Dados a y b, encuentre el valor máximo de a/b sin usar la función de techo.
Ejemplos:
Input : a = 5, b = 4 Output : 2 Explanation: a/b = ceil(5/4) = 2 Input : a = 10, b = 2 Output : 5 Explanation: a/b = ceil(10/2) = 5
El problema se puede resolver usando la función de techo, pero la función de techo no funciona cuando se pasan números enteros como parámetros. Por lo tanto, existen los siguientes 2 enfoques a continuación para encontrar el valor máximo.
Enfoque 1:
ceilVal = (a / b) + ((a % b) != 0)
a/b devuelve el valor de la división de enteros, y ((a % b) != 0) es una condición de verificación que devuelve 1 si nos queda algún resto después de la división de a/b, de lo contrario devuelve 0. El valor de la división de enteros se suma con el valor de verificación para obtener el valor máximo.
A continuación se muestra la ilustración del enfoque anterior:
C++
// C++ program to find ceil(a/b)
// without using ceil() function
#include <cmath>
#include <iostream>
using namespace std;
// Driver function
int main()
{
// taking input 1
int a = 4;
int b = 3;
int val = (a / b) + ((a % b) != 0);
cout << "The ceiling value of 4/3 is "
<< val << endl;
// example of perfect division
// taking input 2
a = 6;
b = 3;
val = (a / b) + ((a % b) != 0);
cout << "The ceiling value of 6/3 is "
<< val << endl;
return 0;
}
Java
// Java program to find
// ceil(a/b) without
// using ceil() function
import java.io.*;
class GFG
{
// Driver Code
public static void main(String args[])
{
// taking input 1
int a = 4;
int b = 3, val = 0;
if((a % b) != 0)
val = (a / b) +
(a % b);
else
val = (a / b);
System.out.println("The ceiling " +
"value of 4/3 is " +
val);
// example of perfect
// division taking input 2
a = 6;
b = 3;
if((a % b) != 0)
val = (a / b) + (a % b);
else
val = (a / b);
System.out.println("The ceiling " +
"value of 6/3 is " +
val);
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
Python3
# Python3 program to find ceil(a/b)
# without using ceil() function
import math
# Driver Code
# taking input 1
a = 4;
b = 3;
val = (a / b) + ((a % b) != 0);
print("The ceiling value of 4/3 is",
math.floor(val));
# example of perfect division
# taking input 2
a = 6;
b = 3;
val = int((a / b) + ((a % b) != 0));
print("The ceiling value of 6/3 is", val);
# This code is contributed by mits
C#
// C# program to find ceil(a/b)
// without using ceil() function
using System;
class GFG
{
// Driver Code
static void Main()
{
// taking input 1
int a = 4;
int b = 3, val = 0;
if((a % b) != 0)
val = (a / b) + (a % b);
else
val = (a / b);
Console.WriteLine("The ceiling " +
"value of 4/3 is " +
val);
// example of perfect
// division taking input 2
a = 6;
b = 3;
if((a % b) != 0)
val = (a / b) + (a % b);
else
val = (a / b);
Console.WriteLine("The ceiling " +
"value of 6/3 is " +
val);
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
PHP
<?php // PHP program to find ceil(a/b) // without using ceil() function // Driver function // taking input 1 $a = 4; $b = 3; $val = ($a / $b) + (($a % $b) != 0); echo "The ceiling value of 4/3 is " , floor($val) ,"\n"; // example of perfect division // taking input 2 $a = 6; $b = 3; $val = ($a / $b) + (($a % $b) != 0); echo "The ceiling value of 6/3 is " , $val ; // This code is contributed by anuj_67. ?>
Javascript
<script>
// javascript program to find
// ceil(a/b) without
// using ceil() function
// Driver Code
// taking input 1
var a = 4;
var b = 3, val = 0;
if ((a % b) != 0)
val = parseInt(a / b) + (a % b);
else
val = parseInt(a / b);
document.write("The ceiling " + "value of 4/3 is " + val+"<br/>");
// example of perfect
// division taking input 2
a = 6;
b = 3;
if ((a % b) != 0)
val = parseInt(a / b) + (a % b);
else
val = parseInt(a / b);
document.write("The ceiling " + "value of 6/3 is " + val);
// This code is contributed by gauravrajput1
</script>
Producción:
The ceiling value of 4/3 is 2 The ceiling value of 6/3 is 2
Enfoque 2:
ceilVal = (a+b-1) / b
Usando matemáticas simples, podemos agregar el denominador al numerador y restarle 1 y luego dividirlo por el denominador para obtener el valor máximo.
