暴力法

暴力模拟枚举即可：

#include <iostream>
#include <cmath>
using namespace std;
int gcd(int a,int b){
    if (b == 0){
        return a;
    }return gcd(b,a%b);
}void to_simp(int &p,int &q){
    if (p == 0){
        q = 1;
        return;
    }int g = gcd(abs(p),abs(q));
    p /= g,q /= g;
    if (q < 0){
        p = -p,q = -q;
    }
}void solve(){
    int a,b,c,d,k = 1,t,r;
    cin >> a >> b >> c;
    d = b*b-4*a*c;
    if (d < 0){//无解
        cout << "NO\n";
        return;
    }if (d == 0){
        int p = -b,q = 2*a;
        to_simp(p,q);
        if(q == 1){
            cout << p << endl;
        }else{
            cout << p << "/" << q << endl;
        }return;
    }t = d;// d = k*k+r
    for (int i = 2;i*i <= t;i++){
        while(t%(i*i) == 0){
            t /= i*i,k *= i;
        }
    }r = t;
    if (r == 1){//r == 1 => sqrt(d) = k 有理数形式 方程解x = (-b+(-)k)/(2*a)
        int p = (a > 0) ? (-b+k) : (-b-k),q = 2*a;
        to_simp(p,q);
        if (q == 1){
            cout << p << endl;
        }else{
            cout << p << "/" << q << endl;
        }
    }else{//无理数 r != 1 sqrt(d) = k*sqrt(r) x = -b/(2*a)+(-)k/(2*a)*sqrt(r) 
        int p1 = -b,q1 = 2*a;
        to_simp(p1,q1);
        int s = (a > 0) ? 1 : -1,p2 = s*k,q2 = 2*a;
        to_simp(p2,q2);
        if (p1 != 0){
            if (q1 == 1){
                cout << p1;
            }else{
                cout << p1 << "/" << q1;
            }cout << "+";
        }if (p2 == 1 && q2 == 1){
            cout << "sqrt(" << r << ")";
        }else if (q2 == 1){
            cout << p2 << "*sqrt(" << r << ")";
        }else if (p2 == 1){
            cout << "sqrt(" << r << ")/" << q2;
        }else{
            cout << p2 << "*sqrt(" << r << ")/" << q2;
        }cout << endl;
    }
}int main(){
    freopen("uqe.in","read",stdin);
    freopen("uqe.out","w",stdout);
    int t,m;
    cin >> t >> m;
    for (int i = 0;i < t;i++){
        solve();
    }fclose(stdin);
    fclose(stdout);
    return 0;
}