题意略:
题解:二次剩余板子题
//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
//#include <bits/extc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define mt make_tuple
//#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define MOD 1000000009
#define ld long double
//#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define sqr(x) ((x)*(x))
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
#define bpc __builtin_popcount
#define base 1000000000000000000ll
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
#define mr mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
//inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
//inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll mul(ll a,ll b,ll c){return (a*b-(ll)((ld)a*b/c)*c+c)%c;}
//inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=mul(ans,a,c);a=mul(a,a,c),b>>=1;}return ans;}
using namespace std;
//using namespace __gnu_pbds;
const ld pi=acos(-1);
const ull ba=233;
const db eps=1e-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=100000+10,maxn=2000000+10,inf=0x3f3f3f3f;
ll mod;
struct T
{
ll p, d;
};
ll w;
//O1乘法取模黑科技
ll mul(ll x,ll y)
{
return (x * y-(ll)(x /(long double)mod * y + 1e-3) * mod + mod) % mod;
}
//二次域乘法
T multi_er(T a, T b)
{
T ans;
ans.p = (mul(a.p, b.p) + mul(mul(a.d,b.d), w)) % mod;
ans.d = (mul(a.p, b.d) + mul(a.d, b.p)) % mod;
return ans;
}
ll quick_mod(ll a, ll b)
{
ll ans = 1;
a %= mod;
while(b)
{
if(b & 1)
{
ans = mul(ans , a);
b--;
}
b >>= 1;
a = mul(a , a);
}
return ans;
}
//二次域上快速幂
T power(T a, ll b)
{
T ans;
ans.p = 1;
ans.d = 0;
while(b)
{
if(b & 1)
{
ans = multi_er(ans, a);
b--;
}
b >>= 1;
a = multi_er(a, a);
}
return ans;
}
//求勒让德符号
ll Legendre(ll a, ll p)
{
return quick_mod(a, (p-1)>>1);
}
ll QuadraticResidue()
{
ll rem = (-3 % mod + mod) % mod;
if(rem == 0)//特判mod==3
return 0;
if(mod == 2)//特判非奇素数
return 1;
if(Legendre(rem, mod) + 1 == mod)//欧拉判别条件 非剩余
return -1;
ll b;
while(1)//找一个非剩余求二次域上的单位w=sqrt(b^2 - rem)
{
b = rand() % mod;
w = (mul(b, b) - rem + mod) % mod;
if(quick_mod(w, (mod - 1)/2) + 1 == mod)//cipolla
break;
}
T tmp;
tmp.p = b;
tmp.d = 1;
T ans = power(tmp, (mod + 1) / 2);
return ans.p;
}
ll a[N];
map<ll,int>ma;
int main()
{
int T;scanf("%d",&T);
while(T--)
{
ll cc = 0;
ma.clear();
int n;ll p;scanf("%d%lld",&n,&p);
mod = p;
ll te=QuadraticResidue();
for(int i=1;i<=n;i++) {
scanf("%lld",&a[i]);
if(a[i]) ma[a[i]]++, cc++;
}
if(p == 2) {
printf("%lld\n", cc * (cc - 1) / 2);
continue;
}
if(te==-1){puts("0");continue;}
ll ans=0;
for(int i=1;i<=n;i++)if(a[i])
{
ll x=mul(a[i],(-te-1+p),p);x=mul(x,qp(2,p-2,p),p);
ll y=mul(a[i],(te-1+p)%p,p);y=mul(y,qp(2,p-2,p),p);
if(x==y)
{
if(a[i]!=x)ans+=ma[x];
else ans+=ma[x]-1;
}
else
{
if(a[i]==x)ans+=ma[x]-1;
else ans+=ma[x];
if(a[i]==y)ans+=ma[y]-1;
else ans+=ma[y];
}
}
printf("%lld\n",ans/2);
}
return 0;
}
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