/* */) * @Link : https://github.com/ * @Version : $Id$ */ #include <bits/stdc++.h> #define LL long long #define PII pair<int ,int> #define MP(x, y) make_pair((x),(y)) #define fi first #define se second #define PB(x) push_back((x)) #define MMG(x) memset((x), -1,sizeof(x)) #define MMF(x) memset((x),0,sizeof(x)) #define MMI(x) memset((x), INF, sizeof(x)) using namespace std; const int INF = 0x3f3f3f3f; const int N = 1e5+20; const double eps = 1e-8; 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; } //二次域上快速冪 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; } //求勒讓德符號 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)//特判非奇素數 return 1; if(Legendre(rem, mod) + 1 == mod)//歐拉判別條件 非剩余 return -1; LL b; while(1)//找一個非剩余求二次域上的單位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; } vector<LL>a; int main() { int T; cin >> T; while(T--) { a.clear(); LL n, p; scanf("%lld%lld", &n, &mod); for(int i = 0; i < n; i++) { LL t; scanf("%lld", &t); if(t > 0)//註意有0的... a.PB(t); } LL cnt = a.size(); sort(a.begin(), a.end()); /////////// LL ans = 0; if(mod == 2)//特殊情況無解 ans = cnt * (cnt - 1) / 2LL; else { LL t = QuadraticResidue(); if(t == -1) { printf("0\n"); continue; } LL inv = (mod + 1) >> 1; LL x = mul((mod + t - 1LL)%mod, inv); LL y = mul((mod - t - 1LL)%mod, inv); for(int i = 0; i < cnt; i++) { LL tmp = mul(x , a[i]); ans += upper_bound(a.begin(), a.begin() + i, tmp) - lower_bound(a.begin(), a.begin() + i, tmp); } if(x != y)//兩個解 { for(int i = 0; i < cnt; i++) { LL tmp = mul(y, a[i]); ans += upper_bound(a.begin(), a.begin() + i, tmp) - lower_bound(a.begin(), a.begin() + i, tmp); } } } printf("%lld\n", ans); } return 0; } /* 99 5 7 1 2 3 4 5 6 7 1 2 3 4 5 6 */

HDU6128 二次剩余/二次域求二次剩余解/LL快速乘法取模