检查第n个项是否为奇数或偶数似Fibonacci序列

考虑一个序列a ₀ ，a ₁ ，…，a _n ，其中a _i = a _i-1 + a _i-2 。给定一个₀ ，一个₁和一个正整数n 。任务是找出_n是奇数还是偶数。

请注意，给定的序列类似于Fibonacci，不同之处在于前两个项可以是任意值，而不是0或1。

例子 ：

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

方法1：想法是使用数组查找序列，并检查第n个元素是偶数还是奇数。

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 
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

输出：

Even

方法2(有效)：
可以看到，第n个项的性质(奇数或偶数)取决于先前的项，而上一个项的性质取决于其先前的项，最后取决于初始值，即₀和₁ 。
因此，对于₀和₁ ，我们有四种可能的情况：
情况1：当一个₀和一个₁是偶数时
在这种情况下，序列中的每个值都将仅是偶数。

情况2：当₀和₁为奇数时
在这种情况下，观察一个₂是偶数，一个₃是奇数，一个₄是奇数，依此类推。因此，我们可以说_我是即使我是形式3 * K – 1，否则奇怪。

情况3：当₀为偶数而₁为奇数时
在这种情况下，观察一个₂为奇数，一个₃为偶数，一个₄和一个₅为奇数，一个₆为偶数，依此类推。因此，我们可以说，一个_我是我就算是多的3，其他奇

情况4：当₀为奇数而₁为偶数时
在这种情况下，观察一个₂和一个₃为奇数，一个₄为偶数，一个₅和一个₆为奇数，一个₇为偶数，依此类推。因此，我们可以说，一个_i是即使i是形式3 * K + 1，K> = 0的，否则奇数。

以下是此方法的实现：

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 
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 eve
        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 eve
        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 eve
        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 eve
        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

输出 ：

Even