Dados dos enteros a y b . La tarea es encontrar el número de dígitos antes del punto decimal en a/b .
Ejemplos:
Entrada: a = 100, b = 4
Salida: 2
100 / 4 = 25 y número de dígitos en 25 = 2.
Entrada: a = 100000, b = 10
Salida: 5
Enfoque ingenuo : divida los dos números y luego encuentre el número de dígitos en la división. Toma el valor absoluto de la división para encontrar el número de dígitos.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the number of digits
// before the decimal in a / b
int countDigits(int a, int b)
{
int count = 0;
// Absolute value of a / b
int p = abs(a / b);
// If result is 0
if (p == 0)
return 1;
// Count number of digits in the result
while (p > 0) {
count++;
p = p / 10;
}
// Return the required count of digits
return count;
}
// Driver code
int main()
{
int a = 100;
int b = 10;
cout << countDigits(a, b);
return 0;
}
Java
// Java implementation of the approach
class GFG {
// Function to return the number of digits
// before the decimal in a / b
static int countDigits(int a, int b)
{
int count = 0;
// Absolute value of a / b
int p = Math.abs(a / b);
// If result is 0
if (p == 0)
return 1;
// Count number of digits in the result
while (p > 0) {
count++;
p = p / 10;
}
// Return the required count of digits
return count;
}
// Driver code
public static void main(String args[])
{
int a = 100;
int b = 10;
System.out.print(countDigits(a, b));
}
}
Python
# Python 3 implementation of the approach # Function to return the number of digits # before the decimal in a / b def countDigits(a, b): count = 0 # Absolute value of a / b p = abs(a // b) # If result is 0 if (p == 0): return 1 # Count number of digits in the result while (p > 0): count = count + 1 p = p // 10 # Return the required count of digits return count # Driver code a = 100 b = 10 print(countDigits(a, b))
C#
// C# implementation of the approach
using System;
class GFG {
// Function to return the number of digits
// before the decimal in a / b
static int countDigits(int a, int b)
{
int count = 0;
// Absolute value of a / b
int p = Math.Abs(a / b);
// If result is 0
if (p == 0)
return 1;
// Count number of digits in the result
while (p > 0) {
count++;
p = p / 10;
}
// Return the required count of digits
return count;
}
// Driver code
public static void Main()
{
int a = 100;
int b = 10;
Console.Write(countDigits(a, b));
}
}
PHP
<?php
// PHP implementation of the approach
// Function to return the number of digits
// before the decimal in a / b
function countDigits($a, $b)
{
$count = 0;
// Absolute value of a / b
$p = abs($a / $b);
// If result is 0
if ($p == 0)
return 1;
// Count number of digits in the result
while ($p > 0) {
$count++;
$p = (int)($p / 10);
}
// Return the required count of digits
return $count;
}
// Driver code
$a = 100;
$b = 10;
echo countDigits($a, $b);
?>
Javascript
<script>
// Javascript implementation of the approach
// Function to return the number of digits
// before the decimal in a / b
function countDigits(a, b)
{
var count = 0;
// Absolute value of a / b
var p = Math.abs(parseInt(a / b));
// If result is 0
if (p == 0)
return 1;
// Count number of digits in the result
while (p > 0) {
count++;
p = parseInt(p / 10);
}
// Return the required count of digits
return count;
}
// Driver code
var a = 100;
var b = 10;
document.write(countDigits(a, b));
// This code is contributed by rrrtnx.
</script>
Producción:
2
Complejidad temporal: O(log 10 (a/ b))
Espacio Auxiliar: O(1)
Enfoque eficiente: para contar el número de dígitos en a/b , podemos usar la fórmula:
piso (log 10 (a) – log 10 (b)) + 1
Aquí ambos números deben ser enteros positivos. Para esto podemos tomar los valores absolutos de a y b .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the number of digits
// before the decimal in a / b
int countDigits(int a, int b)
{
// Return the required count of digits
return floor(log10(abs(a)) - log10(abs(b))) + 1;
}
// Driver code
int main()
{
int a = 100;
int b = 10;
cout << countDigits(a, b);
return 0;
}
Java
// Java implementation of the approach
class GFG {
// Function to return the number of digits
// before the decimal in a / b
public static int countDigits(int a, int b)
{
double digits = Math.log10(Math.abs(a))
- Math.log10(Math.abs(b)) + 1;
// Return the required count of digits
return (int)Math.floor(digits);
}
// Driver code
public static void main(String[] args)
{
int a = 100;
int b = 10;
System.out.print(countDigits(a, b));
}
}
Python
# Python3 implementation of the approach import math # Function to return the number of digits # before the decimal in a / b def countDigits(a, b): # Return the required count of digits return math.floor(math.log10(abs(a)) - math.log10(abs(b))) + 1 # Driver code a = 100 b = 10 print(countDigits(a, b))
C#
// C# implementation of the approach
using System;
class GFG {
// Function to return the number of digits
// before the decimal in a / b
public static int countDigits(int a, int b)
{
double digits = Math.Log10(Math.Abs(a))
- Math.Log10(Math.Abs(b)) + 1;
// Return the required count of digits
return (int)Math.Floor(digits);
}
// Driver code
static void Main()
{
int a = 100;
int b = 10;
Console.Write(countDigits(a, b));
}
}
PHP
<?php
// PHP implementation of the approach
// Function to return the number of digits
// before the decimal in a / b
function countDigits($a, $b)
{
// Return the required count of digits
return floor(log10(abs($a)) -
log10(abs($b))) + 1;
}
// Driver code
$a = 100;
$b = 10;
echo countDigits($a, $b);
?>
Javascript
<script>
// Javascript implementation of the approach
// Function to return the number of digits
// before the decimal in a / b
function countDigits(a, b)
{
// Return the required count of digits
return Math.floor((Math.log(Math.abs(a))/Math.log(10)) - (Math.log(Math.abs(b))/Math.log(10))) + 1;
}
// Driver code
var a = 100;
var b = 10;
document.write(countDigits(a, b));
// This code is contributed by rutvik_56.
</script>
Producción:
2
Complejidad temporal: O(log 10 (a/ b))
Espacio Auxiliar: O(1)