N 可以表示为斐波那契数之和而不重复的方式数

给定一个数字N ，任务是找到整数N可以表示为斐波那契数的总和而不重复任何斐波那契数的方法数。

例子：

Input: N = 13
Output: 3
Explanation: 
The possible ways to select N as 13 are: {13} {8, 5} {8, 3, 2}. Note that it is not possible to select {5 + 5 + 3} because 5 appears twice.

Input: N = 87
Output: 5
Explanation:
The possible ways to select N as 13 are: {55 + 21 + 8 + 3}, {55 + 21 + 8 + 2 + 1}, {55 + 21 + 5 + 3 + 2 + 1}, {55 + 13 + 8 + 5 + 3 + 2 + 1}, {34 + 21 + 13 + 8 + 5 + 3 + 2 + 1}.

天真的方法：天真的想法是写出加起来等于给定数字N 的所有可能组合。检查任何组合是否有重复的整数，然后不要增加计数器，否则每次将计数增加 1。最后返回计数。

时间复杂度： O(N)
辅助空间： O(1)

Efficient Approach：思路是使用Dynamic Programming来优化上面的方法。以下是步骤：

• 让我们在斐波那契代码中表示一个数字。

Imagine Fibonacci coding by following way: the i-th bit of number corresponds to the i-th Fibonacci number. 
For Example: 16 = 13 + 3 will be written as 100100.

• 为每个正数编写斐波那契代码，使得没有两个相邻位为1 。
• 对于所有数字都是如此，因为如果有两个相邻的位是 1 位，那么我们可以通过斐波那契数的属性将其转换为单个1 位。我们称这种表示为规范表示。
• 获取规范表示。生成几个斐波那契数(约90 )，然后尝试以降序减去所有这些数。
• 让我们将给定数字的规范表示的 1 位位置按递增顺序存储到数组 v 中，并将任何1 位分解为两个1 位，如下所示：

Starting canonical representation:                                              1000000001
After decomposing leftmost 1-bit into two smaller 1-bits:           0110000001
After decomposing 2’nd leftmost 1-bit into two smaller 1-bits:   0101100001 
After decomposing 3’rd leftmost 1-bit into two smaller 1-bits:   0101011001
After decomposing 4’th leftmost 1-bit into two smaller 1-bits:   0101010111 

• 经过多次这样的操作，我们将得到下一个1 位(或数字的结尾)。这个1位也可以分解，但只能移位一位。
• 初始化一个 dp 数组dp1[] ， dp1[i]是表示一个数字的多种方法，该数字由数字的i最左边的1 位组成，用于所有剩余 1 位未分解的情况。此外，采用dp2[i] ，它标记了表示由数字的i 个最左边的1 位组成的数字的方式数，用于分解所有剩余1 位的情况。

For Example: N = 87

Canonical form of N = 101010100  
Other 4 possible representations of N are 101010011, 101001111, 100111111, 011111111

下面是相同的插图：

因此，答案是dp1[cnt] + dp2[cnt] ，其中cnt是规范表示中1 位的总数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
long long fib[101], dp1[101];
long long dp2[101], v[101];
 
// Function to generate the
// fibonacci number
void fibonacci()
{
    // First two number of
    // fibonacci sqequence
    fib[1] = 1;
    fib[2] = 2;
 
    for (int i = 3; i <= 87; i++) {
        fib[i] = fib[i - 1] + fib[i - 2];
    }
}
 
// Function to find maximum ways to
// represent num as the sum of
// fibonacci number
int find(int num)
{
    int cnt = 0;
 
    // Generate the Canonical form
    // of given number
    for (int i = 87; i > 0; i--) {
        if (num >= fib[i]) {
            v[cnt++] = i;
            num -= fib[i];
        }
    }
 
    // Reverse the number
    reverse(v, v + cnt);
 
    // Base condition of dp1 and dp2
    dp1[0] = 1;
    dp2[0] = (v[0] - 1) / 2;
 
    // Iterate from 1 to cnt
    for (int i = 1; i < cnt; i++) {
 
        // Calculate dp1[]
        dp1[i] = dp1[i - 1] + dp2[i - 1];
 
        // Calculate dp2[]
        dp2[i] = ((v[i] - v[i - 1]) / 2)
                     * dp2[i - 1]
                 + ((v[i] - v[i - 1] - 1) / 2)
                       * dp1[i - 1];
    }
 
    // Return final ans
    return (dp1[cnt - 1] + dp2[cnt - 1]);
}
 
// Driver Code
int main()
{
    // Function call to generate the
    // fibonacci numbers
    fibonacci();
 
    // Given Number
    int num = 13;
 
    // Function Call
    cout << find(num);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
static long[] fib = new long[101];
static long[] dp1 = new long[101];
static long[] dp2 = new long[101];
static long[] v = new long[101];
 
