Dados tres alfabetos diferentes ‘a’ , ‘b’ y ‘c’ con cierta regla que después de cada 2 segundos cada ‘a’ cambia a una ‘b’ , después de cada 5 segundos cada ‘b’ cambia a una ‘c’ y después de cada 12 segundos , cada ‘c’ cambia nuevamente a dos ‘a’s .
Comenzando con una ‘a’ , la tarea es encontrar la cuenta final de a , b y c después de n segundos dados .
Ejemplos:
Entrada: n = 2
Salida: a = 0, b = 1, c = 0
Inicialmente a = 1, b = 0, c = 0
En n = 1, nada cambiará
En n = 2, todo a cambiará a b es decir a = 0, b = 1, c = 0
Entrada: n = 72
Salida: a = 64, b = 0, c = 0
Enfoque: Se puede observar que los valores de a, b y c formarán un patrón cada 60 segundos (que es el MCM de 2, 5 y 12) de la siguiente manera:
- En n = 60 -> a = 32 1 , b = 0, c = 0
- En n = 120 -> a = 32 2 , b = 0, c = 0
- En n = 180 -> a = 32 3 , b = 0, c = 0 y así sucesivamente.
Si n es un múltiplo de 60 , calcule el resultado de la observación anterior; de lo contrario, calcule el resultado del múltiplo de 60 más cercano a n , digamos x , y luego actualice el resultado para los segundos de x + 1 a n .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ull unsigned long long
// Function to print the count of
// a, b and c after n seconds
void findCount(int n)
{
ull a = 1, b = 0, c = 0;
// Number of multiples of 60 below n
int x = n / 60;
a = (ull)pow(32, x);
// Multiple of 60 nearest to n
x = 60 * x;
for (int i = x + 1; i <= n; i++) {
// Change all a to b
if (i % 2 == 0) {
b += a;
a = 0;
}
// Change all b to c
if (i % 5 == 0) {
c += b;
b = 0;
}
// Change each c to two a
if (i % 12 == 0) {
a += (2 * c);
c = 0;
}
}
// Print the updated values of a, b and c
cout << "a = " << a << ", ";
cout << "b = " << b << ", ";
cout << "c = " << c;
}
// Driver code
int main()
{
int n = 72;
findCount(n);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to print the count of
// a, b and c after n seconds
static void findCount(int n)
{
long a = 1, b = 0, c = 0;
// Number of multiples of 60 below n
int x = n / 60;
a = (long)Math.pow(32, x);
// Multiple of 60 nearest to n
x = 60 * x;
for (int i = x + 1; i <= n; i++)
{
// Change all a to b
if (i % 2 == 0)
{
b += a;
a = 0;
}
// Change all b to c
if (i % 5 == 0)
{
c += b;
b = 0;
}
// Change each c to two a
if (i % 12 == 0)
{
a += (2 * c);
c = 0;
}
}
// Print the updated values of a, b and c
System.out.println("a = " + a + ", b = " +
b + ", c = " + c);
}
// Driver code
public static void main (String[] args)
{
int n = 72;
findCount(n);
}
}
// This code is contributed by mits
Python3
# Python3 implementation of the approach
# Function to print the count of
# a, b and c after n seconds
import math
def findCount(n):
a, b, c = 1, 0, 0;
# Number of multiples of 60 below n
x = (int)(n / 60);
a = int(math.pow(32, x));
# Multiple of 60 nearest to n
x = 60 * x;
for i in range(x + 1, n + 1):
# Change all a to b
if (i % 2 == 0):
b += a;
a = 0;
# Change all b to c
if (i % 5 == 0):
c += b;
b = 0;
# Change each c to two a
if (i % 12 == 0):
a += (2 * c);
c = 0;
# Print the updated values of a, b and c
print("a =", a, end = ", ");
print("b =", b, end = ", ");
print("c =", c);
# Driver code
if __name__ == '__main__':
n = 72;
findCount(n);
# This code is contributed
# by 29AjayKumar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to print the count of
// a, b and c after n seconds
static void findCount(int n)
{
long a = 1, b = 0, c = 0;
// Number of multiples of 60 below n
int x = n / 60;
a = (long)Math.Pow(32, x);
// Multiple of 60 nearest to n
x = 60 * x;
for (int i = x + 1; i <= n; i++)
{
// Change all a to b
if (i % 2 == 0)
{
b += a;
a = 0;
}
// Change all b to c
if (i % 5 == 0)
{
c += b;
b = 0;
}
// Change each c to two a
if (i % 12 == 0)
{
a += (2 * c);
c = 0;
}
}
// Print the updated values of a, b and c
Console.WriteLine("a = " + a + ", b = " + b + ", c = " + c);
}
// Driver code
static void Main()
{
int n = 72;
findCount(n);
}
}
// This code is contributed by mits
PHP
<?php
// PHP implementation of the approach
// Function to print the count of
// a, b and c after n seconds
function findCount($n)
{
$a = 1; $b = 0; $c = 0;
// Number of multiples of 60 below n
$x = $n / 60;
$a = pow(32, $x);
// Multiple of 60 nearest to n
$x = 60 * $x;
for ($i = $x + 1; $i <= $n; $i++)
{
// Change all a to b
if ($i % 2 == 0)
{
$b += $a;
$a = 0;
}
// Change all b to c
if ($i % 5 == 0)
{
$c += $b;
$b = 0;
}
// Change each c to two a
if ($i % 12 == 0)
{
$a += (2 * $c);
$c = 0;
}
}
// Print the updated values of a, b and c
echo("a = " . $a . ", b = " .
$b . ", c = " . $c);
}
// Driver code
$n = 72;
findCount($n);
// This code is contributed
// by Code_Mech.
?>
Javascript
<script>
// JavaScript implementation of the approach
// Function to print the count of
// a, b and c after n seconds
function findCount(n) {
var a = 1, b = 0, c = 0;
// Number of multiples of 60 below n
var x = parseInt(n / 60);
a = Math.pow(32, x);
// Multiple of 60 nearest to n
x = 60 * x;
for (i = x + 1; i <= n; i++) {
// Change all a to b
if (i % 2 == 0) {
b += a;
a = 0;
}
// Change all b to c
if (i % 5 == 0) {
c += b;
b = 0;
}
// Change each c to two a
if (i % 12 == 0) {
a += (2 * c);
c = 0;
}
}
// Print the updated values of a, b and c
document.write("a = " + a + ", b = " + b + ", c = " + c);
}
// Driver code
var n = 72;
findCount(n);
// This code contributed by Rajput-Ji
</script>
a = 64, b = 0, c = 0
Complejidad de tiempo: O(n)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por Shivam.Pradhan y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA