给定三个形成单调递增函数的数字a ， b和c 
形式为a * x 2 + b * x + c以及两个数字A和B ，任务是找到函数的根，即，找到x的值，使得
其中A≤x≤B.
例子：
Input: a = 2, b = -3, c = -2, A = 0, B = 3
Output: 2.0000
Explanation:
f(x) = 2x^2 – 3x – 2 putting the value of x = 2.000
We get f(2.000) = 2(2.000)^2 – 3(2.000) – 2 = 0
Input: a = 2, b = -3, c = -2, A = -2, B = 1
Output: No solution
方法：以下是任何函数的图形表示
：
从上图可以看出，
- 每当f(A)* f(B)≤0时，表示函数的图将在该范围内的某处切割x轴，并表示某处存在使该函数的值为0的点，然后该图变为负y轴。
- 如果f(A)* f(B)> 0 ，则表示在[A，B]范围内， f(A)和f(B)的y值始终保持正值，因此它们从不切割x轴。
因此，从上面的观察中，想法是使用二进制搜索来解决此问题。使用给定的[A，B]范围作为方程式根的上下限，可以通过对该范围进行二进制搜索来找出x 。步骤如下：
- 找到范围[A，B]的中间(例如mid )。
- 如果f(mid)* f(A)≤0为真，则搜索空间减小为[A，mid] ，因为在此段中发生了x轴切割。
- 其他搜索空间缩小为[mid，B]
- 根的最小可能值是高值刚好小于EPS (最小值〜10 -6 )+下限值时，即fabs(高–低)> EPS 。
- 打印根的值。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
#define eps 1e-6
// Given function
double func(double a, double b,
double c, double x)
{
return a * x * x + b * x + c;
}
// Function to find the root of the
// given non-decreasing Function
double findRoot(double a, double b,
double c, double low,
double high)
{
double x;
// To get the minimum possible
// answer for the root
while (fabs(high - low) > eps) {
// Find mid
x = (low + high) / 2;
// Search in [low, x]
if (func(a, b, c, low)
* func(a, b, c, x)
<= 0) {
high = x;
}
// Search in [x, high]
else {
low = x;
}
}
// Return the required answer
return x;
}
// Function to find the roots of the
// given equation within range [a, b]
void solve(double a, double b, double c,
double A, double B)
{
// If root doesn't exists
if (func(a, b, c, A)
* func(a, b, c, B)
> 0) {
cout << "No solution";
}
// Else find the root upto 4
// decimal places
else {
cout << fixed
<< setprecision(4)
<< findRoot(a, b, c,
A, B);
}
}
// Driver Code
int main()
{
// Given range
double a = 2, b = -3, c = -2,
A = 0, B = 3;
// Function Call
solve(a, b, c, A, B);
return 0;
} Java
// Java program for the above approach
import java.util.*;
import java.lang.*;
class GFG{
static final double eps = 1e-6;
// Given function
static double func(double a, double b,
double c, double x)
{
return a * x * x + b * x + c;
}
// Function to find the root of the
// given non-decreasing Function
static double findRoot(double a, double b,
double c, double low,
double high)
{
double x = -1;
// To get the minimum possible
// answer for the root
while (Math.abs(high - low) > eps)
{
// Find mid
x = (low + high) / 2;
// Search in [low, x]
if (func(a, b, c, low) *
func(a, b, c, x) <= 0)
{
high = x;
}
// Search in [x, high]
else
{
low = x;
}
}
// Return the required answer
return x;
}
// Function to find the roots of the
// given equation within range [a, b]
static void solve(double a, double b, double c,
double A, double B)
{
// If root doesn't exists
if (func(a, b, c, A) * func(a, b, c, B) > 0)
{
System.out.println("No solution");
}
// Else find the root upto 4
// decimal places
else
{
System.out.format("%.4f", findRoot(
a, b, c, A, B));
}
}
// Driver Code
public static void main (String[] args)
{
// Given range
double a = 2, b = -3, c = -2,
A = 0, B = 3;
// Function call
solve(a, b, c, A, B);
}
}
// This code is contributed by offbeatPython3
# Python3 program for the above approach
import math
eps = 1e-6
# Given function
def func(a ,b , c , x):
return a * x * x + b * x + c
# Function to find the root of the
# given non-decreasing Function
def findRoot( a, b, c, low, high):
x = -1
# To get the minimum possible
# answer for the root
while abs(high - low) > eps:
# Find mid
x = (low + high) / 2
# Search in [low, x]
if (func(a, b, c, low) *
func(a, b, c, x) <= 0):
high = x
# Search in [x, high]
else:
low = x
# Return the required answer
return x
# Function to find the roots of the
# given equation within range [a, b]
def solve(a, b, c, A, B):
# If root doesn't exists
if (func(a, b, c, A) *
func(a, b, c, B) > 0):
print("No solution")
# Else find the root upto 4
# decimal places
else:
print("{:.4f}".format(findRoot(
a, b, c, A, B)))
# Driver code
if __name__ == '__main__':
# Given range
a = 2
b = -3
c = -2
A = 0
B = 3
# Function call
solve(a, b, c, A, B)
# This code is contributed by jana_sayantanC#
// C# program for the above approach
using System;
class GFG{
static readonly double eps = 1e-6;
// Given function
static double func(double a, double b,
double c, double x)
{
return a * x * x + b * x + c;
}
// Function to find the root of the
// given non-decreasing Function
static double findRoot(double a, double b,
double c, double low,
double high)
{
double x = -1;
// To get the minimum possible
// answer for the root
while (Math.Abs(high - low) > eps)
{
// Find mid
x = (low + high) / 2;
// Search in [low, x]
if (func(a, b, c, low) *
func(a, b, c, x) <= 0)
{
high = x;
}
// Search in [x, high]
else
{
low = x;
}
}
// Return the required answer
return x;
}
// Function to find the roots of the
// given equation within range [a, b]
static void solve(double a, double b, double c,
double A, double B)
{
// If root doesn't exists
if (func(a, b, c, A) * func(a, b, c, B) > 0)
{
Console.WriteLine("No solution");
}
// Else find the root upto 4
// decimal places
else
{
Console.Write("{0:F4}", findRoot(
a, b, c, A, B));
}
}
// Driver Code
public static void Main(String[] args)
{
// Given range
double a = 2, b = -3, c = -2,
A = 0, B = 3;
// Function call
solve(a, b, c, A, B);
}
}
// This code is contributed by 29AjayKumar输出：
2.0000
时间复杂度： O(log(B – A))
辅助空间： O(1)