不可能包AC

方法1：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll k,n,e,d;
int main() {
    cin >> k;
    while (k--) { 
        cin >> n >> e >> d;
        ll m=n+2-e*d,de=m*m-4*n;
        if (de<0) {
            cout << "NO" << endl;
            continue;
        }
        long long sqrt_de = sqrt(de);
        if (sqrt_de * sqrt_de != de) {
            cout << "NO" << endl;
            continue;
        }
        ll p=(m-sqrt_de)/2,q=(m+sqrt_de)/2;
        if (p>0&&q>0&&p*q==n) {
            cout << p << " " << q << endl;
        } else {
            cout << "NO" << endl;
        }
    }
    //彭子洋原创^_^; 
    return 0;
}

方法2：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll k,n,d,e,m,p,q; 
int main(){
	cin >> k;
	while(k--){
		cin >> n >> d >> e;
		m=n+2-e*d;
		ll lf=1,rt=m/2;
		bool f=0;
		while(lf<=rt){
			p=(lf+rt)/2;
			q=m-p;
			if(p*q>n){
				rt=p-1;
			}
			else if(p*q<n) {
				lf=p+1;
			}
			else {
				cout << p << " " << q << "\n";
				f=1;
				break;
			}
		}
		if(!(f)) cout << "NO\n";
	}
	return 0;
}
//

#include <iostream>
#include <cmath>
using namespace std;
typedef long long ll;
int main() {
    int k;
    cin >> k;
    while (k--) {
        ll n, e, d;
        cin >> n >> e >> d;
        ll m = n - e * d + 2;
        ll delta = m * m - 4 * n;
        if (delta < 0) {
            cout << "NO\n";
            continue;
        }
        ll root = sqrt(delta);
        if (root * root != delta) {
            cout << "NO\n";
            continue;
        }
        ll p = (m - root) / 2;
        ll q = (m + root) / 2;
        if (p > 0 && q > 0 && p * q == n) {
            cout << p << " " << q << "\n";
        } else {
            cout << "NO\n";
        }
    }
    return 0;
}

包AC

#include<bits/stdc++.h>
using namespace std;
int k,n,d,e,m,p,q,ans=1;
int main(){
	cin>>k;
	while(k--){
		cin>>n>>d>>e;
		m=n+2-e*d;
		long long lf=1,rt=m/2;
		bool ok=0;
		while(lf<=rt){
		 p=(lf+rt)/2;
		 q=m-p;		
		 if(p*q>n) rt=p-1;
		 else if(p*q<n) lf=p+1;
		 else {
		 	cout<<p<<" "<<q<<endl;
		 	ok=1;
		 	break;
		 }	
		}
		if(ok==0)cout<<"NO"<<endl;
	} 
	return 0;
}