二次剩余算法

解决的问题:

形如：求是否存在x满足 ： x^2 = n (mod p)
https://www.luogu.com.cn/problem/P5491

模板：

ll n,p;
ll w;
struct num{  //负数 
	ll x,y;
};

num mul(num a,num b,ll p){
	num ans = {0,0};
	ans.x = ((a.x*b.x%p+a.y*b.y%p*w%p)%p+p)%p;
	ans.y = ((a.x*b.y%p+a.y*b.x%p)%p+p)%p;
	return ans;
}

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

ll qpwii(num a,ll b,ll p){  //复数的快速幂 
	num ans={1,0};
	while(b){
		if(b&1)ans=mul(ans,a,p);
		a=mul(a,a,p);
		b>>=1;
	}
	return ans.x%p;
}

ll solve(ll n,ll p){  //x^2 = n (mod p)
	n %= p;
	if(p == 2)return n;
	if(qpw(n,(p-1)/2,p) == p-1)return -1;
	ll a;
	while(1){
		a = rand()%p;
		w=((a*a%p-n)%p+p)%p;
		if(qpw(w,(p-1)/2,p) == p-1)break;
	}
	num x = {a,1};
	return qpwii(x,(p+1)/2,p);
}

void work() {
	scanf("%lld%lld",&n,&p);
	if(!n) {
		printf("0\n");
		return ;
	}
	ll ans1 = solve(n,p),ans2;
	if(ans1 == -1)printf("Hola!\n");
	else{
		ans2 = p - ans1;
		if(ans1 > ans2)swap(ans1,ans2);
		if(ans1 == ans2)printf("%lld\n",ans1);
		else printf("%lld %lld\n",ans1,ans2);
	} 
}