Dado un gran número positivo como string, cuente todas las rotaciones del número dado que son divisibles por 8.
Ejemplos:
Input: 8
Output: 1
Input: 40
Output: 1
Rotation: 40 is divisible by 8
04 is not divisible by 8
Input : 13502
Output : 0
No rotation is divisible by 8
Input : 43262488612
Output : 4
Enfoque: para números grandes, es difícil rotar y dividir cada número por 8. Por lo tanto, se usa la propiedad de ‘divisibilidad por 8’ que dice que un número es divisible por 8 si los últimos 3 dígitos del número son divisibles por 8. Aquí en realidad, no rotamos el número y verificamos la divisibilidad de los últimos 8 dígitos, sino que contamos una secuencia consecutiva de 3 dígitos (en forma circular) que son divisibles por 8.
Ilustración:
Consider a number 928160 Its rotations are 928160, 092816, 609281, 160928, 816092, 281609. Now form consecutive sequence of 3-digits from the original number 928160 as mentioned in the approach. 3-digit: (9, 2, 8), (2, 8, 1), (8, 1, 6), (1, 6, 0),(6, 0, 9), (0, 9, 2) We can observe that the 3-digit number formed by the these sets, i.e., 928, 281, 816, 160, 609, 092, are present in the last 3 digits of some rotation. Thus, checking divisibility of these 3-digit numbers gives the required number of rotations.
C++
// C++ program to count all rotations divisible
// by 8
#include <bits/stdc++.h>
using namespace std;
// function to count of all rotations divisible
// by 8
int countRotationsDivBy8(string n)
{
int len = n.length();
int count = 0;
// For single digit number
if (len == 1) {
int oneDigit = n[0] - '0';
if (oneDigit % 8 == 0)
return 1;
return 0;
}
// For two-digit numbers (considering all
// pairs)
if (len == 2) {
// first pair
int first = (n[0] - '0') * 10 + (n[1] - '0');
// second pair
int second = (n[1] - '0') * 10 + (n[0] - '0');
if (first % 8 == 0)
count++;
if (second % 8 == 0)
count++;
return count;
}
// considering all three-digit sequences
int threeDigit;
for (int i = 0; i < (len - 2); i++) {
threeDigit = (n[i] - '0') * 100 +
(n[i + 1] - '0') * 10 +
(n[i + 2] - '0');
if (threeDigit % 8 == 0)
count++;
}
// Considering the number formed by the
// last digit and the first two digits
threeDigit = (n[len - 1] - '0') * 100 +
(n[0] - '0') * 10 +
(n[1] - '0');
if (threeDigit % 8 == 0)
count++;
// Considering the number formed by the last
// two digits and the first digit
threeDigit = (n[len - 2] - '0') * 100 +
(n[len - 1] - '0') * 10 +
(n[0] - '0');
if (threeDigit % 8 == 0)
count++;
// required count of rotations
return count;
}
// Driver program to test above
int main()
{
string n = "43262488612";
cout << "Rotations: "
<< countRotationsDivBy8(n);
return 0;
}
Java
// Java program to count all
// rotations divisible by 8
import java.io.*;
class GFG
{
// function to count of all
// rotations divisible by 8
static int countRotationsDivBy8(String n)
{
int len = n.length();
int count = 0;
// For single digit number
if (len == 1) {
int oneDigit = n.charAt(0) - '0';
if (oneDigit % 8 == 0)
return 1;
return 0;
}
// For two-digit numbers
// (considering all pairs)
if (len == 2) {
// first pair
int first = (n.charAt(0) - '0') *
10 + (n.charAt(1) - '0');
// second pair
int second = (n.charAt(1) - '0') *
10 + (n.charAt(0) - '0');
if (first % 8 == 0)
count++;
if (second % 8 == 0)
count++;
return count;
}
// considering all three-digit sequences
int threeDigit;
for (int i = 0; i < (len - 2); i++)
{
threeDigit = (n.charAt(i) - '0') * 100 +
(n.charAt(i + 1) - '0') * 10 +
(n.charAt(i + 2) - '0');
if (threeDigit % 8 == 0)
count++;
}
// Considering the number formed by the
// last digit and the first two digits
threeDigit = (n.charAt(len - 1) - '0') * 100 +
(n.charAt(0) - '0') * 10 +
(n.charAt(1) - '0');
if (threeDigit % 8 == 0)
count++;
// Considering the number formed by the last
// two digits and the first digit
threeDigit = (n.charAt(len - 2) - '0') * 100 +
(n.charAt(len - 1) - '0') * 10 +
(n.charAt(0) - '0');
if (threeDigit % 8 == 0)
count++;
// required count of rotations
return count;
}
// Driver program
public static void main (String[] args)
{
String n = "43262488612";
System.out.println( "Rotations: "
+countRotationsDivBy8(n));
}
}
// This code is contributed by vt_m.
