HUST 1541 Student’s question

2019-06-06

08:47:59

adhere to! ! !

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     int n;
 6     scanf("%d", &n);
 7     while (n--)
 8     {
 9         int a,b,c;
10         scanf("%d %d %d", &a, &b, &c);
11         int m = b * b - 4 * a * c;
12         if(m < 0)
13             cout << "NO";
14         else if(m == 0)
15         {
16             double x = (double)b * (-1) / (2 * a);
17             printf("%.2lf", x);
18         }
19         else
20         {
21             double x = ((double)b * (-1) - sqrt(m)) / (2 * a);
22             double y = ((double)b * (-1) + sqrt(m)) / (2 * a);
23             printf("%.2lf %.2lf", x, y);
24         }
25         cout << endl;
26     }
27     return 0;
28 }

Other approaches

Veda theorem: x1 + x2 = -b / a, x1 * x2 = c / a to solve x1-x2 = sqrt ((x1 + x2) * (x1 + x2) -4 * x1 * x2), in order to then be out of x1, x2 it! Also remember two decimal places yo!