牛顿迭代法、迭代逼近

/*zstu  求一元三次方程的根

要求1附近的根，只要把x0初始化为1代入，然后牛顿迭代1000次就出结果了

*/

#include<stdio.h>double c[4],d[3];double f(double x){double fac=1,sum=0;int i;for(i=0;i<4;i++){sum+=fac*c[i];fac*=x;}return sum;}double df(double x){double fac=1,sum=0;int i;for(i=0;i<3;i++){sum+=fac*d[i];fac*=x;}return sum;}main(){int i;double x0;while(scanf("%lf%lf%lf%lf",&c[3],&c[2],&c[1],&c[0])!=EOF){d[0]=c[1];d[1]=c[2]*2;d[2]=c[3]*3;x0=1;i=1;while(i<1000){i++;x0=x0-f(x0)/df(x0);}printf("%.2f/n",x0);}}