7-5 Dichotomy for a single root of a polynomial (20 points)

The principle of finding the function root by the dichotomy is: if the continuous function f (x) takes different signs at the two endpoints of the interval [a, b], ie f (a) f (b) <0, then it is within this interval There is at least one root r, that is f (r) = 0.

The steps of the dichotomy are:

Check the length of the interval, if it is less than the given threshold, stop, output the midpoint of the interval (a + b) / 2; otherwise
if f (a) f (b) <0, calculate the value of the midpoint f ((a + b ) / 2);
if f ((a + b) / 2) is exactly 0, then (a + b) / 2 is the required root; otherwise
if f ((a + b) / 2) and f (a) The same sign means that the root is in the interval [(a + b) / 2, b], let a = (a + b) / 2, repeat the cycle;
if f ((a + b) / 2) and f (b) The same sign means that the root is in the interval [a, (a + b) / 2], let b = (a + b) / 2 and repeat the cycle.
This topic requires writing a program to calculate the root of a given 3rd order polynomial f (x) = a ₃ x ³ + a ₂ x ² + a ₁ x + a ₀ within a given interval [a, b].

Input format:
Enter the four coefficients a ₃ , a ₂ , a ₁ , and a ₀ that sequentially give the polynomial in the first row, and give the interval endpoints a and b in the second row. The problem is to ensure that the polynomial has a single root in a given interval.

Output format:
output the root of the polynomial in the interval in one line, accurate to 2 decimal places.

Sample input:

3 -1 -3 1
-0.5 0.5

Sample output:

0.33

Without further ado, go to the code first

#define _CRT_SECURE_NO_DEPRECATE 0
#include<stdio.h>
#include<stdlib.h>
double  a0, a1, a2, a3;
double target(double n){
	double f;
	f = a3*(n*n*n) + a2*(n*n) + a1*n + a0;			//计算函数值
	return f;
}
int main(){
	double  a, b;
	int i = 0;
	scanf("%lf %lf %lf %lf", &a3, &a2, &a1, &a0);
	scanf("%lf %lf", &a, &b);
	double f1, f2, n, f;
	f1 = target(a);
	f2 = target(b);
	if (f1*f2 < 0){
		n = (a + b) / 2;
		f = target(n);
		while (1){
			i++;
			if (f == 0){			//f正好为0，则n就是要求的根
				printf("%.2lf", n);
				break;
			}
			if (i == 100){			//当出现求不尽的情况时
				printf("%.2lf", n);
				break;
			}
			else if (f*f1 < 0){		//如果f与f1异号，则说明根在区间[a,n]，令b=n，重复循环
				b = n;
				n = (n + a) / 2;
			}
			else if (f*f2 < 0){		//如果f与f2异号，则说明根在区间[n,b]，令a=n，重复循环
				a = n;
				n = (n + b) / 2;
			}
			f = target(n);
		}
	}
	else if (f1 == 0){			//区间端点是根的情况
		printf("%.2lf", a);
	}
	else if (f2 == 0){			//区间端点是根的情况
		printf("%.2lf", b);
	}
	system("pause");
	return 0;
}