Python3
# Python3 program to count all
# rotations divisible by 8
# function to count of all
# rotations divisible by 8
def countRotationsDivBy8(n):
l = len(n)
count = 0
# For single digit number
if (l == 1):
oneDigit = int(n[0])
if (oneDigit % 8 == 0):
return 1
return 0
# For two-digit numbers
# (considering all pairs)
if (l == 2):
# first pair
first = int(n[0]) * 10 + int(n[1])
# second pair
second = int(n[1]) * 10 + int(n[0])
if (first % 8 == 0):
count+=1
if (second % 8 == 0):
count+=1
return count
# considering all
# three-digit sequences
threeDigit=0
for i in range(0,(l - 2)):
threeDigit = (int(n[i]) * 100 +
int(n[i + 1]) * 10 +
int(n[i + 2]))
if (threeDigit % 8 == 0):
count+=1
# Considering the number
# formed by the last digit
# and the first two digits
threeDigit = (int(n[l - 1]) * 100 +
int(n[0]) * 10 +
int(n[1]))
if (threeDigit % 8 == 0):
count+=1
# Considering the number
# formed by the last two
# digits and the first digit
threeDigit = (int(n[l - 2]) * 100 +
int(n[l - 1]) * 10 +
int(n[0]))
if (threeDigit % 8 == 0):
count+=1
# required count
# of rotations
return count
# Driver Code
if __name__=='__main__':
n = "43262488612"
print("Rotations:",countRotationsDivBy8(n))
# This code is contributed by mits.
C#
// C# program to count all
// rotations divisible by 8
using System;
class GFG {
// function to count of all
// rotations divisible by 8
static int countRotationsDivBy8(String n)
{
int len = n.Length;
int count = 0;
// For single digit number
if (len == 1)
{
int oneDigit = n[0] - '0';
if (oneDigit % 8 == 0)
return 1;
return 0;
}
// For two-digit numbers
// (considering all pairs)
if (len == 2)
{
// first pair
int first = (n[0] - '0') *
10 + (n[1] - '0');
// second pair
int second = (n[1] - '0') *
10 + (n[0] - '0');
if (first % 8 == 0)
count++;
if (second % 8 == 0)
count++;
return count;
}
// considering all three -
// digit sequences
int threeDigit;
for (int i = 0; i < (len - 2); i++)
{
threeDigit = (n[i] - '0') * 100 +
(n[i + 1] - '0') * 10 +
(n[i + 2] - '0');
if (threeDigit % 8 == 0)
count++;
}
// Considering the number formed by the
// last digit and the first two digits
threeDigit = (n[len - 1] - '0') * 100 +
(n[0] - '0') * 10 +
(n[1] - '0');
if (threeDigit % 8 == 0)
count++;
// Considering the number formed
// by the last two digits and
// the first digit
threeDigit = (n[len - 2] - '0') * 100 +
(n[len - 1] - '0') * 10 +
(n[0] - '0');
if (threeDigit % 8 == 0)
count++;
// required count of rotations
return count;
}
// Driver Code
public static void Main ()
{
String n = "43262488612";
Console.Write("Rotations: "
+countRotationsDivBy8(n));
}
}
// This code is contributed by Nitin Mittal.
