找出给定立方方程的积分根

给定5个整数，分别表示A，B，C，D和E ，它们表示三次方程$f(x) = A*x^{3} + B*x^{2} + C*x + D = E$ ，任务是找到该方程的积分解。如果不存在任何积分解，则打印“ NA” 。

例子：

Input: A = 1, B = 0, C = 0, D = 0, E = 27
Output: 3

Input: A = 1, B = 0, C = 0, D = 0, E = 16
Output: NA

方法：想法是使用二进制搜索。步骤如下：

1. 将开始变量和结束变量分别初始化为0和10 ⁵ 。
2. 查找开始和结束检查的中间(例如mid )值是否满足给定方程式。
3. 如果当前中值满足给定方程式，则打印中值。
4. 否则，如果f(x)的值小于E，则更新开始为mid + 1 。
5. 其他更新在1月中旬结束。
6. 如果我们找不到上述方程式的任何积分解，请打印“ -1” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
  
// Function to find the value at x of
// the given equation
long long int check(int A, int B, int C,
                    int D, long long int x)
{
  
    long long int ans;
  
    // Find the value equation at x
    ans = (A * x * x * x
           + B * x * x
           + C * x
           + D);
  
    // Return the value of ans
    return ans;
}
  
// Function to find the integral
// solution of the given equation
void findSolution(int A, int B, int C,
                  int D, int E)
{
  
    // Initialise start and end
    int start = 0, end = 100000;
  
    long long int mid, ans;
  
    // Implement Binary Search
    while (start <= end) {
  
        // Find mid
        mid = start + (end - start) / 2;
  
        // Find the value of f(x) using
        // current mid
        ans = check(A, B, C, D, mid);
  
        // Check if current mid satisfy
        // the equation
        if (ans == E) {
  
            // Print mid and return
            cout << mid << endl;
            return;
        }
  
        if (ans < E)
            start = mid + 1;
        else
            end = mid - 1;
    }
  
    // Print "NA" if not found
    // any integral solution
    cout << "NA";
}
  
// Driver Code
int main()
{
    int A = 1, B = 0, C = 0;
    int D = 0, E = 27;
  
    // Function Call
    findSolution(A, B, C, D, E);
}

Java

// Java program for the above approach
import java.util.*;
class GFG{
  
// Function to find the value at x of
// the given equation
static long check(int A, int B, int C,
                         int D, long x)
{
    long ans;
  
    // Find the value equation at x
    ans = (A * x * x * x + 
           B * x * x + C * x + D);
  
    // Return the value of ans
    return ans;
}
  
// Function to find the integral
// solution of the given equation
static void findSolution(int A, int B, int C,
                         int D, int E)
{
  
    // Initialise start and end
    long start = 0, end = 100000;
  
    long mid, ans;
  
    // Implement Binary Search
    while (start <= end) 
    {
  
        // Find mid
        mid = start + (end - start) / 2;
  
        // Find the value of f(x) using
        // current mid
        ans = check(A, B, C, D, mid);
  
        // Check if current mid satisfy
        // the equation
        if (ans == E)
        {
  
            // Print mid and return
            System.out.println(mid);
            return;
        }
  
        if (ans < E)
            start = mid + 1;
        else
            end = mid - 1;
    }
  
    // Print "NA" if not found
    // any integral solution
    System.out.println("NA");
}
  
// Driver Code
public static void main(String args[])
{
    int A = 1, B = 0, C = 0;
    int D = 0, E = 27;
  
    // Function Call
    findSolution(A, B, C, D, E);
}
}
  
// This code is contributed by Code_Mech

Python3

# Python3 program for the above approach
  
# Function to find the value at x of
# the given equation
def check(A, B, C, D, x) :
  
    ans = 0;
  
    # Find the value equation at x
    ans = (A * x * x * x + 
           B * x * x + C * x + D);
  
    # Return the value of ans
    return ans;
  
# Function to find the integral
# solution of the given equation
def findSolution(A, B, C, D, E) :
  
    # Initialise start and end
    start = 0; end = 100000;
  
    mid = 0;
    ans = 0;
  
    # Implement Binary Search
    while (start <= end) :
  
        # Find mid
        mid = start + (end - start) // 2;
  
        # Find the value of f(x) using
        # current mid
        ans = check(A, B, C, D, mid);
  
        # Check if current mid satisfy
        # the equation
        if (ans == E) :
  
            # Print mid and return
            print(mid);
            return;
  
        if (ans < E) :
            start = mid + 1;
        else :
            end = mid - 1;
  
    # Print "NA" if not found
    # any integral solution
    print("NA");
  
# Driver Code
if __name__ == "__main__" :
  
    A = 1; B = 0; C = 0;
    D = 0; E = 27;
  
    # Function Call
    findSolution(A, B, C, D, E);
      
# This code is contributed by AnkitRai01

C#

// C# program for the above approach
using System;
class GFG{
  
// Function to find the value at x of
// the given equation
static long check(int A, int B, int C,
                         int D, long x)
{
    long ans;
  
    // Find the value equation at x
    ans = (A * x * x * x + 
           B * x * x + C * x + D);
  
    // Return the value of ans
    return ans;
}
  
// Function to find the integral
// solution of the given equation
static void findSolution(int A, int B, int C,
                         int D, int E)
{
  
    // Initialise start and end
    long start = 0, end = 100000;
  
    long mid, ans;
  
    // Implement Binary Search
    while (start <= end) 
    {
  
        // Find mid
        mid = start + (end - start) / 2;
  
        // Find the value of f(x) using
        // current mid
        ans = check(A, B, C, D, mid);
  
        // Check if current mid satisfy
        // the equation
        if (ans == E)
        {
  
            // Print mid and return
            Console.WriteLine(mid);
            return;
        }
  
        if (ans < E)
            start = mid + 1;
        else
            end = mid - 1;
    }
  
    // Print "NA" if not found
    // any integral solution
    Console.Write("NA");
}
  
// Driver Code
public static void Main()
{
    int A = 1, B = 0, C = 0;
    int D = 0, E = 27;
  
    // Function Call
    findSolution(A, B, C, D, E);
}
}
  
// This code is contributed by Code_Mech

输出：

3

时间复杂度： O(log N)