HDU - 6128

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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