Dada una array con n números. La tarea es imprimir el número de ceros consecutivos al final después de multiplicar todo el número n.
Ejemplos:
Input : arr[] = {100, 10}
Output : 3
Explanation : 100 x 10 = 1000,
3 zero's at the end.
Input : arr[] = {100, 10, 5, 25, 35, 14}
Output : 4
Explanation :
100 x 10 x 5 x 25 x 35 x 14 = 61250000,
4 zero's at the end
Enfoque ingenuo:
primero multiplique todo el número y guárdelo en la string (porque si el número multiplicativo es tan grande como 2 ^ 64, entonces da mal, así que almacene cada múltiplo en la string) y luego cuente el número de ceros.
Enfoque eficiente:
primero cuente el factor de 2 de n números y luego cuente el factor de 5 de todos los n números y luego imprima el más pequeño.
Por ejemplo –
n_number's | 2's factor | 5's factor 100 | 2 | 2 10 | 1 | 1 5 | 0 | 1 25 | 0 | 2 35 | 0 | 1 14 | 1 | 0 Total | 4 | 7 we can take a minimum so there number of zero's is 4
A continuación se muestra la implementación del enfoque anterior:
C++
// CPP program to find the number of consecutive zero
// at the end after multiplying n numbers
#include<iostream>
using namespace std;
// Function to count two's factor
int two_factor(int n)
{
// Count number of 2s present in n
int twocount = 0;
while (n % 2 == 0)
{
twocount++;
n = n / 2;
}
return twocount;
}
// Function to count five's factor
int five_factor(int n)
{
int fivecount = 0;
while (n % 5 == 0)
{
fivecount++;
n = n / 5;
}
return fivecount;
}
// Function to count number of zeros
int find_con_zero(int arr[], int n)
{
int twocount = 0;
int fivecount = 0;
for (int i = 0; i < n; i++) {
// Count the two's factor of n number
twocount += two_factor(arr[i]);
// Count the five's factor of n number
fivecount += five_factor(arr[i]);
}
// Return the minimum
if (twocount < fivecount)
return twocount;
else
return fivecount;
}
// Driver Code
int main()
{
int arr[] = { 100, 10, 5, 25, 35, 14 };
int n = 6;
cout << find_con_zero(arr, n);
}
Java
// Java program to find the number
// of consecutive zero at the end
// after multiplying n numbers
public class GfG{
// Function to count two's factor
static int two_factor(int n)
{
// Count number of 2s
// present in n
int twocount = 0;
while (n % 2 == 0)
{
twocount++;
n = n / 2;
}
return twocount;
}
// Function to count five's
// factor
static int five_factor(int n)
{
int fivecount = 0;
while (n % 5 == 0)
{
fivecount++;
n = n / 5;
}
return fivecount;
}
// Function to count number of zeros
static int find_con_zero(int arr[], int n)
{
int twocount = 0;
int fivecount = 0;
for (int i = 0; i < n; i++) {
// Count the two's factor
// of n number
twocount += two_factor(arr[i]);
// Count the five's factor
// of n number
fivecount += five_factor(arr[i]);
}
// Return the minimum
if (twocount < fivecount)
return twocount;
else
return fivecount;
}
// driver function
public static void main(String s[])
{
int arr[] = { 100, 10, 5, 25, 35, 14 };
int n = 6;
System.out.println(find_con_zero(arr, n));
}
}
// This code is contributed by Gitanjali
Python3
# Python3 code to find the number of consecutive zero # at the end after multiplying n numbers # Function to count two's factor def two_factor( n ): # Count number of 2s present in n twocount = 0 while n % 2 == 0: twocount+=1 n =int( n / 2) return twocount # Function to count five's factor def five_factor( n ): fivecount = 0 while n % 5 == 0: fivecount+=1 n = int(n / 5) return fivecount # Function to count number of zeros def find_con_zero( arr, n ): twocount = 0 fivecount = 0 for i in range(n): # Count the two's factor of n number twocount += two_factor(arr[i]) # Count the five's factor of n number fivecount += five_factor(arr[i]) # Return the minimum if twocount < fivecount: return twocount else: return fivecount # Driver Code arr = [ 100, 10, 5, 25, 35, 14 ] n = 6 print(find_con_zero(arr, n)) # This code is contributed by "Sharad_Bhardwaj".
