Given 5 integers say A, B, C, D, and E which represents the cubic equation 
, the task is to find the integral solution for this equation. If there doesn’t exist any integral solution then print “NA”.
Examples:
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
Approach: The idea is to use binary search. Below are the steps:
- Initialise the start and end variable as 0 & 105 respectively.
- Find the middle(say mid) value of start and end check if it satisfy the given equation or not.
- If current mid satisfy the given equation then print the mid value.
- Else if the value of f(x) is less than E then update start as mid + 1.
- Else Update end as mid – 1.
- If we can’t find any integral solution for the above equation then print “-1”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the value at x of// the given equationlong 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 equationvoid 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 Codeint 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 approachimport java.util.*;class GFG{// Function to find the value at x of// the given equationstatic 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 equationstatic 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 Codepublic 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 equationdef 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 equationdef 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 Codeif __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 approachusing System;class GFG{// Function to find the value at x of// the given equationstatic 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 equationstatic 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 Codepublic 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 |
Javascript
<script>// Javascript program for the above approach// Function to find the value at x of// the given equationfunction check(A, B, C, D, x){ var 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 equationfunction findSolution(A, B, C, D, E){ // Initialise start and end var start = 0, end = 100000; var mid, ans; // Implement Binary Search while (start <= end) { // Find mid mid = parseInt(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 document.write(mid); return; } if (ans < E) start = mid + 1; else end = mid - 1; } // Print "NA" if not found // any integral solution document.write("NA");}// Driver Codevar A = 1, B = 0, C = 0;var D = 0, E = 27;// Function CallfindSolution(A, B, C, D, E);// This code is contributed by Ankita saini</script> |
Output:
3
Time Complexity: O(log N)
Auxiliary Space: O(1) as constant space for variables is being used
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