数论理论基础学习

唯一分解定理

  任何大于1的自然数,都可以唯一分解成有限个质数的乘积.

$n=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}...p_k^{{\alpha }_k}(p_i均为质数,指数{\alpha }_k为正整数)$

欧拉函数

$表示小于等于n且和n互质的数的个数$
$\varphi (1)=1$
$1.当n是质数时,\varphi (n)=n-1$
$2.gcd(a,b)=1,\varphi (a\times b)=\varphi (a)\times \varphi (b)$
$3.若n=p^k,其中p是质数,那么\varphi (n)=p^k-p^{k-1}$
$4.由唯一分解定理,设n={\prod }_{i=1}^s{p}_i^{k_i},其中p_i是质数,有\varphi (n)=n\times {\prod }_{i=1}^s\frac{p_i-1}{p_i}$

裴蜀定理

$设a,b是不全为零的整数,则存在整数x,y,使得ax+by=gcd(a,b)$
$如果a,b互质(即gcd(a,b)=1),那么一定存在两个整数x与y,使得ax+by=1$

费马小定理

$若p为素数,gcd(a,p)=1,则a^{p-1}\equiv 1(\mathrm{mod}p)$
$对于任意整数a,有a^p\equiv a(\mathrm{mod}p)$

欧拉定理

$若gcd(a,m)=1,则a^{\varphi (m)}\equiv 1(\mathrm{mod}m)$

乘法逆元

typedef long long int ll;
const int maxn = 3e6 + 7;
ll n, p, inv[maxn];

ll qpow(ll a,ll b,ll m){
    
    
	b %= m;
	ll ret = 1,x = a;
	while (b){
    
    
		if (b & 1)ret = (ret * x) % m;
		x = (x * x) % m;
		b >>= 1;
	}
	return ret;
}

//(a,a mod b) ==> (b,a - kb) ==> (b,a - b a/b) 
//ax + by = bx' + (a - kb) * y'
//ax + by = bx' + ay' - a/b b y'
//a(x - y') + b(y - (x - a/b * y')) = 0
ll exgcd(ll a, ll b, ll &x, ll &y){
    
    
	if (a < b) return exgcd(b, a, y, x);
	if (b == 0){
    
    
		x = 1;
		y = 0;
		return a;
	}
	else{
    
    
		ll x1;
		ll d = exgcd(b, a % b, x1, x);
		y = x1 - a / b * x;
		return d;
	}
}

int main()
{
    
    
	cin >> n >> p;
	inv[1] = 1;

	ll x, y;
	for (int i = 2; i <= n; i++){
    
    
		//p = k * i + r;
		//k * i + r =- 0 mod p
		//kr_1 + r_1=- 0 mod p
		//i_1 = -p/i * (p mod i)_1 mod p
		inv[i] = (p - p / i)* inv[p % i] % p;		//递推
		//inv[i] = qpow(i, p - 2, p);				//快速幂	
		//exgcd(i, p, x, y);						//扩展欧几里得
		//inv[i] = (x % p + p) % p;
	}
	for (int i = 1; i <= n; i++)printf("%d\n", inv[i]);
	return 0;
}

快速GCD

int binary_zc(int a){
    
    
	int c = a & -a,ans = 0;
	while (c >>= 1)ans++;
	return ans;
}
int gcd(int a, int b)
{
    
    
	int az = __builtin_ctz(a), bz = __builtin_ctz(b);
	int z = min(az, bz);
	int dif;
	b >>= bz;
	while (a)
	{
    
    
		a >>= az;
		dif = b - a;
		az = __builtin_ctz(dif);
		if (a<b) b = a;
		if (dif<0) a = -dif;
		else a = dif;
	}
	return b << z;
}