2019牛客暑期多校训练营（第九场） - B - Quadratic equation - 二次剩余

https://ac.nowcoder.com/acm/contest/889/B

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<bits/stdc++.h>
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);
        }
    }
}