Noip草稿紙1

擴展歐幾里得

$ax_1+by_1 = gcd(a,b)...1$

$gcd(a,b)=gcd(b,a\%b)...2$

$bx_2+(a\%b)y_2=gcd(b,a\%b)...3$

$ax_1+by_1=bx_2+(a\%b)y_2...4$

$a\%b=a-a/b*b...5$

$ax_1+by_1=bx_2+(a-a/b*b)y_2...6$

$ax_1+by_1=ay_2+b(x_2-a/b*y_2)...7$

$x_1=y_2$

$y_1=x_2-a/b*y2$

int exgcd(int a,int b,int &x,int &y) {
	if(b == 0) {
		x = 1;
		y = 0;
		return a;
	}
	int x2,y2;
	int g = exgcd(b,a%b,x2,y2);
	x = y2;
	y = x2-a/b*y2;
	return g;
}

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中國剩餘定理

$x \equiv a_i(\%m_i)...1$

$M = \prod_{i = 1}^{n}{m_i}...2$

令$\frac{M}{m_i}t_i\equiv1(\%m_i)...3$

$x = \sum_{i=1}^{n}\frac{M}{m_i}t_ia_i+k*M...4$

最小非負整數解爲$(x\%M+M)\%M$

可以用擴展歐幾里得求逆元.

int crt(int a[],int m[],int n) {
	int M = 1,ans = 0;
	for(int i = 0;i < n;++i) M *= m[i];
	for(int i = 0;i < n;++i) {
		int Mi = lcm/m[i];
		exgcd(Mi,m[i],x,y);
		ans = (ans + Mi%M*x%M*a[i]%M+M)%M;
	}
	if(ans < 0 ) ans += M;
	return ans;
}