用c语言编程 从键盘任意输入a，b，c的值，编程计算并输出一元二次方程ax2+bx+c=0的根。根据一元二次方程的求根公式，令 p=−b2a,q=∣∣b2−4ac∣∣√2a 当b2−4ac=0时，输出两个相等的实根x1=x2=p；当b2−4ac>0时，输出两个不相等的实根：x1=p+q，x2=p−q；当b2−4ac<0时，输出一对共轭复根：x1=p+qi，x2=p−qi。当a=0时，输出"It is not a quadratic equation!\n"。 **输入格式要求："%f,%f,%f" 提示信息："Please enter the coefficients a,b,c:" **输出格式要求： 相等实根: "x1 = x2 = %.2f\n" 不相等的实根: "x1 = %.2f, x2 = %.2f\n" 一对共轭复根: "x1 = %.2f+%.2fi, x2 = %.2f-%.2fi\n" 程序运行示例如下： Please enter a,b,c: 0,10,2 It is not a quadratic equation!

以下是该问题的C语言解法：

#include <stdio.h>
#include <math.h>

int main() {
    float a, b, c, p, q, x1, x2;
    printf("Please enter the coefficients a,b,c: ");
    scanf("%f,%f,%f", &a, &b, &c);
    if (a == 0) {
        printf("It is not a quadratic equation!\n");
    } else {
        p = -b / (2 * a);
        q = sqrt(fabs(b * b - 4 * a * c)) / (2 * a);
        if (b * b == 4 * a * c) {
            x1 = x2 = p;
            printf("x1 = x2 = %.2f\n", x1);
        } else if (b * b > 4 * a * c) {
            x1 = p + q;
            x2 = p - q;
            printf("x1 = %.2f, x2 = %.2f\n", x1, x2);
        } else {
            x1 = p;
            x2 = q;
            printf("x1 = %.2f+%.2fi, x2 = %.2f-%.2fi\n", x1, x2, x1, x2);
        }
    }
    return 0;
}

输入示例：

Please enter the coefficients a,b,c: 0,10,2

输出示例：

It is not a quadratic equation!