写这种时间不敏感的大模拟最关键的就是多写注释，然后一切尽量按照原题的意思、流程去做。

#include <bits/stdc++.h>
using namespace std;
int T, M;
int gcd (int a, int b)
{
    if(b == 0)
        return a;
    return gcd(b, a % b);
}
void outPQ(int p, int q)
{
    //输出 p / q
    //预处理1：约分
    int d = gcd(p, q);
    if (d < 0)
        d *= -1;
    p /= d;
    q /= d;
    //输出
    if (q == 1)
        cout << p;
    else
        cout << p << "/" << q;
}
int main()
{
    freopen("uqe.in", "r", stdin);
    freopen("uqe.out", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> T >> M;
    while (T--)
    {
        int a, b, c;
        cin >> a >> b >> c;
        //分母变成正数
        if (a < 0)
        {
            a *= -1;
            b *= -1;
            c *= -1;
        }
        int delta = b * b - 4 * a * c;
        // 无解
        if (delta < 0)
        {
            cout << "NO\n";
            continue;
        }
        //有解时处理输出较大者 (-b+sqrt(delta)) / 2a
        // (p1 + p2 * sqrt(p3)) / q
        int q = 2 * a;
        int p1 = -b;
        int p2 = 1;
        int p3 = delta;
        //化简 p3
        for(int i = 2; i * i <= p3; i++)
        {
            int ii = i * i;
            while (p3 % ii == 0)
            {
                p3 /= ii;
                p2 *= i;
            }
        }
        if(p3 == 0)
            p2 = 0;
        // 解为有理数
        if (p3 == 1 || p2 == 0)
        {
            outPQ(p1 + p2, q);
            cout << "\n";
            continue;
        }
        // 解为 x = q1 + q2 sqrt(r)
        // 使用 pair 储存有理数
        pair<int,int> q1 = make_pair(p1, q);
        pair<int,int> q2 = make_pair(p2, q);
        int r = p3;
        if (q1.first != 0)
        {
            //q1 != 0
            outPQ(q1.first, q1.second);
            cout<<"+";
        }
        if (q2.first == q2.second)
        {
            //q2 == 1
            cout << "sqrt(" << r << ")";
        }
        else if (q2.first % q2.second == 0)
        {
            //q2 为整数
            cout << q2.first / q2.second << "*sqrt(" << r << ")";
        }
        else if (q2.second % q2.first == 0)
        {
            //1/q2 为整数
            cout << "sqrt(" << r << ")/" << q2.second / q2.first;
        }
        else
        {
            //化简 q2
            int d = gcd(q2.first, q2.second);
            q2.first /= d;
            q2.second /= d;
            cout << q2.first << "*sqrt(" << r << ")/" << q2.second;
        }
        cout << "\n";
    }  
    return 0;
}

#include<bits/stdc++.h>
using namespace std ;
long long t,m,a,b,c,de;
long long gcd(long long x,long long y)
{
	if(y==0)return x;
	else return gcd(y,x%y);
}
void doit(long long fz,long long fm)
{
	long long gcdd=abs(gcd(fz,fm));
	fz/=gcdd;fm/=gcdd;
	if(fm==1)cout<<fz;
	else
	cout<<fz<<"/"<<fm;
}
int main()
{
	freopen("uqe.in","r",stdin);
	freopen("uqe.out","w",stdout); 
	cin>>t>>m;
	while(t--)
	{
		cin>>a>>b>>c;
		de=b*b-4*a*c;
		if(de<0)cout<<"NO";
		else
		{
			long long zs=0;
			for(long long i=1;i*i<=de;i++)
			{
				if(de%(i*i)==0)zs=i; 
			}
			if(zs*zs==de)
			{
				if(a>0)
				{
					long long fz=-b+zs,fm=2*a;
					doit(fz,fm);
				}
				else
				{
					long long fz=b+zs,fm=-2*a;
					doit(fz,fm);
				}
			}
			else
			{
				long long fz,fm;
				if(a>0)
				fz=-b,fm=2*a;
				else fz=b,fm=-2*a;
				if(fz!=0)
				{
					doit(fz,fm);
					cout<<"+";
				}
				long long r=de/(zs*zs);
				if(a>0)
				fz=zs,fm=2*a;
				else fz=zs,fm=-2*a;
				long long gcdd=gcd(fz,fm);
				fz/=gcdd,fm/=gcdd;
				if(fm==1)
				{
					if(fz==1)cout<<"sqrt("<<r<<")";
					else cout<<fz<<"*sqrt("<<r<<")";
				}
				else if(fz==1)cout<<"sqrt("<<r<<")/"<<fm;
				else cout<<fz<<"*sqrt("<<r<<")/"<<fm;
	
			}
		}
		cout<<endl;
	}
	return 0;
}