7-29 二分法求多项式单根

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#include <stdio.h>

double f(double *p, double x);
int main(void)
{

    double an[4];
    double a, b;
    for (int i = 3; i >= 0; i--)
    {
        scanf("%lf", &an[i]);
    }

    scanf("%lf %lf", &a, &b);

    int flag = 0;       //标志位
    while (b - a > 1e-3)
    {
        double m = (a + b) / 2;
        double ret_m = f(an, m);
        if (ret_m == 0)             //判断浮点数等于0
        {
            printf("%.2f
", m);
            flag = 1;
            break;
        }

        double ret_a = f(an, a);
        double ret_b = f(an, b);

        if (ret_a * ret_m > 0)      //判断同号
        {
            a = m;
        }
        if (ret_b * ret_m > 0)
        {
            b = m;
        }
    }

    if (flag == 0)
    {
        printf("%.2f
", (a + b) / 2);
    }
    return 0;
}
double f(double *p, double x)
{
    double ret;

    ret = p[3] * x * x * x + p[2] * x * x + p[1] * x + p[0];

    return ret;
}