正题

题目链接:https://www.luogu.com.cn/problem/P5491

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题目大意

求解$$x^2=N(modP)$$

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解题思路

若$a$在模$p$意义下可以开根那么$a$就是$p$的二次剩余，定义
$$\left(\genfrac{}{}{0ex}{}{a}{p}\right)=\{\begin{array}{c}1(a是p的二次剩余)\\ -1(a是p的二次非剩余)\\ 0(p|a)\end{array}$$

那么如果$gcd(a,p)=1$就有$\left(\genfrac{}{}{0ex}{}{a}{p}\right)=a^{\frac{p-1}{2}}$。

如果我们求$x^2=N(\mathrm{m}\mathrm{o}\mathrm{d}p)$那么我们考虑先求一个$(a^2-N{)}^{\frac{p-1}{2}}=-1$，然后定义$\omega =\sqrt{a^2-N}$，那么就有定理$x=(a+\omega {)}^{\frac{p+1}{2}}$，这里我们找$a$的时候只要随机就好了，因为$p$的二次有$\frac{p+1}{2}$个，期望很快就能找到一个合法的$a$。

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$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
ll T,n,p,k,w;
struct complex{complex(ll xx=0,ll yy=0){x=xx;y=yy;}ll x,y;
};
complex operator+(complex &a,complex &b)
{return complex((a.x+b.x)%p,(a.y+b.y)%p);}
complex operator-(complex &a,complex &b)
{return complex((a.x-b.x+p)%p,(a.y-b.y+p)%p);}
complex operator*(complex &a,complex &b)
{return complex((a.x*b.x%p+a.y*b.y%p*w%p+p)%p,(a.x*b.y+a.y*b.x)%p);}
complex poweri(complex x,ll b){complex ans=complex(1,0);while(b){if(b&1)ans=ans*x;x=x*x;b>>=1; }return ans;
}
ll power(ll x,ll b){ll ans=1;while(b){if(b&1)ans=ans*x%p;x=x*x%p;b>>=1;}return ans;
}
bool check(ll a)
{return power(((a*a%p-n)%p+p)%p,k)==(p-1);}
int main()
{srand(31958);scanf("%lld",&T);while(T--){scanf("%lld%lld",&n,&p);if(!n){printf("0\n");continue;}if(p==2){printf("%lld\n",n);continue;}k=(p-1)>>1;if(power(n,k)==p-1){printf("Hola!\n");continue;}ll a=rand()%p;while(!check(a))a=rand()%p;w=((a*a%p-n)%p+p)%p;ll x1=poweri(complex(a,1),(p+1)/2).x;ll x2=p-x1;if(x1>x2)swap(x1,x2);if(x1!=x2)printf("%lld %lld\n",x1,x2);else printf("%lld\n",x1);}
}