题意：

传送门
已知\(0 <= x <= y < p, p = 1e9 + 7\)且有
\((x+y) = b\mod p\)
\((x\times y)=c\mod p\)
求解任意一对\(x,y\)，不存在输出\(-1\ -1\)。

思路：

由两式变化可得\((y - x)^2 = (b^2 -4c + p) \% p \mod p\)，那么可以应用二次剩余定理解得\(y - x\)的值，我们可以知道\((x+y) = b\)或者\((x+y) = b + p\)，那么直接求解即可。

代码：

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 5e4 + 5;
const int INF = 0x3f3f3f3f;
const ull seed = 131;
const ll MOD = 1e9 + 7;
using namespace std;ll ppow(ll a, ll b, ll mod){ll ret = 1;a = a % mod;while(b){if(b & 1) ret = ret * a % mod;a = a * a % mod;b >>= 1;}return ret;
}
struct TT{ll p, d;
};
ll w;
TT mul_er(TT a, TT b, ll mod){TT ans;ans.p = (a.p * b.p % mod + a.d * b.d % mod * w % mod) % mod;ans.d = (a.p * b.d % mod + a.d * b.p % mod) % mod;return ans;
}
TT power(TT a, ll b, ll mod){TT ret;ret.p = 1, ret.d = 0;while(b){if(b & 1) ret = mul_er(ret, a, mod);a = mul_er(a, a, mod);b >>= 1;}return ret;
}
ll legendre(ll a, ll p){return ppow(a, (p - 1) >> 1, p);
}
ll modulo(ll a, ll mod){a %= mod;if(a < 0) a += mod;return a;
}
ll solve(ll n, ll p){   //x^2 = n mod pif(n == 0) return 0;if(n == 1) return 1;if(p == 2) return 1;if(legendre(n, p) + 1 == p) return -1;  //无解ll a = -1, t;while(true){a = rand() % p;t = a * a - n;w = modulo(t, p);if(legendre(w, p) + 1 == p) break;}TT temp;temp.p = a;temp.d = 1;TT ans = power(temp, (p + 1) >> 1, p);return ans.p;
}
bool getans(ll sum, ll dec, ll &x, ll &y){if((sum + dec) % 2 == 0){y = (sum + dec) / 2;x = y - dec;if(x >= 0 && x + y == sum && y < MOD) return true;else return false;}else return false;
}
int main(){int T;scanf("%d", &T);while(T--){ll b, c;scanf("%lld%lld", &b, &c);ll d = solve((b * b % MOD - 4 * c % MOD + MOD) % MOD, MOD);if(d == -1){printf("-1 -1\n");continue;}ll x, y;if(getans(b, d, x, y)){printf("%lld %lld\n", x, y);}else if(getans(b + MOD, d, x, y)){printf("%lld %lld\n", x, y);}else if(getans(b, MOD - d, x, y)){printf("%lld %lld\n", x, y);}else if(getans(b + MOD, MOD - d, x, y)){printf("%lld %lld\n", x, y);}}return 0;
}