NOIP2001 three-time equation solution [derivative + Newton iterative method]

Title Description

Tangible such as: ax3+bx2+cx+d=0 such a one-dimensional three-time equation. The coefficients of each of the equations (A,b,c,d are real) are given, and the equation has three different real roots (the range of the roots between 100 to 100), and the absolute value of the difference between root and root is >=1. It is required to output these three real roots (spaces between root and root) in the same row from small to large, and accurate to 2 digits after the decimal point.

Hint: Remember equation f (x) = 0, if there are 2 numbers x1 and x2, and x1<x2,f (x1) *f (x2) <0, then there must be a root between (X1,X2).

Input/output format

Input format:

One row, 4 real a,b,c,d.

Output format:

One line, three real roots, and accurate to 2 digits after the decimal point.

Input and Output Sample input example # #:

1-5-4 20

Sample # # of output:

-2.00 2.00 5.00

Data size is too small to be casual violence

But in order to prove that I have been studying calculus for a few days, using an advanced method
First f (x) =ax3+bx2+cx+d derivative get df/dx=3ax2+2bx+c
Find the 0 point of the derivative (that is, the two-time function to find the root formula) to get the maximum value of f (x).
There must be an F (x) 0 point for the three intervals consisting of the maximum points, and the 0 points can be obtained by Newton iterative method.
Newton's iterative method is the constant use of a point tangent to fit the curve, the derivative of that point is the tangent slope

And so on, we can get an iterative method to find 0 points of the higher order function:
To find the n-th function 0 points, need the extreme points to divide the interval, it is necessary to find its derivative (n-1 function) of 0 points, sequentially iterate to n=2 directly through the formula (of course n=3 or 4 can also)
The final complexity depends on the complex reading of the 0-point algorithm.
No one seems to have published it, so it's called Candy iterative method .
But does this have an advantage over the three-part approach to extremum?

////main.cpp//unary three-time equation////Created by Candy on 2016/12/10.//copyright©2016 year Candy. All rights reserved.//#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespacestd;Const Doubleeps=1e-3;DoubleA,b,c,d;inlineDoubleFDoublex) {return((a*x+b) *x+c) *x+D;} InlineDoubledfDoublex) {return(3*a*x+2*B) *x+C;}DoubleSolDoubleLDoubleR) {//printf ("Sol%lf%lf\n", l,r);    intstep= -;Doublex= (L+R)/2;  while(step--) {x=x-f (x)/DF (x); }    returnx;}intMainintargcConst Char*argv[]) {scanf ("%LF%LF%LF%LF",&a,&b,&c,&d); DoubleP1= (-sqrt (b*b-3*A*C)-B)/(3*a), p2= (+sqrt (b*b-3*A*C)-B)/(3*a); printf ("%.2f%.2f%.2f\n", Sol (- -, p1), Sol (P1,P2), Sol (P2, -)); return 0;}

NOIP2001 three-time equation solution [derivative + Newton iterative method]