Dados tres números b, x, n. La tarea es encontrar los valores de ‘a’ en la ecuación (a+b) <= n tal que a+b sea divisible por x. Si tales valores no son posibles, imprima -1.
Ejemplos:
Input: b = 10, x = 6, n = 40 Output: 2 8 14 20 26 Input: b = 10, x = 1, n = 10 Output: -1
Enfoque: Uno puede encontrar el menor valor posible (b/x + 1)*x – b. Luego aumentamos la respuesta en x hasta que no sea mayor que n. Aquí (b/x + 1)*x es el valor mínimo posible que es divisible por x.
A continuación se muestra la implementación del enfoque anterior:
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 code
int 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 code
int 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;
$n = 40;
// function call
PossibleValues($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>
Producción:
2 8 14 20 26
Publicación traducida automáticamente
Artículo escrito por pawan_asipu y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA