Given three integers N, A, and B, the task is to find a pair of positive integers (a, b) such that Aa + Bb = N.If no such pair exists, print -1.
Examples:
Input: N = 106, A = 3, B = 5
Output: 4 2
Explanation: Pair (4, 2) satisfies the answer i.e., 34+52 is equal to 106Input: N = 60467200, A = 6, B = 4
Output: 10 5
Explanation: Pair (10, 5) satisfies the answer i.e., 610+45 is equal to 60467200
Approach: The idea is to calculate logAN and logBN and check for every pair (i, j) (0 ≤ i ≤ logAN and 0 ≤ j ≤ logBN ), whether Ai + Bj is equal to N or not. Follow the steps below to solve the problem:
- First, find the minimum power of A and B that is greater than N and store it in the variables X and Y respectively.
- Check each pair (i, j) such that 0≤i≤X and 0≤j≤Y, and if Ai + Bj is equal to N, print the pair (i, j).
- Print -1, if no such pair is found.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate the minimum// power of A and B greater than Nint power(long long int A, long long int N){ // Stores the power of A which // is greater than N int count = 0; if (A == 1) return 0; while (N) { // Increment count by 1 count++; // Divide N by A N /= A; } return count;}// Function to find a pair (a, b)// such that A^a + B^b = Nvoid Pairs(long long int N, long long int A, long long int B){ int powerA, powerB; // Calculate the minimum power // of A greater than N powerA = power(A, N); // Calculate the minimum power // of B greater than N powerB = power(B, N); // Make copy of A and B long long int intialB = B, intialA = A; // Traverse for every pair (i, j) A = 1; for (int i = 0; i <= powerA; i++) { B = 1; for (int j = 0; j <= powerB; j++) { // Check if B^j + A^i = N // To overcome the overflow problem // use B=N-A rather than B+A=N if (B == N - A) { cout << i << " " << j << endl; return; } // Increment power B by 1 B *= intialB; } // Increment power A by 1 A *= intialA; } // Finally print -1 if no pair // is found cout << -1 << endl; return;}// Driver Codeint main(){ // Given A, B and N long long int N = 106, A = 3, B = 5; // Function Call Pairs(N, A, B); return 0;} |
Java
// Java program for the above approach import java.io.*;class GFG{ // Function to calculate the minimum// power of A and B greater than Nstatic int power(int A, int N){ // Stores the power of A which // is greater than N int count = 0; if (A == 1) return 0; while (N > 0) { // Increment count by 1 count++; // Divide N by A N /= A; } return count;}// Function to find a pair (a, b)// such that A^a + B^b = Nstatic void Pairs(int N, int A, int B){ int powerA, powerB; // Calculate the minimum power // of A greater than N powerA = power(A, N); // Calculate the minimum power // of B greater than N powerB = power(B, N); // Make copy of A and B int intialB = B, intialA = A; // Traverse for every pair (i, j) A = 1; for(int i = 0; i <= powerA; i++) { B = 1; for(int j = 0; j <= powerB; j++) { // Check if B^j + A^i = N // To overcome the overflow problem // use B=N-A rather than B+A=N if (B == N - A) { System.out.println(i + " " + j); return; } // Increment power B by 1 B *= intialB; } // Increment power A by 1 A *= intialA; } // Finally print -1 if no pair // is found System.out.println("-1"); return;}// Driver Codepublic static void main(String args[]){ // Given A, B and N int N = 106, A = 3, B = 5; // Function Call Pairs(N, A, B);}}// This code is contributed by 18bhupenderyadav18 |
Python3
# Python program for the above approach# Function to calculate the minimum# power of A and B greater than Ndef power(A, N): # Stores the power of A which # is greater than N count = 0; if (A == 1): return 0; while (N > 0): # Increment count by 1 count += 1; # Divide N by A N //= A; return int(count);# Function to find a pair (a, b)# such that A^a + B^b = Ndef Pairs(N, A, B): powerA, powerB = 0, 0; # Calculate the minimum power # of A greater than N powerA = power(A, N); # Calculate the minimum power # of B greater than N powerB = power(B, N); # Make copy of A and B intialB = B; intialA = A; # Traverse for every pair (i, j) A = 1; for i in range(powerA + 1): B = 1; for j in range(powerB + 1): # Check if B^j + A^i = N # To overcome the overflow problem # use B=N-A rather than B+A=N if (B == N - A): print(i , " " , j); return; # Increment power B by 1 B *= intialB; # Increment power A by 1 A *= intialA; # Finally pr-1 if no pair # is found print("-1"); return;# Driver Codeif __name__ == '__main__': # Given A, B and N N = 106; A = 3; B = 5; # Function Call Pairs(N, A, B);# This code is contributed by 29AjayKumar |
C#
// C# program for the above approach using System;class GFG{ // Function to calculate the minimum// power of A and B greater than Nstatic int power(int A, int N){ // Stores the power of A which // is greater than N int count = 0; if (A == 1) return 0; while (N > 0) { // Increment count by 1 count++; // Divide N by A N /= A; } return count;}// Function to find a pair (a, b)// such that A^a + B^b = Nstatic void Pairs(int N, int A, int B){ int powerA, powerB; // Calculate the minimum power // of A greater than N powerA = power(A, N); // Calculate the minimum power // of B greater than N powerB = power(B, N); // Make copy of A and B int intialB = B, intialA = A; // Traverse for every pair (i, j) A = 1; for(int i = 0; i <= powerA; i++) { B = 1; for(int j = 0; j <= powerB; j++) { // Check if B^j + A^i = N // To overcome the overflow problem // use B=N-A rather than B+A=N if (B == N - A) { Console.WriteLine(i + " " + j); return; } // Increment power B by 1 B *= intialB; } // Increment power A by 1 A *= intialA; } // Finally print -1 if no pair // is found Console.WriteLine("-1"); return;}// Driver Codepublic static void Main(String []args){ // Given A, B and N int N = 106, A = 3, B = 5; // Function Call Pairs(N, A, B);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for the above approach// Function to calculate the minimum// power of A and B greater than Nfunction power(A, N){ // Stores the power of A which // is greater than N let count = 0; if (A == 1) return 0; while (N > 0) { // Increment count by 1 count++; // Divide N by A N /= A; } return count;} // Function to find a pair (a, b)// such that A^a + B^b = Nfunction Pairs(N, A, B){ let powerA, powerB; // Calculate the minimum power // of A greater than N powerA = power(A, N); // Calculate the minimum power // of B greater than N powerB = power(B, N); // Make copy of A and B let letialB = B, letialA = A; // Traverse for every pair (i, j) A = 1; for(let i = 0; i <= powerA; i++) { B = 1; for(let j = 0; j <= powerB; j++) { // Check if B^j + A^i = N // To overcome the overflow problem // use B=N-A rather than B+A=N if (B == N - A) { document.write(i + " " + j); return; } // Increment power B by 1 B *= letialB; } // Increment power A by 1 A *= letialA; } // Finally print -1 if no pair // is found document.write("-1"); return;} // driver function // Given A, B and N let N = 106, A = 3, B = 5; // Function Call Pairs(N, A, B); // This code is contributed by splevel62.</script> |
Output:
4 2
Time Complexity: O((logAN)*(logBN))
Auxiliary Space: O(1)
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