題目大意：

給出一個$m$，求出它的所有原根，不存在輸出$-1$。 $m ≤ 10000$

分析：

設$n={p_1}^{c1}{p2}^{c2}..{p_x}^{c_x}$ 則有$φ(n)=n(1-\frac{1}{p1})(1-\frac{1}{p2})..(1-\frac{1}{p_x})$ 然後直接求出來$φ(m)$ 然後列舉$[1,m]$暴力判斷是否是原根即可 時間複雜度：$O$

程式碼：

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#define N 10005

using namespace std;

int prime[N], a[N], n, phi, cmt;

int Gcd(int x, int y)
{
	if (x < y) swap(x, y);
	while (1)
	{
	   int r = x % y;
	   x = y, y = r;	
       if (r == 0) break;
	}
	return x;
}

int main()
{
	freopen("math.in", "r", stdin);
	freopen("math.out", "w", stdout);
	scanf("%d", &n);
	phi = n, cmt = n;
	for (int i = 2; i * i <= n; i++)
		if (cmt % i == 0) 
		{
		    phi = phi - phi / i;
		    while (cmt % i == 0) cmt /= i;
        }
    if (cmt > 1) phi = phi - phi / cmt;
	bool dop = 1; 
    for (int i = 1; i <= n; i++)
	{
        if (Gcd(i, n) != 1) continue;     
        bool flag = 0; int gg = 1;
		for (int j = 1; j < phi; j++)
		{
			gg = gg * i % n;
			if (gg == 1) { flag = 1; break; }
		}
		if (!flag) { dop = 0; printf("%d\n", i); }
    }
    if (dop) printf("-1\n");
	return 0;
}