ZR#1004

ZR#1004

解法：

对于 $ (x^2 + y)^2 \equiv (x^2 - y)^2 + 1 \pmod p $

化简并整理得 $ 4x^2y \equiv 1 \pmod p $

即 $ 4x^2 $ 和 $ y $ 互为逆元时统计答案即可

CODE:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>

using namespace std;

#define LL long long
const int N = 1e5 + 100;

inline LL fast_pow(LL a,LL b,LL p) {
	LL ans = 1;
	while(b) {
		if(b & 1) ans = ans * a % p;
		a = a * a % p;
		b >>= 1;
	}
	return ans % p;
}
map<LL,LL> vis;
LL n,p,ans,a[N];

int main() {
	scanf("%lld%lld",&n,&p);
	for(int i = 1 ; i <= n ; i++) {
		scanf("%lld",&a[i]);
		if(4 * a[i] % p * a[i] % p == 0) continue;
		else vis[4 * a[i] % p * a[i] % p]++;
	}
	for(int i = 1 ; i <= n ; i++) {
		LL inv = fast_pow(a[i],p - 2,p);
		if(inv == 0) continue;
		ans += vis[inv];
		if(4 * a[i] % p * a[i] % p == inv) ans--;
	}
	printf("%lld\n",ans);
	//system("pause");
	return 0;
}