二次剩余

https://blog.csdn.net/acdreamers/article/details/10182281
http://blog.miskcoo.com/2014/08/quadratic-residue
对于奇素数$p\mathrm{m}\mathrm{o}\mathrm{d}4=3$，$\sqrt{n}\equiv n^{\frac{p+1}{4}}(\mathrm{m}\mathrm{o}\mathrm{d}p)$
对于（任意）奇素数$p$可以找一个非二次剩余（用随机方法找）$\omega \equiv a^2-n(\mathrm{m}\mathrm{o}\mathrm{d}p)$，
那么$\sqrt{n}\equiv (a+\sqrt{\omega }{)}^{\frac{p+1}{2}}(\mathrm{m}\mathrm{o}\mathrm{d}p)$

$$(a+\sqrt{\omega }{)}^{\frac{p+1}{2}}\equiv \sqrt{(a+\sqrt{\omega })(a+\sqrt{\omega }{)}^p}$$
$$(a+\sqrt{\omega }{)}^p=\sum _{i=0}^p(\genfrac{}{}{0ex}{}{p}{i})a^i(\sqrt{\omega }{)}^{p-i}\equiv \left(\genfrac{}{}{0ex}{}{p}{0}\right)(\sqrt{\omega }{)}^p+(\genfrac{}{}{0ex}{}{p}{p})a^p\equiv {\omega }^{\frac{p-1}{2}}\sqrt{\omega }+a\equiv a-\sqrt{\omega }(\mathrm{m}\mathrm{o}\mathrm{d}p)$$
$$(a+\sqrt{\omega }{)}^{\frac{p+1}{2}}\equiv \sqrt{(a+\sqrt{\omega })(a-\sqrt{\omega })}\equiv \sqrt{a^2-\omega }\equiv \sqrt{n}(\mathrm{m}\mathrm{o}\mathrm{d}p)$$

Luogu模板cipolla：
AC Code:

#include
using namespace std;

int mod,n;

int Pow(int base,int k){
	int ret = 1;
	for(;k;k>>=1,base=1ll*base*base%mod)
		if(k&1)
			ret=1ll*ret*base%mod;
	return ret;
}

int w;
struct cplx{
	int r,i;
	cplx(int r=0,int i=0):r(r),i(i){}
	cplx operator *(const cplx &B)const{ return cplx((1ll*r*B.r+1ll*i*B.i%mod*w)%mod,(1ll*i*B.r+1ll*r*B.i)%mod); } 
}A;

cplx Pow(cplx base,int k){
	cplx ret = 1;
	for(;k;k>>=1,base=base*base)
		if(k&1)
			ret=ret*base;
	return ret;
}

int main(){
	int T;
	for(scanf("%d",&T);T--;){
		scanf("%d%d",&n,&mod);
		
		if(n == 0){
			puts("0");
			continue;
		}
		
		srand(19260817);
		int a = 0;
		do{
			a = rand() % (mod-1) + 1;
			w = ((1ll * a * a - n) % mod + mod) % mod; 
		}while(Pow(w,(mod-1)/2) == 1);
		A = Pow(cplx(a , 1) , (mod+1)/2);
		
		if(A.i) puts("Hola!");
		else{
			vector<int>ans;
			ans.push_back((A.r+mod)%mod);
			ans.push_back((mod-A.r)%mod);
			sort(ans.begin(),ans.end());
			ans.resize(unique(ans.begin(),ans.end())-ans.begin());
			for(int i=0;i<ans.size();i++)
				printf("%d%c",ans[i]," \n"[i==ans.size()-1]);
		}
	}
}

对于奇素数的幂$p^q$，先求出$t\equiv \sqrt{n}(\mathrm{m}\mathrm{o}\mathrm{d}p)$
那么$t^2-n\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}p)$
$(t^2-n{)}^q\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}{p}^q)$
即$(t-\sqrt{n}{)}^q(t+\sqrt{n}{)}^q\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}{p}^q)$
可以发现若设$(t+\sqrt{n}{)}^q=a+b\sqrt{n}$
则$(t-\sqrt{n}{)}^q=a-b\sqrt{n}$
那么$(t^2-n{)}^q=a^2-b^{2}n\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}{p}^q)$
那么就可以发现$\sqrt{n}\equiv \frac{a}{b}(\mathrm{m}\mathrm{o}\mathrm{d}{p}^q)$，直接在这个域内算就行了。