Encuentre el entero positivo mínimo x tal que a(x^2) + b(x) + c >= k

Dados cuatro enteros a , b , c y k . La tarea es encontrar el valor mínimo positivo de x tal que ax ² + bx + c ≥ k .
Ejemplos: 
 

Entrada: a = 3, b = 4, c = 5, k = 6 
Salida: 1 
Para x = 0, a * 0 + b * 0 + c = 5 < 6 
Para x = 1, a * 1 + b * 1 + c = 3 + 4 + 5 = 12 > 6
Entrada: a = 2, b = 7, c = 6, k = 3 
Salida: 0 
 

Enfoque: La idea es utilizar la búsqueda binaria . El límite inferior para nuestra búsqueda será 0 ya que x tiene que ser el número entero positivo mínimo.
A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the minimum positive
// integer satisfying the given equation
int MinimumX(int a, int b, int c, int k)
{
    int x = INT_MAX;
 
    if (k <= c)
        return 0;
 
    int h = k - c;
    int l = 0;
 
    // Binary search to find the value of x
    while (l <= h) {
        int m = (l + h) / 2;
        if ((a * m * m) + (b * m) > (k - c)) {
            x = min(x, m);
            h = m - 1;
        }
        else if ((a * m * m) + (b * m) < (k - c))
            l = m + 1;
        else
            return m;
    }
 
    // Return the answer
    return x;
}
 
// Driver code
int main()
{
    int a = 3, b = 2, c = 4, k = 15;
    cout << MinimumX(a, b, c, k);
 
    return 0;
}

// Java implementation of the approach
class GFG
{
     
// Function to return the minimum positive
// integer satisfying the given equation
static int MinimumX(int a, int b, int c, int k)
{
    int x = Integer.MAX_VALUE;
 
    if (k <= c)
        return 0;
 
    int h = k - c;
    int l = 0;
 
    // Binary search to find the value of x
    while (l <= h)
    {
        int m = (l + h) / 2;
        if ((a * m * m) + (b * m) > (k - c))
        {
            x = Math.min(x, m);
            h = m - 1;
        }
        else if ((a * m * m) + (b * m) < (k - c))
            l = m + 1;
        else
            return m;
    }
 
    // Return the answer
    return x;
}
 
// Driver code
public static void main(String[] args)
{
    int a = 3, b = 2, c = 4, k = 15;
    System.out.println(MinimumX(a, b, c, k));
}
}
 
// This code is contributed by Code_Mech.

# Python3 implementation of the approach
 
# Function to return the minimum positive
# integer satisfying the given equation
def MinimumX(a, b, c, k):
 
    x = 10**9
 
    if (k <= c):
        return 0
 
    h = k - c
    l = 0
 
    # Binary search to find the value of x
    while (l <= h):
        m = (l + h) // 2
        if ((a * m * m) + (b * m) > (k - c)):
            x = min(x, m)
            h = m - 1
 
        elif ((a * m * m) + (b * m) < (k - c)):
            l = m + 1
        else:
            return m
 
    # Return the answer
    return x
 
# Driver code
a, b, c, k = 3, 2, 4, 15
print(MinimumX(a, b, c, k))
 
# This code is contributed by mohit kumar

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the minimum positive
// integer satisfying the given equation
static int MinimumX(int a, int b, int c, int k)
{
    int x = int.MaxValue;
 
    if (k <= c)
        return 0;
 
    int h = k - c;
    int l = 0;
 
    // Binary search to find the value of x
    while (l <= h)
    {
        int m = (l + h) / 2;
        if ((a * m * m) + (b * m) > (k - c))
        {
            x = Math.Min(x, m);
            h = m - 1;
        }
        else if ((a * m * m) + (b * m) < (k - c))
            l = m + 1;
        else
            return m;
    }
 
    // Return the answer
    return x;
}
 
// Driver code
public static void Main()
{
    int a = 3, b = 2, c = 4, k = 15;
    Console.Write(MinimumX(a, b, c, k));
}
}
 
// This code is contributed by Akanksha Rai

<?php
// PHP implementation of the approach
 
// Function to return the minimum positive
// integer satisfying the given equation
function MinimumX($a, $b, $c, $k)
{
    $x = PHP_INT_MAX;
 
    if ($k <= $c)
        return 0;
 
    $h = $k - $c;
    $l = 0;
 
    // Binary search to find the value of x
    while ($l <= $h)
    {
        $m = floor(($l + $h) / 2);
        if (($a * $m * $m) +
            ($b * $m) > ($k - $c))
        {
            $x = min($x, $m);
            $h = $m - 1;
        }
        else if (($a * $m * $m) +
                 ($b * $m) < ($k - $c))
            $l = $m + 1;
        else
            return $m;
    }
 
    // Return the answer
    return $x;
}
 
// Driver code
$a = 3; $b = 2; $c = 4; $k = 15;
 
echo MinimumX($a, $b, $c, $k);
 
// This code is contributed by Ryuga
?>

<script>
// Javascript implementation of the approach
 
// Function to return the minimum positive
// integer satisfying the given equation
function MinimumX(a,b,c,k)
{
    let x = Number.MAX_VALUE;
   
    if (k <= c)
        return 0;
   
    let h = k - c;
    let l = 0;
   
    // Binary search to find the value of x
    while (l <= h)
    {
        let m = Math.floor((l + h) / 2);
        if ((a * m * m) + (b * m) > (k - c))
        {
            x = Math.min(x, m);
            h = m - 1;
        }
        else if ((a * m * m) + (b * m) < (k - c))
            l = m + 1;
        else
            return m;
    }
   
    // Return the answer
    return x;
}
 
// Driver code
let a = 3, b = 2, c = 4, k = 15;
document.write(MinimumX(a, b, c, k));
 
// This code is contributed by patel2127
</script>

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