题解 | #计算一元二次方程#
计算一元二次方程
https://www.nowcoder.com/practice/7da524bb452441b2af7e64545c38dc26
#include <stdio.h>
#include <math.h>
int main() {
double a, b, c;
double x1 = 0.00, x2 = 0.00; //方程的根
double x =0.00,y=0.00;//x表示实部，y表示虚部
while (scanf("%lf %lf %lf", &a, &b, &c) != EOF) {
if (a == 0) {
printf("Not quadratic equation\n");
} else {
if (b * b - 4 * a * c == 0) { //等于零单实根
if (b==0&&c==0) {
printf("x1=x2=0.00\n");
} else {
x1 = -1.0 * b / (2.0 * a);
printf("x1=x2=%.2lf\n", x1);
}
} else if (b * b - 4 * a * c > 0) { //大于零双实根
x1 = (-b - sqrt(b * b - 4 * a * c)) / 2.0 / a;
x2 = (-b + sqrt(b * b - 4 * a * c)) / 2.0 / a;
printf("x1=%.2lf;x2=%.2lf\n", x1, x2);
} else { //小于零虚根
x = -b/(2.0*a);
y = sqrt(4 * a * c - b * b)/(2.0*a);
printf("x1=%.2lf-%.2lfi;x2=%.2lf+%.2lfi\n",x,y,x,y );
}
}
}
return 0;
}
