铁子们，怎么用c++写

Question

Answer 1

仅供参考：

//  牛顿法解多项式的根
//  输入：多项式系数c[]，多项式度数n，求在[a,b]间的根
//  输出：根
//  要求保证[a,b]间有根
#include <stdio.h>
#include <stdlib.h>
#define MAXN1 10
double C[MAXN1];
double ROOT;
double fabs( double x ) {
    return (x<0)? -x : x;
}
double f(int m, double c[], double x) {
    int i;
    double p = c[m];

    for (i=m; i>0; i--) p = p*x + c[i-1];
    return p;
}
int newton(double x0, double *r,double c[], double cp[], int n,double a, double b, double eps) {
    int MAX_ITERATION = 1000;
    int i = 1;
    double x1, x2, fp, eps2 = eps/10.0;

    x1 = x0;
    while (i < MAX_ITERATION) {
        x2 = f(n, c, x1);
        fp = f(n-1, cp, x1);
        if ((fabs(fp)<0.000000001) && (fabs(x2)>1.0)) return 0;
        x2 = x1 - x2/fp;
        if (fabs(x1-x2)<eps2) {
            if (x2<a || x2>b) return 0;
            *r = x2;
            return 1;
        }
        x1 = x2;
        i++;
    }
    return 0;
}
double Polynomial_Root(double c[], int n, double a, double b, double eps) {
    double *cp;
    int i;
    double root;

    cp = (double *)calloc(n, sizeof(double));
    for (i=n-1; i>=0; i--) {
        cp[i] = (i+1)*c[i+1];
    }
    if (a>b) {
        root = a; a = b; b = root;
    }
    if ((!newton(a        , &root, c, cp, n, a, b, eps))
     && (!newton(b        , &root, c, cp, n, a, b, eps))) {
          newton((a+b)*0.5, &root, c, cp, n, a, b, eps);
    }
    free(cp);
    if (fabs(root)<eps) return fabs(root);
    else return root;
}
void main() {
    C[0]=-0.5;
    C[1]=0.0;
    C[2]=1;
    ROOT=Polynomial_Root(C, 2, -1.0, 0.0, 0.0001);
    printf("%g\n",ROOT);//x*x-0.5=0在[-1,0]之间的根精确到0.0001约是-0.707107
}