满足给定方程的斐波那契对数

给定一个包含Q 个正整数和两个数字A和B的数组arr[] ，任务是在数组arr 中找到每个数字N的有序对(x, y)的数量，使得：

• 它们满足方程A*x + B*y = N ，并且
• x 和 y 是斐波那契数列。

例子：

Input: arr[] = {15, 25}, A = 1, B = 2 
Output: 2 1 
Explanation: 
For 15: There are 2 ordered pairs (x, y) = (13, 1) & (5, 5) such that 1*x + 2*y = 15 
For 25: There is only one ordered pair (x, y) = (21, 2) such that 1*x + 2*y = 25
Input: arr[] = {50, 150}, A = 5, B = 10 
Output: 1 0 
 

天真方法：这个问题的天真方法是：

• 对于数组中的每个查询N ，计算最多 N 的斐波那契数
• 如果满足给定的条件A*x + B*y = N ，则从此斐波那契数检查所有可能的对。

时间复杂度： O(N ² )
有效的方法：

• 预先计算所有斐波那契数并将其存储在数组中。
• 现在，迭代斐波那契数和所有可能的组合，更新范围 [1, max(arr)] 中每个数字的有序对数，并将其存储在另一个数组中。
• 现在，对于每个查询 N，可以在恒定时间内回答有序对的数量。

下面是上述方法的实现：

C++

// C++ program to find the count of
// Fibonacci pairs (x, y) which
// satisfy the equation Ax+By=N
 
#include 
#define size 10001
using namespace std;
 
// Array to store the Fibonacci numbers
long long fib[100010];
 
// Array to store the number of ordered pairs
int freq[100010];
 
// Function to find if a number
// is a perfect square
bool isPerfectSquare(int x)
{
    int s = sqrt(x);
    return (s * s == x);
}
 
// Function that returns 1
// if N is non-fibonacci number else 0
int isFibonacci(int n)
{
    // N is Fibinacci if one of
    // 5*n*n + 4 or 5*n*n - 4 or both
    // are perferct square
    if (isPerfectSquare(5 * n * n + 4)
        || isPerfectSquare(5 * n * n - 4))
        return 1;
    return 0;
}
 
// Function to store the fibonacci numbers
// and their frequency in form a * x + b * y
void compute(int a, int b)
{
    // Storing the Fibonacci numbers
    for (int i = 1; i < 100010; i++) {
        fib[i] = isFibonacci(i);
    }
 
    // For loop to find all the possible
    // combinations of the Fibonacci numbers
    for (int x = 1; x < 100010; x++) {
        for (int y = 1; y < size; y++) {
 
            // Finding the number of ordered pairs
            if (fib[x] == 1 && fib[y] == 1
                && a * x + b * y < 100010) {
                freq[a * x + b * y]++;
            }
        }
    }
}
 
// Driver code
int main()
{
    int Q = 2, A = 5, B = 10;
    compute(A, B);
    int arr[Q] = { 50, 150 };
 
    // Find the ordered pair for every query
    for (int i = 0; i < Q; i++) {
        cout << freq[arr[i]] << " ";
    }
 
    return 0;
}

Java

// Java program to find the count of
// Fibonacci pairs (x, y) which
// satisfy the equation Ax+By=N
class GFG{
     
static final int size = 10001;
 
// Array to store the Fibonacci numbers
static long []fib = new long[100010];
  
// Array to store the number of ordered pairs
static int []freq = new int[100010];
  
// Function to find if a number
// is a perfect square
static boolean isPerfectSquare(int x)
{
    int s = (int) Math.sqrt(x);
    return (s * s == x);
}
  
// Function that returns 1
// if N is non-fibonacci number else 0
static int isFibonacci(int n)
{
    // N is Fibinacci if one of
    // 5*n*n + 4 or 5*n*n - 4 or both
    // are perferct square
    if (isPerfectSquare(5 * n * n + 4)
        || isPerfectSquare(5 * n * n - 4))
        return 1;
    return 0;
}
  
// Function to store the fibonacci numbers
// and their frequency in form a * x + b * y
static void compute(int a, int b)
{
    // Storing the Fibonacci numbers
    for (int i = 1; i < 100010; i++) {
        fib[i] = isFibonacci(i);
    }
  
    // For loop to find all the possible
    // combinations of the Fibonacci numbers
    for (int x = 1; x < 100010; x++) {
        for (int y = 1; y < size; y++) {
  
            // Finding the number of ordered pairs
            if (fib[x] == 1 && fib[y] == 1
                && a * x + b * y < 100010) {
                freq[a * x + b * y]++;
            }
        }
    }
}
  
// Driver code
public static void main(String[] args)
{
    int Q = 2, A = 5, B = 10;
    compute(A, B);
    int arr[] = { 50, 150 };
  
    // Find the ordered pair for every query
    for (int i = 0; i < Q; i++) {
        System.out.print(freq[arr[i]]+ " ");
    }
}
}
 
// This code is contributed by PrinciRaj1992

Python 3

# Python program to find the count of
# Fibonacci pairs (x, y) which
# satisfy the equation Ax+By=N
import math
size = 101
 
