modulo 7.
Write a program which given any odd prime 3 <= p < 65536 outputs the number of primitive roots modulo p.

Input
Each line of the input contains an odd prime numbers p. Input is terminated by the end-of-file seperator.
Output
For each p, print a single number that gives the number of primitive roots in a single line.
Sample Input

23
31
79

Sample Output

10
8
24

题意：

求模素数p的原根个数

分析：

结论：如果p是素数，则模p必有原根，其个数为 $\phi(p-1)$

code:

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 1e5+10;
int phi[maxn];
void Euler(){
    for(int i = 1; i < maxn; i++){
        phi[i] = i;
    }
    for(int i = 2; i < maxn; i++){
        if(phi[i] == i){
            for(int j = i; j < maxn; j += i){
                phi[j] = phi[j] / i * (i - 1);
            }
        }
    }
}
int main(){
    Euler();
    int p;
    while(~scanf("%d",&p)){
        printf("%d\n",phi[p-1]);
    }
    return 0;
}