牛顿迭代法解非线性方程 x ^ 3 = a, 即求a的开三次方

执行环境：VC6.0

//牛顿迭代法解非线性方程 x ^ 3 = a, 即求a的开三次方
//---------------------Include Files-------------
#include
//#include
//-----------------end Include Files-------------
#define M 10//迭代次数

double f(double x, double a);//f(x)函数
double df(double x);//f(x)的导数

int main()
{
    double x[100];
    double a;
    int i = 1;

    x[0] = 1;
    printf("Please input a: ");
    scanf("%lf",&a);

//    for(; i < M; i++)
//    {

//        x[i] = x[i-1] - f(x[i-1], a) / df(x[i-1]);

//        printf("x[%d] = %.7lf/n", i, x[i]);
//    }

//    while(fabs(f(x[i-1], a)) > 0.0000005)
//    while((f(x[i-1], a) >= -0.0000005) && (f(x[i-1], a) <= 0.0000005))
    while(f(x[i-1], a) > 0 ? f(x[i-1], a) : -f(x[i-1], a) >= 0.0000005)
        //0.0000005是运算精度
    {
        x[i] = x[i-1] - f(x[i-1], a) / df(x[i-1]);

        printf("x[%d] = %.7lf/n", i, x[i]);

        i++;
    }

    return(0);
}

double f(double x, double a)
{
    return (x * x * x - a);
}

double df(double x)
{
    return (3 * x * x);
}