数论模板

判断质数

bool isprime (int n){
	for (int i = 2;i * i <= n;i ++)
		if (n % i == 0)
			return 0;
	return 1;
}

整数的唯一分解

void factorize(int n){
	for (int i = 2;i * i <= n;i ++)
		if (n % i == 0){
			p[++ k] = i;
			while (n % i == 0)
				n /= i,w[k] ++;
		}
	if (n != 1)
		p[++ k] = i,w[k] = 1;
}

埃拉托斯特尼筛法

void sieve(int n){
	for (int i = 2;i <= n;i ++){
		if (!vis[i]){
			prime[++ pn] = i;
			for (int j = i * i;j <= n;j += i)
				vis[j] = 1;
		}
	}
}

筛法求约数个数，约数和

for (int i = 1;i <= n;++ i)
	for (int j = i;j <= n;j += i)
		d[]j ++,s[j] += i;

欧拉筛法

void sieve(int n){
	for (int i = 2;i <= n;i ++){
		if (!vis[i])
			prime[++ pn] = i;
		for (int j = 1;j <= pn&& i * prime[j] <= n;j ++){
			vis[i * prime[j]] = 1;
			if (i % prime[j] == 0)
				break;
		}
	}
}

欧拉筛法求约数及其和

不经常用，略...

update7/26:考试居然考了（虽然数据水，没用过了），我很震惊....所以我还是写吧。

/*约数个数*/
for (int i = 2;i <= b;i ++){
	if (!vis[i]){
		pri[++ cnt] = i;
		num[i] = 1;
		d[i] = 2;
	}
	for (int j = 1;j <= cnt && 1ll  * i * pri[j] <= b;j ++){
		vis[pri[j] * i] = 1;
		if (i % pri[j] == 0){
			num[pri[j] * i] = num[i] + 1;
			d[pri[j] * i] = d[i] / (num[i] + 1) * (num[i] + 2);
			break;
		}
		d[pri[j] * i] = d[i] * 2;
		num[i] = 1;
	}
}

/*约数和*/
s[1] = 1;
for(int i = 2;i <= n;i ++){
	if (!vis[i]){
		pri[++ cnt] = i;
		s[i] = i + 1;
		psum[i] = i + 1;
	}
	for (int j = 1;j <= cnt && 1ll * i * pri[j] <= n;j ++){
		vis[pri[j] * i] = 1;
		if (i % pri[j] == 0){
			psum[pri[j] * i] = psum[i] * pri[j] + 1;
			s[pri[j] * i] = s[i] / psum[i] * psum[pri[j] * i];
			break;
		}
		s[pri[j] * i] = s[i] * (pri[j] + 1);
		psum[pri[j] * i] = pri[j] + 1;
	}
}

辗转相除法

int gcd(int a,int b){
	if (b == 0)
		return a;
	return gcd(b,a % b);
}
int lcm(int a,int b){
	return a / gcd(a,b) * b;
}

欧拉函数

求1到n中与n互质的数

int phi(int n){
	int res = n;
	for (int i = 2;i * i <= n;i ++){
		if (n % i == 0){
			res = res / i * (i - 1);
			while (n % i == 0) n /= i;
		}
	}
	if (n > 1)
		res = res / n * (n - 1);
	return res;
}

void sieve(int n){
	phi[1] = 1;
	for (int i = 2;i <= n;i ++){
		if (!vis[i]){
			pri[++ pn] = i;
			phi[i] = i - 1;
		}
		for (int j = 1;j <= pn&& i * pri[j] <= n;j ++){
			vis[i * pri[j]] = 1;
			if (i % pri[j] == 0){
				phi[i * pri[j]] = phi[i] * pri[j];
				break;
			}
			else 
				phi[i * pri[j]] = phi[i] * (pri[j] - 1);
		}
	}
}

Miller_Rabin

判断素数

long long qkpow(long long x,long long y,long long mod){
    long long ans = 1;
    while (y){
        if (y % 2 == 1)
            ans = ans * x % mod;
        x = x * x % mod;
        y /=2;
    }
    return ans;
}
bool Miller_Rabin(int x){
    if (x == 2 || x == 3 || x == 5 || x == 7)
        return 1;
    if (x % 2 == 0 || x % 3 == 0 || x % 5 == 0 || x % 7 == 0)
        return 0;
    long long  m = x - 1,k = 0;
    while (m % 2 == 0){
        k ++;
        m /= 2;
    }
    for (int i = 1;i <= 10;i ++){
        long long a = rand() % (n - 1) + 1;
        long long xx = qkpow(a,m,x);
        long long y;
        for (int j = 1;j <= k ;j ++){
            y = xx * xx % x;
            if (y == 1 && xx != 1 && xx != x - 1)
                return 0;
            xx = y;
        }
        if (y != 1)
            return 0;
    }
    return 1;
}

拓展欧几里得

求解二元一次不定方程一组特解

int exgcd(int a,int b,int &x,int &y){
	if (!b){
		x = 1,y = 0;
		return a;
	}
	else {
		r = exgcd(b,a % b,y,x);
		y -= x * (a - b);
		return r;
	}
}

逆元

ab %p == 1，称a,b在模p意义下互为逆元。

高斯消元

解方程模板

异或线性基

不会...没打过...略...

杨辉三角

组合数，计算系数

void initial (){
	int i,j;
	fr (int i = 0;i <= maxn;++ i)
		c[i][0] = 1;
	for (int i = 1;i <= maxn;++ i)
		for (int j = 1;j <= maxn;++ j)
			c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
}

预处理阶乘

阶乘及其逆元。模P意义下

void prepare(){
	fac[0] = rfac[0] = fac[1] = rfac[1] = 1;
	for (int i = 2;i <= n;i ++){
		fac[i] = 1ll * i * fac[i - 1] % p;
		rfac[i] = 1ll * rfac[p % i] * (p - p / i) % p;
	}
	for (int i = 2;i <= n;i ++)
		rfac[i] = 1ll * rfac[i] * rfac[i - 1] % p;
}
int c(int n,int m){
	return 1ll * fac[n] * rfac[n - m] % p * rfac[m] % p;
}

Lucas

组合数C（n,m）%p，适用于n,m很大，p在10w左右

int Lucas(int n,int m){
	if (m == 0 || n == m)
		return 1;
	else 
		return 1ll * C(n % p,m % p) * Lucas(n / p,m / p) % p;
}

Legendre

这不是超神（legendary）定理

求阶乘的质因数分解

for (int i = 1;i <= cnt;i ++)
		for (long long j = pri[i];j <= n;j *= pri[i])
			s[i] += n / j;