pat-"c/c++/java/pascal Program design Basics" Problem sets-loop -08 (mooc3-1) Two-part method for finding polynomial single

First, stick to the topic:

The principle of the dichotomy method is: if the continuous function f (x) in the interval [a, b] two endpoints of the value of the difference, that is F (a) f (b) <0, then it has at least 1 root R in this interval, that is, f (r) = 0.

The steps of the dichotomy are:

Check the interval length, if it is less than the given threshold, stop, the output interval midpoint (a+b)/2;
If f (a) f (b) <0, the value of the midpoint is computed F ((A+B)/2);
If f ((A+B)/2) is exactly 0, then (a+b)/2 is the root of the requirement;
If f ((A+B)/2) is the same as F (a), then the root is in the interval [(A+B)/2, b], making a= (A+B)/2, repeating the cycle;
If f ((A+B)/2) is the same as F (b), then the root is in the interval [A, (A+B)/2], making b= (A+B)/2, repeating the cycle;

This topic requires a program to calculate the root of a given 3-order polynomial f (x) =a3x3+a2x2+a1x+a0 within a given interval [a, b].

Input format:

Enter the 4 coefficients A3, A2, A1, A0 of the polynomial in the order given in the 1th row, giving the interval endpoints a and B sequentially in line 2nd. The topic guarantees that the polynomial has a unique single root within a given interval.

Output format:

The root of the polynomial within the interval is output in a row, which is exactly 2 digits after the decimal point.

Input Sample:
3-1-3 1
-0.5 0.5
Sample output:
0.33

Precautions:

1. The title is required to be accurate to 2 digits after the decimal point, so a constant of # define or const, #define EPS 1e-3 or const double eps=1e-3. Use this constant to determine the size of the interval. That is, when (a, b) interval length b-a is less than or equal to 1e-3, terminate the loop.

2. During the cycle, if f ((A+B)/2) equals 0, the loop is terminated prematurely and the result is output. Judging from the outside of the loop if F ((A+B)/2) is not equal to 0, it means that there is no result output in the loop body, but because the interval length is too small (less than or equal to 1e-3) so that the loop terminates, at which time to output (A + B)/2.

3. Function calculation, using x * (x * (A3 * x + A2) + A1) + a0 is better than a3*x*x*x+a2*x*x+a1*x+a0!

Because this problem is relatively simple, the notice has been explained in advance, so the direct adhesive code:

1#include <iostream>2#include <cstdio>3 4 #defineEPS (1e-3)//Precautions 15 using namespacestd;6 7 DoubleA3, A2, A1, A0;8 DoubleFuncDoublex);9 intMain ()Ten { One     DoubleA, B; ACIN >> A3 >> A2 >> A1 >>A0; -Cin >> a >>b; -  the     DoubleFA =func (a); -     DoubleFB =func (b); -      while(B-a > EPS)//Precautions 1 -     { +         DoubleMid = (A + b)/2; -         DoubleFmid =func (mid); +         if(Fmid = =0)//Precautions 2 A         { atprintf"%.2LF", mid); -              Break; -         } -         if(FA * Fmid >0) -         { -A =mid; inFA =func (a); -             Continue; to         } +         if(FB * fmid >0) -         { theb =mid; *FB =func (b); $             Continue;Panax Notoginseng         } -     } the     if(Func (A + B)/2) !=0) +printf"%.2f\n", (A + B)/2);//Precautions 2 A  the     return 0; + } - DoubleFuncDoublex) $ { $     returnX * (x * (A3 * x + A2) + A1) + a0;//Precautions 3 -}

Stick to the usual results of AC:

pat-"c/c++/java/pascal Program design Basics" Problem sets-loop -08 (mooc3-1) Two-part method for finding polynomial single