Considere una secuencia a 0 , a 1 , …, a n , donde a i = a i-1 + a i-2 . Dado un 0 , un 1 y un entero positivo n . La tarea es encontrar si una n es par o impar.
Tenga en cuenta que la secuencia dada es como Fibonacci con la diferencia de que los primeros dos términos pueden ser cualquier cosa en lugar de 0 o 1.
Ejemplos:
Input : a0 = 2, a1 = 4, n =3 Output : Even a2 = 6, a3 = 10 And a3 is even. Input : a0 = 1, a1 = 9, n = 2 Output : Even
Método 1: la idea es encontrar la secuencia usando la array y verificar si el elemento n es par o impar.
C++
// CPP Program to check if the nth is odd or even in a
// sequence where each term is sum of previous two term
#include <bits/stdc++.h>
using namespace std;
#define MAX 100
// Return if the nth term is even or odd.
bool findNature(int a, int b, int n)
{
int seq[MAX] = { 0 };
seq[0] = a;
seq[1] = b;
for (int i = 2; i <= n; i++)
seq[i] = seq[i - 1] + seq[i - 2];
// Return true if odd
return (seq[n] & 1);
}
// Driven Program
int main()
{
int a = 2, b = 4;
int n = 3;
(findNature(a, b, n) ? (cout << "Odd"
<< " ")
: (cout << "Even"
<< " "));
return 0;
}
Java
// Java Program to check if
// the nth is odd or even
// in a sequence where each
// term is sum of previous
// two term
// Return if the nth
// term is even or odd.
class GFG
{
public static int findNature(int a,
int b, int n)
{
int[] seq = new int[100];
seq[0] = a;
seq[1] = b;
for (int i = 2; i <= n; i++)
seq[i] = seq[i - 1] + seq[i - 2];
// Return true if odd
if((seq[n] & 1) != 0)
return 1;
else
return 0;
}
// Driver Code
public static void main(String[] args)
{
int a = 2, b = 4;
int n = 3;
if(findNature(a, b, n) == 1)
System.out.println("Odd ");
else
System.out.println("Even ");
}
}
// This code is contributed
// by mits
Python3
# Python3 Program to check if
# the nth is odd or even in a
# sequence where each term is
# sum of previous two term
MAX = 100;
# Return if the nth
# term is even or odd.
def findNature(a, b, n):
seq = [0] * MAX;
seq[0] = a;
seq[1] = b;
for i in range(2, n + 1):
seq[i] = seq[i - 1] + seq[i - 2];
# Return true if odd
return (seq[n] & 1);
# Driver Code
a = 2;
b = 4;
n = 3;
if(findNature(a, b, n)):
print("Odd");
else:
print("Even");
# This code is contributed by mits
C#
// C# Program to check if the
// nth is odd or even in a
// sequence where each term
// is sum of previous two term
using System;
// Return if the nth term
// is even or odd.
class GFG
{
public static int findNature(int a,
int b, int n)
{
int[] seq = new int[100];
seq[0] = a;
seq[1] = b;
for (int i = 2; i <= n; i++)
seq[i] = seq[i - 1] +
seq[i - 2];
// Return true if odd
if((seq[n] & 1)!=0)
return 1;
else
return 0;
}
// Driver Code
public static void Main()
{
int a = 2, b = 4;
int n = 3;
if(findNature(a, b, n) == 1)
Console.Write("Odd ");
else
Console.Write("Even ");
}
}
// This code is contributed
// by mits
PHP
<?php
// PHP Program to check if the
// nth is odd or even in a
// sequence where each term is
// sum of previous two term
$MAX = 100;
// Return if the nth
// term is even or odd.
function findNature($a, $b, $n)
{
global $MAX;
$seq = array_fill(0,$MAX, 0);
$seq[0] = $a;
$seq[1] = $b;
for ($i = 2; $i <= $n; $i++)
$seq[$i] = $seq[$i - 1] +
$seq[$i - 2];
// Return true if odd
return ($seq[$n] & 1);
}
// Driver Code
$a = 2;
$b = 4;
$n = 3;
if(findNature($a, $b, $n))
echo "Odd";
else
echo "Even";
// This code is contributed by mits
?>
Javascript
<script>
// Javascript Program to check if the
// nth is odd or even in a
// sequence where each term is sum
// of previous two term
var MAX = 100;
// Return if the nth term is even or odd.
function findNature(a, b, n)
{
var seq = Array(MAX).fill(0);
seq[0] = a;
seq[1] = b;
for (var i = 2; i <= n; i++)
seq[i] = seq[i - 1] + seq[i - 2];
// Return true if odd
return (seq[n] & 1);
}
// Driven Program
var a = 2, b = 4;
var n = 3;
(findNature(a, b, n) ? (document.write( "Odd"
+ " "))
: (document.write( "Even"
+ " ")));
</script>
Producción:
Even
Método 2 (eficiente):
observe que la naturaleza (par o impar) del término n depende de los términos anteriores, y la naturaleza del término anterior depende de sus términos anteriores y finalmente depende del valor inicial, es decir, un 0 y un 1 .
Entonces, tenemos cuatro escenarios posibles para un 0 y un 1 :
Caso 1: Cuando un 0 y un 1 son pares
En este caso, cada uno de los valores en la secuencia será solo par.
Caso 2: Cuando un 0 y un 1 son impares
En este caso, observe que un 2 es par, un 3 es impar, un 4 es impar y así sucesivamente. Entonces, podemos decir uni es par si i es de la forma 3*k – 1, de lo contrario impar.