PHP
<?php
// PHP program to count all
// rotations divisible by 8
// function to count of all
// rotations divisible by 8
function countRotationsDivBy8($n)
{
$len = strlen($n);
$count = 0;
// For single digit number
if ($len == 1)
{
$oneDigit = $n[0] - '0';
if ($oneDigit % 8 == 0)
return 1;
return 0;
}
// For two-digit numbers
// (considering all pairs)
if ($len == 2)
{
// first pair
$first = ($n[0] - '0') * 10 +
($n[1] - '0');
// second pair
$second = ($n[1] - '0') * 10 +
($n[0] - '0');
if ($first % 8 == 0)
$count++;
if ($second % 8 == 0)
$count++;
return $count;
}
// considering all
// three-digit sequences
$threeDigit;
for ($i = 0; $i < ($len - 2); $i++)
{
$threeDigit = ($n[$i] - '0') * 100 +
($n[$i + 1] - '0') * 10 +
($n[$i + 2] - '0');
if ($threeDigit % 8 == 0)
$count++;
}
// Considering the number
// formed by the last digit
// and the first two digits
$threeDigit = ($n[$len - 1] - '0') * 100 +
($n[0] - '0') * 10 +
($n[1] - '0');
if ($threeDigit % 8 == 0)
$count++;
// Considering the number
// formed by the last two
// digits and the first digit
$threeDigit = ($n[$len - 2] - '0') * 100 +
($n[$len - 1] - '0') * 10 +
($n[0] - '0');
if ($threeDigit % 8 == 0)
$count++;
// required count
// of rotations
return $count;
}
// Driver Code
$n = "43262488612";
echo "Rotations: " .
countRotationsDivBy8($n);
// This code is contributed by mits.
?>
Javascript
<script>
// Javascript program to count all
// rotations divisible by 8
// Function to count of all
// rotations divisible by 8
function countRotationsDivBy8(n)
{
let len = n.length;
let count = 0;
// For single digit number
if (len == 1)
{
let oneDigit = n[0] - '0';
if (oneDigit % 8 == 0)
return 1;
return 0;
}
// For two-digit numbers
// (considering all pairs)
if (len == 2)
{
// first pair
let first = (n[0] - '0') * 10 +
(n[1] - '0');
// second pair
let second = (n[1] - '0') * 10 +
(n[0] - '0');
if (first % 8 == 0)
count++;
if (second % 8 == 0)
count++;
return count;
}
// Considering all
// three-digit sequences
let threeDigit;
for(let i = 0; i < (len - 2); i++)
{
threeDigit = (n[i] - '0') * 100 +
(n[i + 1] - '0') * 10 +
(n[i + 2] - '0');
if (threeDigit % 8 == 0)
count++;
}
// Considering the number
// formed by the last digit
// and the first two digits
threeDigit = (n[len - 1] - '0') * 100 +
(n[0] - '0') * 10 +
(n[1] - '0');
if (threeDigit % 8 == 0)
count++;
// Considering the number
// formed by the last two
// digits and the first digit
threeDigit = (n[len - 2] - '0') * 100 +
(n[len - 1] - '0') * 10 +
(n[0] - '0');
if (threeDigit % 8 == 0)
count++;
// Required count
// of rotations
return count;
}
// Driver Code
let n = "43262488612";
document.write("Rotations: " +
countRotationsDivBy8(n));
// This code is contributed by _saurabh_jaiswal
</script>
Producción:
Rotations: 4
Complejidad de tiempo: O(n), donde n es el número de dígitos en el número de entrada.
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por ayushjauhari14 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA