H-目标是成为数论大师

题目链接：
https://codeforces.com/gym/102174/problem/H
注意事项

1. \(\sqrt{ax} + b = x \Rightarrow \sqrt{ax} = x - b\)
   需满足\(ax \geq 0 , x - b \geq 0\) 两个条件.

2. 注意该题的设定为数论题，题目保证至少有一解，故\(\Delta \geq 0\), 注意从就行考虑.

AC代码：

//md以后解方程好好解
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

int a, b;

int main()
{
    int T;
    scanf("%d", &T);
    while(T --)
    {
        int ans1, ans2;
        scanf("%d%d", &a, &b);
        int delta = 4 * a * b + a * a;
        //cout << delta << endl;
        if(delta == 0){
            int sdel = sqrt(delta);
            //cout << sdel << endl;
            ans1 = (2 * b + a) / 2;
            //cout << ans1 << endl;
            if(ans1 * a >= 0){
                puts("1");
                printf("%d\n", ans1);
            }
        }
        else if(delta > 0){
            int cnt = 0;
            bool flag1 = false, flag2 = false;
            int sdel = sqrt(delta);
            //cout << sdel << endl;
            if((a + sdel) % 2 == 0){
                ans1 = (2 * b + a + sdel) / 2;
                //cout << ans1 << ' ' << sqrt(a * ans1) + b << endl;
                if(ans1 * a >= 0 && ans1 - b >= 0){
                    flag1 = true;
                    cnt ++;
                }
            }
            if((a - sdel) % 2 == 0){
                ans2 = (2 * b + a - sdel) / 2;
                if(ans2 * a >= 0 && ans2 - b >= 0){
                    flag2 = true;
                    cnt ++;
                }
            }
            if(cnt == 1 && flag1 == true) printf("%d\n%d\n", cnt, ans1);
            else if(cnt == 1 && flag2 == true) printf("%d\n%d\n", cnt, ans2);
            else if(cnt == 2) printf("%d\n%d %d\n", cnt, ans2, ans1);
        }
    }
    return 0;
}