C#
// C# program to find the number
// of consecutive zero at the end
// after multiplying n numbers
using System;
public class GfG {
// Function to count two's factor
static int two_factor(int n)
{
// Count number of 2s
// present in n
int twocount = 0;
while (n % 2 == 0)
{
twocount++;
n = n / 2;
}
return twocount;
}
// Function to count five's
// factor
static int five_factor(int n)
{
int fivecount = 0;
while (n % 5 == 0)
{
fivecount++;
n = n / 5;
}
return fivecount;
}
// Function to count number of zeros
static int find_con_zero(int []arr, int n)
{
int twocount = 0;
int fivecount = 0;
for (int i = 0; i < n; i++) {
// Count the two's factor
// of n number
twocount += two_factor(arr[i]);
// Count the five's factor
// of n number
fivecount += five_factor(arr[i]);
}
// Return the minimum
if (twocount < fivecount)
return twocount;
else
return fivecount;
}
// driver function
public static void Main()
{
int []arr = { 100, 10, 5, 25, 35, 14 };
int n = 6;
Console.WriteLine(find_con_zero(arr, n));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program to find the number
// of consecutive zero at the end
// after multiplying n numbers
// Function to count two's factor
function two_factor($n)
{
// Count number of
// 2s present in n
$twocount = 0;
while ($n % 2 == 0)
{
$twocount++;
$n = (int)($n / 2);
}
return $twocount;
}
// Function to count
// five's factor
function five_factor($n)
{
$fivecount = 0;
while ($n % 5 == 0)
{
$fivecount++;
$n = (int)($n / 5);
}
return $fivecount;
}
// Function to count
// number of zeros
function find_con_zero($arr, $n)
{
$twocount = 0;
$fivecount = 0;
for ($i = 0; $i < $n; $i++)
{
// Count the two's
// factor of n number
$twocount += two_factor($arr[$i]);
// Count the five's
// factor of n number
$fivecount += five_factor($arr[$i]);
}
// Return the minimum
if ($twocount < $fivecount)
return $twocount;
else
return $fivecount;
}
// Driver Code
$arr= array(100, 10, 5,
25, 35, 14);
$n = 6;
echo find_con_zero($arr, $n);
// This code is contributed by mits
?>
Javascript
<script>
// JavaScript program to find the number
// of consecutive zero at the end
// after multiplying n numbers
// Function to count two's factor
function two_factor(n)
{
// Count number of 2s
// present in n
let twocount = 0;
while (n % 2 == 0)
{
twocount++;
n = n / 2;
}
return twocount;
}
// Function to count five's
// factor
function five_factor(n)
{
let fivecount = 0;
while (n % 5 == 0)
{
fivecount++;
n = n / 5;
}
return fivecount;
}
// Function to count number of zeros
function find_con_zero(arr, n)
{
let twocount = 0;
let fivecount = 0;
for (let i = 0; i < n; i++)
{
// Count the two's factor
// of n number
twocount += two_factor(arr[i]);
// Count the five's factor
// of n number
fivecount += five_factor(arr[i]);
}
// Return the minimum
if (twocount < fivecount)
return twocount;
else
return fivecount;
}
// Driver code
let arr = [ 100, 10, 5, 25, 35, 14 ];
let n = 6;
document.write(find_con_zero(arr, n));
// This code is contributed by code_hunt.
</script>
Producción :
4
Publicación traducida automáticamente
Artículo escrito por DevanshuAgarwal y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA