Steffensen

#include <stdio.h>
#include<math.h>
double cal(double x,int option){
if(option==1)
return (3*x+1)/(x*x);
else if(option==2)
return (x*x*x-1)/3.0;
else if (option==3)
return pow(3*x+1,1.0/3.0);
else if(option==4)
return 1.0/(x*x-3);
else if(option==5)
return sqrt(3+1.0/x);
else 
return x-1.0/3.0*(x*x*x-3*x-1)/(x*x-1);
}
double judge(double y,double z,double x,double x0,double flag,int n,int N,int option)
{
printf("               迭代结果如下：\n");
printf("K         Xk          Yk          Zk \n");
while(n<=N)
{
y=cal(x0,option);
z=cal(y,option);
printf("%d    %lf      %lf   %lf\n",n,x0,y,z);
x=x0-(y-x0)*(y-x0)/(z-2*y+x0)*1.0;
if(fabs(x-x0)<flag)
{
return x;
}
++n;
x0=x;
}
return 0; 
}
int main()
{
double y=0,z=0,x=0,x0=0,flag=0;
int N=0,n=0,option=0;
for(;1>0;)
{
printf("请输入相应的计算格式：\n");
        printf(" ① x = (3x+1)/(x^2)② x = (x^3-1)/3 \n");
      printf("③ x = sqrt(3x+1) ④ x = 1/(x^2-3) \n");
   printf(" ⑤ x = pow((3+1/x),1/3) ⑥ x = x-(1/3)*(x^3-3x-1)/(x^2-1) \n");
   printf("\n请输入题号（0 退出）： ");
scanf("%d",&option);
if(option)
{
printf("请输入初始值(x0)，精度要求(flag)，最大的迭代次数(N)：\n");
scanf("%lf %lf %d",&x0,&flag,&N);
if(fabs(x=judge(y,z,x,x0,flag,n,N,option))>0)
printf("方程的近似解为:%lf\n",x);
else printf("不收敛!\n");
}
else return 0;
}
return 0;
}