C++14(g++5.4) 解法, 执行用时: 74ms, 内存消耗: 2012K, 提交时间: 2018-08-08 19:55:36

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll gcd(ll a,ll b)
{
    return b?gcd(b,a%b):a;
}
void solve()
{
    ll a,b,c;scanf("%lld%lld%lld",&a,&b,&c);
    ll deta=b*b-4*a*c;
    if(deta<0)
    {
        printf("%lld %lld %lld\n",a,b,c);return;
    }
    ll sqr=deta;
    deta=(ll)sqrt(deta);
    if(sqr!=deta*deta){
            printf("%lld %lld %lld\n",a,b,c);return;
    }
    ll s=2*a,t=b+deta;
    ll u=2*a,v=b-deta;
    ll gcd1=gcd(s,t),gcd2=gcd(u,v);
    s/=gcd1;t/=gcd1;u/=gcd2;v/=gcd2;
    ll M=gcd1*gcd2/4/a;
    if(s<0){s*=-1;t*=-1;M*=-1;}
    if(u<0){u*=-1;v*=-1;M*=-1;}
    if(s<u){swap(s,u);swap(t,v);}
    else if(s==u&&t<v)swap(t,v);
    if(M<0){M*=-1;u*=-1;v*=-1;}
    s*=M;t*=M;
    printf("%lld %lld %lld %lld\n",s,t,u,v);
}
int main()
{
    int T;scanf("%d",&T);
    while(T--)solve();
    return 0;
}

C++(g++ 7.5.0) 解法, 执行用时: 241ms, 内存消耗: 3472K, 提交时间: 2022-11-19 13:01:01

#include<bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
    int a,b,c;
    cin>>a>>b>>c;   
    int res=b*b-4*a*c;
    if(res<0){cout<<a<<" "<<b<<" "<<c<<endl;return;}
    
    int res1=sqrt(res);
    if(res1*res1!=res){cout<<a<<" "<<b<<" "<<c<<endl;return;}
    int s=2*a,t=b+res1;
    int u=2*a,v=b-res1;
    int g=__gcd(abs(s),abs(t));
    s/=g,t/=g;
    g=__gcd(abs(u),abs(v));
    u/=g,v/=g;
    if(s<0)    s=-s,t=-t;
    if(u<0)    u=-u,v=-v;
    
    if(a % (s*u)!=0){cout<<a<<" "<<b<<" "<<c<<endl;return;}
    a/=s;
    a/=u;
    if(s<u || (s==u && t<v))    swap(s,u),swap(t,v);
    s*=abs(a);
    t*=abs(a);
    if(a<0)    u=-u,v=-v;
    cout<<s<<" "<<t<<" "<<u<<" "<<v<<endl;
    return ;
}
    
signed main()
{
    int t;
    cin>>t;
    while(t--)solve();
    return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 84ms, 内存消耗: 3448K, 提交时间: 2018-10-09 13:52:35

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b)
{
	return b?gcd(b,a%b):a;
} 
int T,a,b,c;
ll u,v,s,t;
int main()
{
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d%d",&a,&b,&c);
		ll d=1ll*b*b-4ll*a*c,e=0;
		if(d>=0)e=(ll)sqrt(d);
		while(e*e<d)e++;
		if(e*e!=d)printf("%d %d %d\n",a,b,c);
		else
		{
			t=b+e,v=b-e,s=u=2*a;
			ll g=gcd(abs(s),abs(t));
			s/=g,t/=g;
			g=gcd(abs(u),abs(v));
			u/=g,v/=g;
			if(s<0)s=-s,t=-t;
			if(u<0)u=-u,v=-v;
			if(a%((ll)s*u)!=0)printf("%d %d %d\n",a,b,c);
			else
			{
				a/=s,a/=u;
				if(s<u||s==u&&t<v)swap(s,u),swap(t,v);
				s*=abs(a),t*=abs(a);
				if(a<0)u=-u,v=-v;
				printf("%lld %lld %lld %lld\n",s,t,u,v);
			}
		}
	}
	return 0;
}