假如我们能够求出 \(x-y\) 在模p意义的值，那么就可以和 \(x+y\) 联立解出来了。

由于 \((x-y)^2=(x+y)^2-4xy=b^2-4c\) ，那么设 \(n=b^2-4c\) ，就是要求解出一个二次剩余。

#include
    
     using namespace std;typedef long long ll;const int p = 1e9 + 7;struct hh {    ll x, y;    hh() {};    hh(ll _x, ll _y) {        x = _x;        y = _y;    }};ll w;hh mul(hh a, hh b, ll p) {    hh ans;    ans.x = (a.x * b.x % p + a.y * b.y % p * w % p) % p;    ans.y = (a.x * b.y % p + a.y * b.x % p) % p;    return ans;}hh quick1(hh a, ll b, ll p) {    hh ans = hh(1, 0);    while(b) {        if(b & 1)            ans = mul(ans, a, p);        a = mul(a, a, p);        b >>= 1;    }    return ans;}ll quick2(ll a, ll b, ll p) {    ll ans = 1;    while(b) {        if(b & 1)            ans = (ans * a) % p;        b >>= 1;        a = (a * a) % p;    }    return ans;}ll solve(ll a, ll p) { //求解 x^2=a(mod p) 的x的值    a %= p; //注意这句话    if(a == 0)        return 0;//注意这句话    if(p == 2)        return a;    if(quick2(a, (p - 1) / 2, p) == p - 1)        return -1;    ll b, t;    while(1) {        b = rand() % p;        t = b * b - a;        w = (t % p + p) % p;        if(quick2(w, (p - 1) / 2, p) == p - 1)            break;    }    hh ans = hh(b, 1);    ans = quick1(ans, (p + 1) / 2, p);    return ans.x;}ll qpow(ll x, ll n, ll p) {    ll res = 1;    while(n) {        if(n & 1)            res = res * x % p;        x = x * x % p;        n >>= 1;    }    return res;}ll x, y;void solvexy(ll b, ll d) {    y = (b + d) % p * qpow(2, p - 2, p) % p;    x = (b - y + p) % p;    if(x > y)        swap(x, y);    return;}int main() {#ifdef Yinku    freopen("Yinku.in", "r", stdin);#endif // Yinku    int t;    scanf("%d", &t);    while (t--) {        ll b, c;        scanf("%lld%lld", &b, &c);        ll n = b * b - 4ll * c;        n = (n % p + p) % p;        ll d = solve(n, p);        if (d == -1)            printf("-1 -1\n");        else {            solvexy(b, d);            printf("%lld %lld\n", x, y);        }    }}