# Array to store the Fibonacci numbers
fib = [0]*100010
 
# Array to store the number of ordered pairs
freq = [0]*(100010)
 
# Function to find if a number
# is a perfect square
def isPerfectSquare(x):
    s = int(math.sqrt(x))
     
    return (s * s) == x
 
# Function that returns 1
# if N is non-fibonacci number else 0
def isFibonacci(n):
     
    # N is Fibinacci if one of
    # 5*n*n + 4 or 5*n*n - 4 or both
    # are perferct square
    if (isPerfectSquare(5 * n * n + 4) or isPerfectSquare(5 * n * n - 4)):
        return 1;
    return 0;
 
# Function to store the fibonacci numbers
# and their frequency in form a * x + b * y
def compute( a, b):
 
    # Storing the Fibonacci numbers
    for i in range(1, 100010):
        fib[i] = isFibonacci(i)
 
    # For loop to find all the possible
    # combinations of the Fibonacci numbers
    for x in range(1, 100010):
        for y in range(1, size):
 
            # Finding the number of ordered pairs
            if (fib[x] == 1 and fib[y] == 1 and a * x + b * y < 100010):
                freq[a * x + b * y] += 1
             
# Driver code
 
Q = 2
A = 5
B = 10
compute(A, B);
arr = [ 50, 150 ]
 
# Find the ordered pair for every query
for i in range(Q):
        print(freq[arr[i]], end=" ")
         
# This code is contributed by ANKITKUMAR34

C#

// C# program to find the count of
// Fibonacci pairs (x, y) which
// satisfy the equation Ax+By=N
using System;
 
class GFG{
      
static readonly int size = 10001;
  
// Array to store the Fibonacci numbers
static long []fib = new long[100010];
   
// Array to store the number of ordered pairs
static int []freq = new int[100010];
   
// Function to find if a number
// is a perfect square
static bool isPerfectSquare(int x)
{
    int s = (int) Math.Sqrt(x);
    return (s * s == x);
}
   
// Function that returns 1
// if N is non-fibonacci number else 0
static int isFibonacci(int n)
{
    // N is Fibinacci if one of
    // 5*n*n + 4 or 5*n*n - 4 or both
    // are perferct square
    if (isPerfectSquare(5 * n * n + 4)
        || isPerfectSquare(5 * n * n - 4))
        return 1;
    return 0;
}
   
// Function to store the fibonacci numbers
// and their frequency in form a * x + b * y
static void compute(int a, int b)
{
    // Storing the Fibonacci numbers
    for (int i = 1; i < 100010; i++) {
        fib[i] = isFibonacci(i);
    }
   
    // For loop to find all the possible
    // combinations of the Fibonacci numbers
    for (int x = 1; x < 100010; x++) {
        for (int y = 1; y < size; y++) {
   
            // Finding the number of ordered pairs
            if (fib[x] == 1 && fib[y] == 1
                && a * x + b * y < 100010) {
                freq[a * x + b * y]++;
            }
        }
    }
}
   
// Driver code
public static void Main(String[] args)
{
    int Q = 2, A = 5, B = 10;
    compute(A, B);
    int []arr = { 50, 150 };
   
    // Find the ordered pair for every query
    for (int i = 0; i < Q; i++) {
        Console.Write(freq[arr[i]]+ " ");
    }
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript program to find the count of
// Fibonacci pairs (x, y) which
// satisfy the equation Ax+By=N
var size = 10001
 
// Array to store the Fibonacci numbers
var fib = Array(100010).fill(0);
 
// Array to store the number of ordered pairs
var freq = Array(100010).fill(0);
 
// Function to find if a number
// is a perfect square
function isPerfectSquare(x)
{
    var s = parseInt(Math.sqrt(x));
    return (s * s == x);
}
 
// Function that returns 1
// if N is non-fibonacci number else 0
function isFibonacci(n)
{
 
    // N is Fibinacci if one of
    // 5*n*n + 4 or 5*n*n - 4 or both
    // are perferct square
    if (isPerfectSquare(5 * n * n + 4)
        || isPerfectSquare(5 * n * n - 4))
        return 1;
    return 0;
}
 
// Function to store the fibonacci numbers
// and their frequency in form a * x + b * y
function compute(a, b)
{
 
    // Storing the Fibonacci numbers
    for (var i = 1; i < 100010; i++)
    {
        fib[i] = isFibonacci(i);
    }
 
    // For loop to find all the possible
    // combinations of the Fibonacci numbers
    for (var x = 1; x < 100010; x++)
    {
        for (var y = 1; y < size; y++)
        {
 
            // Finding the number of ordered pairs
            if (fib[x] == 1 && fib[y] == 1
                && a * x + b * y < 100010)
            {
                freq[a * x + b * y]++;
            }
        }
    }
}
 
// Driver code
var Q = 2, A = 5, B = 10;
compute(A, B);
var arr = [ 50, 150 ];
 
// Find the ordered pair for every query
for (var i = 0; i < Q; i++)
{
    document.write(freq[arr[i]] + " ");
}
 
// This code is contributed by rutvik_56.
 

输出：

1 0