阶

阶

定义：满足 $a^x \equiv1\;(mod\;m)$ 最小的 $x$ 称为 $a$ 模 $m$ 的阶

性质：$x\mid \phi(m)$

求阶：

​ 先将 $\phi( m)$ 标准分解，$phi(m)=p_1^{e_1}*p_2^{e_2}*...p_k^{e_k}$

​ 先设阶为 $d= \phi(m)$, 每次除以一个素因子 $p_i$

​ 当且仅当 $p_i\mid d$ 且 $a^{\frac d{p_i}} \equiv1\;(mod\;m)$ 时 $d$ 可再除以一个 $p_i$

int solve(int a)
{
	int d = p - 1;
	for (int i = 1; i <= t; i++)
		while(d % divisor[i] == 0 && qmi(a, d / divisor[i], p) == 1) d /= divisor[i];
	return d;
}

阶 - 题目 - Daimayuan Online Judge

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;
typedef long long ll;
const int N = 100;
int p, t;
int divisor[N];

ll qmi(ll a, ll b, ll p)
{
	ll ans = 1;
	while(b)
	{
		if (b & 1) ans = ans * a % p;
		b >>= 1;
		a = a * a % p;
	}
	return ans % p;
}

int solve(int a)
{
	int d = p - 1;
	for (int i = 1; i <= t; i++)
		while(d % divisor[i] == 0 && qmi(a, d / divisor[i], p) == 1) d /= divisor[i];
	return d;
}
int main()
{
	int T;
	scanf("%d%d", &p, &T);
	int m = p - 1;
	for (int i = 2; i <= m / i; i++)
	{
		if (m % i) continue;
		divisor[++t] = i;
		while(m % i == 0) m /= i;
	}
	if (m > 1) divisor[++t] = m;
	
	while(T--)
	{
		int a;
		scanf("%d", &a);
		printf("%d\n", solve(a));
	}
	return 0;
}