Given three numbers b, x, n. The task is to find the values of ‘a’ in equation (a+b) <= n such that a+b is divisible by x. If no such values are possible then print -1.
Examples:
Input: b = 10, x = 6, n = 40 Output: 2 8 14 20 26 Input: b = 10, x = 1, n = 10 Output: -1
Approach: One can find the least possible value (b/x + 1)*x – b. Then we increase answer by x until it is not greater than n. Here (b/x + 1)*x is the least possible value which is divisible by x.
Below is the implementation of above approach:
C++
// CPP program to Find values of a, in equation// (a+b)<=n and a+b is divisible by x.#include <bits/stdc++.h>using namespace std;// function to Find values of a, in equation// (a+b)<=n and a+b is divisible by x.void PossibleValues(int b, int x, int n){ // least possible which is divisible by x int leastdivisible = (b / x + 1) * x; int flag = 1; // run a loop to get required answer while (leastdivisible <= n) { if (leastdivisible - b >= 1) { cout << leastdivisible - b << " "; // increase value by x leastdivisible += x; // answer is possible flag = 0; } else break; } if (flag) cout << -1;}// Driver codeint main(){ int b = 10, x = 6, n = 40; // function call PossibleValues(b, x, n); return 0;} |
C
// C program to Find values of a, in equation// (a+b)<=n and a+b is divisible by x.#include <stdio.h>// function to Find values of a, in equation// (a+b)<=n and a+b is divisible by x.void PossibleValues(int b, int x, int n){ // least possible which is divisible by x int leastdivisible = (b / x + 1) * x; int flag = 1; // run a loop to get required answer while (leastdivisible <= n) { if (leastdivisible - b >= 1) { printf("%d ",leastdivisible - b); // increase value by x leastdivisible += x; // answer is possible flag = 0; } else break; } if (flag) printf("%d",-1);}// Driver codeint main(){ int b = 10, x = 6, n = 40; // function call PossibleValues(b, x, n); return 0;}// This code is contributed by kothavvsaakash. |
Java
// Java program to Find values of a, in equation// (a+b)<=n and a+b is divisible by x.import java.io.*;class GFG { // function to Find values of a, in equation// (a+b)<=n and a+b is divisible by x.static void PossibleValues(int b, int x, int n){ // least possible which is divisible by x int leastdivisible = (b / x + 1) * x; int flag = 1; // run a loop to get required answer while (leastdivisible <= n) { if (leastdivisible - b >= 1) { System.out.print( leastdivisible - b + " "); // increase value by x leastdivisible += x; // answer is possible flag = 0; } else break; } if (flag>0) System.out.println(-1);}// Driver code public static void main (String[] args) { int b = 10, x = 6, n = 40; // function call PossibleValues(b, x, n); }}// This code is contributed// by shs |
Python3
# Python3 program to Find values of a, in equation # (a+b)<=n and a+b is divisible by x. # function to Find values of a, in equation # (a+b)<=n and a+b is divisible by x. def PossibleValues(b, x, n) : # least possible which is divisible by x leastdivisible = int(b / x + 1) * x flag = 1 # run a loop to get required answer while (leastdivisible <= n) : if (leastdivisible - b >= 1) : print(leastdivisible - b ,end= " ") # increase value by x leastdivisible += x # answer is possible flag = 0 else : break if (flag != 0) : print(-1) # Driver code if __name__=='__main__': b = 10 x = 6 n = 40# function call PossibleValues(b, x, n) # This code is contributed by# Smitha Dinesh Semwal |
C#
// C# program to Find values of a,// in equation (a+b)<=n and a+b // is divisible by x. using System;class GFG { // function to Find values// of a, in equation (a+b)<=n // and a+b is divisible by x. static void PossibleValues(int b, int x, int n) { // least possible which // is divisible by x int leastdivisible = (b / x + 1) * x; int flag = 1; // run a loop to get required answer while (leastdivisible <= n) { if (leastdivisible - b >= 1) { Console.Write( leastdivisible - b + " "); // increase value by x leastdivisible += x; // answer is possible flag = 0; } else break; } if (flag > 0) Console.WriteLine(-1); } // Driver code public static void Main () { int b = 10, x = 6, n = 40; // function call PossibleValues(b, x, n); } } // This code is contributed by Shubadeep |
PHP
<?php// PHP program to Find values of a, // in equation (a+b)<=n and a+b is // divisible by x.// function to Find values of a, // in equation (a+b)<=n and a+b// is divisible by x.function PossibleValues($b, $x, $n){ // least possible which is // divisible by x $leastdivisible = (intval($b / $x) + 1) * $x; $flag = 1; // run a loop to get required answer while ($leastdivisible <= $n) { if ($leastdivisible - $b >= 1) { echo $leastdivisible - $b . " "; // increase value by x $leastdivisible += $x; // answer is possible $flag = 0; } else break; } if ($flag) echo "-1";}// Driver code$b = 10;$x = 6;$n = 40;// function callPossibleValues($b, $x, $n);// This code is contributed// by ChitraNayal?> |
Javascript
<script>// Javascript program to Find values of a, in equation// (a+b)<=n and a+b is divisible by x. // function to Find values of a, in equation// (a+b)<=n and a+b is divisible by x. function PossibleValues(b,x,n) { // least possible which is divisible by x let leastdivisible = (Math.floor(b / x) + 1) * x; let flag = 1; // run a loop to get required answer while (leastdivisible <= n) { if (leastdivisible - b >= 1) { document.write( leastdivisible - b + " "); // increase value by x leastdivisible += x; // answer is possible flag = 0; } else break; } if (flag>0) document.write(-1+"<br>"); } // Driver code let b = 10, x = 6, n = 40; // function call PossibleValues(b, x, n); // This code is contributed by rag2127</script> |
Output:
2 8 14 20 26
Auxiliary Space: O(1)
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