题解 | #求解立方根#

f(x)=x^3-n

f'(x)=3x^2

所以有xn+1=xn-(xn^3-n)/(3xn^2)

cin>>x;
int b=1.0,a=0.0;
while(abs(b-a)>0.001){
	a=b;
  	b=a-(a*a*a-x)/(3*a*a);
}
printf("%.1f",b);