将数字表示为 k 个斐波那契数之和的多种方法

给定两个数字 N 和 K。找出将 N 表示为 K 个斐波那契数之和的方法数。
例子：

Input : n = 12, k = 1 
Output : 0

Input : n = 13, k = 3
Output : 2
Explanation : 2 + 3 + 8, 3 + 5 + 5.  

方法：斐波那契数列是 f(0)=1, f(1)=2 和 f(i)=f(i-1)+f(i-2) 对 i>1。让我们假设 F(x, k, n) 是使用 f(0), f(1), …f(n-1) 中的 k 个数字来形成和 x 的方法数。要找到 F(x, k, n) 的循环，请注意有两种情况：总和中是否为 f(n-1)。

• 如果 f(n-1) 不在总和中，则 x 使用 f(0), f(1), …, f(n-2) 中的 k 个数字形成一个总和。
• 如果 f(n-1) 在总和中，则剩余的 xf(n-1) 由 f(0), f(1), …, f(n-1) 中的 k-1 个数字构成。 (请注意 f(n-1) 仍然包括在内，因为允许重复数字。)。

所以递归关系将是：

F(x, k, n)= F(x, k, n-1)+F(x-f(n-1), k-1, n)

基本情况：

• 如果 k=0，则系列中的数字为零，因此总和只能为 0。因此，F(0, 0, n)=1。
• F(x, 0, n)=0，如果 x 不等于 0。

此外，还有其他情况使 F(x, k, n)=0，如下所示：

• 如果 k>0 且 x=0，因为至少有一个正数必然导致正和。
• 如果 k>0 且 n=0，因为没有可能剩下的数字选择。
• 如果 x<0，因为无法使用有限数量的非负数形成负和。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// to store fibonacci numbers
// 42 second number in fibonacci series
// largest possible integer
int fib[43] = { 0 };
 
// Function to generate fibonacci series
void fibonacci()
{
    fib[0] = 1;
    fib[1] = 2;
    for (int i = 2; i < 43; i++)
        fib[i] = fib[i - 1] + fib[i - 2];
}
 
// Recursive function to return the
// number of ways
int rec(int x, int y, int last)
{
    // base condition
    if (y == 0) {
        if (x == 0)
            return 1;
        return 0;
    }
    int sum = 0;
    // for recursive function call
    for (int i = last; i >= 0 and fib[i] * y >= x; i--) {
        if (fib[i] > x)
            continue;
        sum += rec(x - fib[i], y - 1, i);
    }
    return sum;
}
 
// Driver code
int main()
{
    fibonacci();
    int n = 13, k = 3;
    cout << "Possible ways are: "
         << rec(n, k, 42);
 
    return 0;
}

Java

//Java implementation of above approach
public class AQW {
 
    //to store fibonacci numbers
    //42 second number in fibonacci series
    //largest possible integer
    static int fib[] = new int[43];
 
    //Function to generate fibonacci series
    static void fibonacci()
    {
     fib[0] = 1;
     fib[1] = 2;
     for (int i = 2; i < 43; i++)
         fib[i] = fib[i - 1] + fib[i - 2];
    }
 
    //Recursive function to return the
    //number of ways
    static int rec(int x, int y, int last)
    {
     // base condition
     if (y == 0) {
         if (x == 0)
             return 1;
         return 0;
     }
     int sum = 0;
     // for recursive function call
     for (int i = last; i >= 0 && fib[i] * y >= x; i--) {
         if (fib[i] > x)
             continue;
         sum += rec(x - fib[i], y - 1, i);
     }
     return sum;
    }
 
    //Driver code
    public static void main(String[] args) {
         
        fibonacci();
         int n = 13, k = 3;
         System.out.println("Possible ways are: "+ rec(n, k, 42));
 
    }
 
}

Python3

# Python3 implementation of the above approach
 
# To store fibonacci numbers 42 second
# number in fibonacci series largest
# possible integer
fib = [0] * 43
 
# Function to generate fibonacci
# series
def fibonacci():
 
    fib[0] = 1
    fib[1] = 2
    for i in range(2, 43):
        fib[i] = fib[i - 1] + fib[i - 2]
 
# Recursive function to return the
# number of ways
def rec(x, y, last):
 
    # base condition
    if y == 0:
        if x == 0:
            return 1
        return 0
     
    Sum, i = 0, last
     
    # for recursive function call
    while i >= 0 and fib[i] * y >= x:
        if fib[i] > x:
            i -= 1
            continue
        Sum += rec(x - fib[i], y - 1, i)
        i -= 1
         
    return Sum
 
# Driver code
if __name__ == "__main__":
 
    fibonacci()
    n, k = 13, 3
    print("Possible ways are:", rec(n, k, 42))
 
# This code is contributed
# by Rituraj Jain

C#

// C# implementation of above approach
using System;
   
class GFG
{
    // to store fibonacci numbers
    // 42 second number in fibonacci series
    // largest possible integer
    static int[] fib = new int[43];
       
    // Function to generate fibonacci series
    public static void fibonacci()
    {
        fib[0] = 1;
        fib[1] = 2;
        for (int i = 2; i < 43; i++)
            fib[i] = fib[i - 1] + fib[i - 2];
    }
       
    // Recursive function to return the 
    // number of ways 
    public static int rec(int x, int y, int last)
    {
        // base condition
        if (y == 0) {
            if (x == 0)
                return 1;
            return 0;
        }
        int sum = 0;
        // for recursive function call
        for (int i = last; i >= 0 && fib[i] * y >= x; i--) {
            if (fib[i] > x)
                continue;
            sum += rec(x - fib[i], y - 1, i);
        }
        return sum;
    }
       
    // Driver code
    static void Main()
    {
        for(int i = 0; i < 43; i++)
            fib[i] = 0;
        fibonacci();
        int n = 13, k = 3;
        Console.Write("Possible ways are: " + rec(n, k, 42));
    }
    //This code is contributed by DrRoot_
}

PHP

= 0 and
         $fib[$i] * $y >= $x; $i--)
    {
        if ($fib[$i] > $x)
            continue;
        $sum += rec($x - $fib[$i],
                    $y - 1, $i);
    }
    return $sum;
}
 
// Driver code
fibonacci();
$n = 13;
$k = 3;
echo "Possible ways are: " .
            rec($n, $k, 42);
 
// This code is contributed by mits
?>

Javascript

//Javascript implementation of above approach
 
 
     
    //to store fibonacci numbers
    //42 second number in fibonacci series
    //largest possible integer
    let fib=new Array(43);
     
    //Function to generate fibonacci series
    function fibonacci()
    {
        fib[0] = 1;
     fib[1] = 2;
     for (let i = 2; i < 43; i++)
         fib[i] = fib[i - 1] + fib[i - 2];
    }
     
     
    //Recursive function to return the
    //number of ways
    function rec(x,y,last)
    {
        // base condition
     if (y == 0) {
         if (x == 0)
             return 1;
         return 0;
      }
     let sum = 0;
     // for recursive function call
     for (let i = last; i >= 0 && fib[i] * y >= x; i--) {
         if (fib[i] > x)
             continue;
         sum += rec(x - fib[i], y - 1, i);
     }
     return sum;
    }
     
    //Driver code
    fibonacci();
     let n = 13, k = 3;
     document.write("Possible ways are: "+ rec(n, k, 42));
     
     
     
// This code is contributed by rag2127

输出：

Possible ways are: 2

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