Given an integer N representing a non-linear equation of the form 2X + 5Y = N, the task is to find an integral pair (X, Y) that satisfies the given equation. If multiple solutions exist, then print any one of them. Otherwise, print -1.
Examples:
Input: N = 29
Output: X = 2, Y = 2
Explanation:
Since, 22 + 52 = 29
Therefore, X = 2 and Y = 2 satisfy the given equation.Input: N = 81
Output: -1
Approach: Follow the steps below to solve the problem:
- Initialize a variable, Say xMax to store the maximum possible value of X.
- Update xMax to log2N.
- Initialize a variable, say yMax to store the maximum possible of Y.
- Update yMax to log5N
- Iterate over all possible values of X and Y and for each value of X and Y and for each pair, check if it satisfies the given equation or not. If found to be true, then print the corresponding value of X and Y.
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to find the value // of power(X, N) long long power(long long x, long long N) { // Stores the value // of (X ^ N) long long res = 1; // Calculate the value of // power(x, N) while (N > 0) { // If N is odd if (N & 1) { // Update res res = (res * x) ; } // Update x x = (x * x) ; // Update N N = N >> 1; } return res; } // Function to find the value of // X and Y that satisfy the condition void findValX_Y(long long N) { // Base Case if (N <= 1) { cout<<-1<<endl; return; } // Stores maximum possible // of X. int xMax; // Update xMax xMax = log2(N); // Stores maximum possible // of Y. int yMax; // Update yMax yMax = (log2(N) / log2(5.0)); // Iterate over all possible // values of X for (long long i = 1; i <= xMax; i++) { // Iterate over all possible // values of Y for (long long j=1; j <= yMax; j++) { // Stores value of 2^i long long a = power(2, i); // Stores value of 5^j long long b = power(5, j); // If the pair (i, j) // satisfy the equation if (a + b == N) { cout<<i<<" " <<j<<endl; return; } } } // If no solution exists cout<<-1<<endl; } // Driver Code int main() { long long N = 129; findValX_Y(N); return 0; } |
Java
// Java program to implement // the above approach import java.util.*;class GFG{ // Function to find the value // of power(X, N) static int power(int x, int N) { // Stores the value // of (X ^ N) int res = 1; // Calculate the value of // power(x, N) while (N > 0) { // If N is odd if ((N & 1) != 0) { // Update res res = (res * x); } // Update x x = (x * x); // Update N N = N >> 1; } return res; } // Function to find the value of // X and Y that satisfy the condition static void findValX_Y(int N) { // Base Case if (N <= 1) { System.out.println(-1); return; } // Stores maximum possible // of X. int xMax; // Update xMax xMax = (int)Math.log(N); // Stores maximum possible // of Y. int yMax; // Update yMax yMax = (int)(Math.log(N) / Math.log(5.0)); // Iterate over all possible // values of X for(int i = 1; i <= xMax; i++) { // Iterate over all possible // values of Y for(int j = 1; j <= yMax; j++) { // Stores value of 2^i int a = power(2, i); // Stores value of 5^j int b = power(5, j); // If the pair (i, j) // satisfy the equation if (a + b == N) { System.out.print(i + " " + j); return; } } } // If no solution exists System.out.println("-1");} // Driver Code public static void main(String args[]) { int N = 129; findValX_Y(N); } } // This code is contributed by bgangwar59 |
Python3
# Python3 program to implement# the above approachfrom math import log2 # Function to find the value# of power(X, N)def power(x, N): # Stores the value # of (X ^ N) res = 1 # Calculate the value of # power(x, N) while (N > 0): # If N is odd if (N & 1): # Update res res = (res * x) # Update x x = (x * x) # Update N N = N >> 1 return res # Function to find the value of# X and Y that satisfy the conditiondef findValX_Y(N): # Base Case if (N <= 1): print(-1) return # Stores maximum possible # of X xMax = 0 # Update xMax xMax = int(log2(N)) # Stores maximum possible # of Y yMax = 0 # Update yMax yMax = int(log2(N) / log2(5.0)) # Iterate over all possible # values of X for i in range(1, xMax + 1): # Iterate over all possible # values of Y for j in range(1, yMax + 1): # Stores value of 2^i a = power(2, i) # Stores value of 5^j b = power(5, j) # If the pair (i, j) # satisfy the equation if (a + b == N): print(i, j) return # If no solution exists print(-1) # Driver Codeif __name__ == '__main__': N = 129 findValX_Y(N)# This code is contributed by mohit kumar 29 |
C#
// C# program to implement // the above approach using System;class GFG{ // Function to find the// value of power(X, N) static int power(int x, int N) { // Stores the value // of (X ^ N) int res = 1; // Calculate the value // of power(x, N) while (N > 0) { // If N is odd if ((N & 1) != 0) { // Update res res = (res * x); } // Update x x = (x * x); // Update N N = N >> 1; } return res; } // Function to find the // value of X and Y that // satisfy the condition static void findValX_Y(int N) { // Base Case if (N <= 1) { Console.WriteLine(-1); return; } // Stores maximum // possible of X. int xMax; // Update xMax xMax = (int)Math.Log(N); // Stores maximum possible // of Y. int yMax; // Update yMax yMax = (int)(Math.Log(N) / Math.Log(5.0)); // Iterate over all possible // values of X for(int i = 1; i <= xMax; i++) { // Iterate over all possible // values of Y for(int j = 1; j <= yMax; j++) { // Stores value of 2^i int a = power(2, i); // Stores value of 5^j int b = power(5, j); // If the pair (i, j) // satisfy the equation if (a + b == N) { Console.Write(i + " " + j); return; } } } // If no solution exists Console.WriteLine("-1");} // Driver Code public static void Main() { int N = 129; findValX_Y(N); } } // This code is contributed by surendra_gangwar |
Javascript
<script>// JavaScript program to implement the above approach// Function to find the value // of power(X, N) function power(x, N) { // Stores the value // of (X ^ N) let res = 1; // Calculate the value of // power(x, N) while (N > 0) { // If N is odd if ((N & 1) != 0) { // Update res res = (res * x); } // Update x x = (x * x); // Update N N = N >> 1; } return res; } // Function to find the value of // X and Y that satisfy the condition function findValX_Y(N) { // Base Case if (N <= 1) { document.write(-1); return; } // Stores maximum possible // of X. let xMax; // Update xMax xMax = Math.log(N); // Stores maximum possible // of Y. let yMax; // Update yMax yMax = Math.log(N) / Math.log(5.0); // Iterate over all possible // values of X for(let i = 1; i <= xMax; i++) { // Iterate over all possible // values of Y for(let j = 1; j <= yMax; j++) { // Stores value of 2^i let a = power(2, i); // Stores value of 5^j let b = power(5, j); // If the pair (i, j) // satisfy the equation if (a + b == N) { document.write(i + " " + j); return; } } } // If no solution exists document.write("-1");} // Driver Code let N = 129; findValX_Y(N); // This code is contributed by susmitakundugoaldanga.</script> |
Output:
2 3
Time Complexity: O(log2N * log5N)
Auxiliary Space: O(1)
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