Caso 3: Cuando un 0 es par y un 1 es impar
En este caso, observe que un 2 es impar, un 3 es par, un 4 y un 5 son impares, un 6 es par y así sucesivamente. Entonces, podemos decir, a i es par si i es múltiplo de 3, caso contrario impar
4: Cuando un 0 es impar y un 1 es par
En este caso, observe un 2 y un 3 es impar, un 4 es par, un 5 y un 6 es impar, un 7es par y así sucesivamente. Entonces, podemos decir que a i es par si i es de la forma 3*k + 1, k >= 0, de lo contrario impar.
A continuación se muestra la implementación de este enfoque:
C++
// CPP Program to check if the nth is odd or even in a
// sequence where each term is sum of previous two term
#include <bits/stdc++.h>
using namespace std;
// Return if the nth term is even or odd.
bool findNature(int a, int b, int n)
{
if (n == 0)
return (a & 1);
if (n == 1)
return (b & 1);
// If a is even
if (!(a & 1)) {
// If b is even
if (!(b & 1))
return false;
// If b is odd
else
return (n % 3 != 0);
}
// If a is odd
else {
// If b is odd
if (!(b & 1))
return ((n - 1) % 3 != 0);
// If b is even
else
return ((n + 1) % 3 != 0);
}
}
// Driven Program
int main()
{
int a = 2, b = 4;
int n = 3;
(findNature(a, b, n) ? (cout << "Odd"
<< " ")
: (cout << "Even"
<< " "));
return 0;
}
Java
// Java Program to check if the nth is odd or even in a
// sequence where each term is sum of previous two term
class GFG{
// Return if the nth term is even or odd.
static boolean findNature(int a, int b, int n)
{
if (n == 0)
return (a & 1)==1?true:false;
if (n == 1)
return (b & 1)==1?true:false;
// If a is even
if ((a & 1)==0) {
// If b is even
if ((b & 1)==0)
return false;
// If b is odd
else
return (n % 3 != 0);
}
// If a is odd
else {
// If b is odd
if ((b & 1)==0)
return ((n - 1) % 3 != 0);
// If b is even
else
return ((n + 1) % 3 != 0);
}
}
// Driven Program
public static void main(String[] args)
{
int a = 2, b = 4;
int n = 3;
if(findNature(a, b, n))
System.out.println("Odd");
else
System.out.println("Even");
}
}
// This Code is contributed by mits
Python3
# Python3 Program to check if the
# nth is odd or even in a
# sequence where each term is
# sum of previous two term
# Return if the nth
# term is even or odd.
def findNature(a, b, n):
if (n == 0):
return (a & 1);
if (n == 1):
return (b & 1);
# If a is even
if ((a & 1) == 0):
# If b is even
if ((b & 1) == 0):
return False;
# If b is odd
else:
return True if(n % 3 != 0) else False;
# If a is odd
else:
# If b is odd
if ((b & 1) == 0):
return True if((n - 1) % 3 != 0) else False;
# If b is even
else:
return True if((n + 1) % 3 != 0) else False;
# Driver Code
a = 2;
b = 4;
n = 3;
if (findNature(a, b, n) == True):
print("Odd", end = " ");
else:
print("Even", end = " ");
# This code is contributed by mits
C#
// C# Program to check if the nth is odd or even in a
// sequence where each term is sum of previous two term
class GFG{
// Return if the nth term is even or odd.
static bool findNature(int a, int b, int n)
{
if (n == 0)
return (a & 1)==1?true:false;
if (n == 1)
return (b & 1)==1?true:false;
// If a is even
if ((a & 1)==0) {
// If b is even
if ((b & 1)==0)
return false;
// If b is odd
else
return (n % 3 != 0);
}
// If a is odd
else {
// If b is odd
if ((b & 1)==0)
return ((n - 1) % 3 != 0);
// If b is even
else
return ((n + 1) % 3 != 0);
}
}
// Driven Program
static void Main()
{
int a = 2, b = 4;
int n = 3;
if(findNature(a, b, n))
System.Console.WriteLine("Odd");
else
System.Console.WriteLine("Even");
}
}
// This Code is contributed by mits
PHP
<?php
// PHP Program to check if the
// nth is odd or even in a
// sequence where each term is
// sum of previous two term
// Return if the nth
// term is even or odd.
function findNature($a, $b, $n)
{
if ($n == 0)
return ($a & 1);
if ($n == 1)
return ($b & 1);
// If a is even
if (!($a & 1))
{
// If b is even
if (!($b & 1))
return false;
// If b is odd
else
return ($n % 3 != 0);
}
// If a is odd
else
{
// If b is odd
if (!($b & 1))
return (($n - 1) % 3 != 0);
// If b is even
else
return (($n + 1) % 3 != 0);
}
}
// Driver Code
$a = 2; $b = 4;
$n = 3;
if (findNature($a, $b, $n) == true)
echo "Odd"," ";
else
echo "Even"," ";
// This code is contributed by ajit
?>
Javascript
<script>
// Javascript Program to check if the nth is odd or even in a
// sequence where each term is sum of previous two term
// Return if the nth term is even or odd.
function findNature(a, b, n)
{
if (n == 0)
return (a & 1)==1?true:false;
if (n == 1)
return (b & 1)==1?true:false;
// If a is even
if ((a & 1)==0) {
// If b is even
if ((b & 1)==0)
return false;
// If b is odd
else
return (n % 3 != 0);
}
// If a is odd
else {
// If b is odd
if ((b & 1)==0)
return ((n - 1) % 3 != 0);
// If b is even
else
return ((n + 1) % 3 != 0);
}
}
let a = 2, b = 4;
let n = 3;
if(findNature(a, b, n))
document.write("Odd");
else
document.write("Even");
// This code is contributed by suresh07.
</script>
Producción :
Even