A continuación se muestra la ilustración del enfoque anterior:
C++
// C++ program to find ceil(a/b)
// without using ceil() function
#include <cmath>
#include <iostream>
using namespace std;
// Driver function
int main()
{
// taking input 1
int a = 4;
int b = 3;
int val = (a + b - 1) / b;
cout << "The ceiling value of 4/3 is "
<< val << endl;
// example of perfect division
// taking input 2
a = 6;
b = 3;
val = (a + b - 1) / b;
cout << "The ceiling value of 6/3 is "
<< val << endl;
return 0;
}
Java
// Java program to find ceil(a/b)
// without using ceil() function
class GFG {
// Driver Code
public static void main(String args[])
{
// taking input 1
int a = 4;
int b = 3;
int val = (a + b - 1) / b;
System.out.println("The ceiling value of 4/3 is "
+ val);
// example of perfect division
// taking input 2
a = 6;
b = 3;
val = (a + b - 1) / b;
System.out.println("The ceiling value of 6/3 is "
+ val );
}
}
// This code is contributed by Jaideep Pyne
Python3
# Python3 program to find
# math.ceil(a/b) without
# using math.ceil() function
import math
# Driver Code
# taking input 1
a = 4;
b = 3;
val = (a + b - 1) / b;
print("The ceiling value of 4/3 is ",
math.floor(val));
# example of perfect division
# taking input 2
a = 6;
b = 3;
val = (a + b - 1) / b;
print("The ceiling value of 6/3 is ",
math.floor(val));
# This code is contributed by mits
C#
// C# program to find ceil(a/b)
// without using ceil() function
using System;
class GFG {
// Driver Code
public static void Main()
{
// taking input 1
int a = 4;
int b = 3;
int val = (a + b - 1) / b;
Console.WriteLine("The ceiling"
+ " value of 4/3 is " + val);
// example of perfect division
// taking input 2
a = 6;
b = 3;
val = (a + b - 1) / b;
Console.WriteLine("The ceiling"
+ " value of 6/3 is " + val );
}
}
// This code is contributed by anuj_67.
PHP
<?php // PHP program to find ceil(a/b) // without using ceil() function // Driver function // taking input 1 $a = 4; $b = 3; $val = ($a + $b - 1) /$b; echo "The ceiling value of 4/3 is " , floor($val) ,"\n"; // example of perfect division // taking input 2 $a = 6; $b = 3; $val = ($a + $b - 1) / $b; echo "The ceiling value of 6/3 is " ,floor($val) ; // This code is contributed by anuj_67. ?>
Javascript
<script>
// javascript program to find ceil(a/b)
// without using ceil() function
// Driver Code
// taking input 1
var a = 4;
var b = 3;
var val = (a + b - 1) / b;
document.write("The ceiling value of 4/3 is " + val+"<br/>");
// example of perfect division
// taking input 2
a = 6;
b = 3;
val = parseInt((a + b - 1) / b);
document.write("The ceiling value of 6/3 is " + val);
// This code contributed by Rajput-Ji
</script>
Producción:
The ceiling value of 4/3 is 2 The ceiling value of 6/3 is 2
Tiempo Complejidad: O(1)
Espacio Auxiliar: O(1)