// Function to generate the
// fibonacci number
static void fibonacci()
{
     
    // First two number of
    // fibonacci sqequence
    fib[1] = 1;
    fib[2] = 2;
 
    for(int i = 3; i <= 87; i++)
    {
        fib[i] = fib[i - 1] + fib[i - 2];
    }
}
 
// Function to find maximum ways to
// represent num as the sum of
// fibonacci number
static long find(int num)
{
    int cnt = 0;
 
    // Generate the Canonical form
    // of given number
    for(int i = 87; i > 0; i--)
    {
        if (num >= fib[i])
        {
            v[cnt++] = i;
            num -= fib[i];
        }
    }
 
    // Reverse the number
    for(int i = 0; i < cnt / 2; i++)
    {
        long t = v[i];
        v[i] = v[cnt - i - 1];
        v[cnt - i - 1] = t;
    }
     
    // Base condition of dp1 and dp2
    dp1[0] = 1;
    dp2[0] = (v[0] - 1) / 2;
 
    // Iterate from 1 to cnt
    for(int i = 1; i < cnt; i++)
    {
         
        // Calculate dp1[]
        dp1[i] = dp1[i - 1] + dp2[i - 1];
 
        // Calculate dp2[]
        dp2[i] = ((v[i] - v[i - 1]) / 2) *
                 dp2[i - 1] +
                 ((v[i] - v[i - 1] - 1) / 2) *
                 dp1[i - 1];
    }
 
    // Return final ans
    return (dp1[cnt - 1] + dp2[cnt - 1]);
}
 
// Driver code
public static void main (String[] args)
{
     
    // Function call to generate the
    // fibonacci numbers
    fibonacci();
     
    // Given number
    int num = 13;
     
    // Function call
    System.out.print(find(num));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program for the above approach
fib = [0] * 101
dp1 = [0] * 101
dp2 = [0] * 101
v = [0] * 101
 
# Function to generate the
# fibonacci number
def fibonacci():
 
    # First two number of
    # fibonacci sqequence
    fib[1] = 1
    fib[2] = 2
 
    for i in range(3, 87 + 1):
        fib[i] = fib[i - 1] + fib[i - 2]
 
# Function to find maximum ways to
# represent num as the sum of
# fibonacci number
def find(num):
 
    cnt = 0
 
    # Generate the Canonical form
    # of given number
    for i in range(87, 0, -1):
        if(num >= fib[i]):
            v[cnt] = i
            cnt += 1
            num -= fib[i]
 
    # Reverse the number
    v[::-1]
 
    # Base condition of dp1 and dp2
    dp1[0] = 1
    dp2[0] = (v[0] - 1) // 2
 
    # Iterate from 1 to cnt
    for i in range(1, cnt):
 
        # Calculate dp1[]
        dp1[i] = dp1[i - 1] + dp2[i - 1]
 
        # Calculate dp2[]
        dp2[i] = (((v[i] - v[i - 1]) // 2) *
                  dp2[i - 1] +
                  ((v[i] - v[i - 1] - 1) // 2) *
                  dp1[i - 1])
 
    # Return final ans
    return dp1[cnt - 1] + dp2[cnt - 1]
 
# Driver Code
 
# Function call to generate the
# fibonacci numbers
fibonacci()
 
# Given number
num = 13
 
# Function call
print(find(num))
 
# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
 
class GFG{
 
static long[] fib = new long[101];
static long[] dp1 = new long[101]; 
static long[] dp2 = new long[101];
static long[] v = new long[101]; 
   
// Function to generate the 
// fibonacci number 
static void fibonacci() 
{
     
    // First two number of 
    // fibonacci sqequence 
    fib[1] = 1; 
    fib[2] = 2; 
   
    for(int i = 3; i <= 87; i++)
    { 
        fib[i] = fib[i - 1] + fib[i - 2]; 
    } 
} 
   
// Function to find maximum ways to 
// represent num as the sum of 
// fibonacci number 
static long find(long num) 
{ 
    int cnt = 0; 
   
    // Generate the Canonical form 
    // of given number 
    for(int i = 87; i > 0; i--) 
    { 
        if (num >= fib[i])
        { 
            v[cnt++] = i; 
            num -= fib[i]; 
        } 
    } 
   
    // Reverse the number 
    for(int i = 0; i < cnt / 2; i++)
    { 
        long t = v[i]; 
        v[i] = v[cnt - i - 1]; 
        v[cnt - i - 1] = t; 
    }
       
    // Base condition of dp1 and dp2 
    dp1[0] = 1; 
    dp2[0] = (v[0] - 1) / 2; 
   
    // Iterate from 1 to cnt 
    for(int i = 1; i < cnt; i++)
    { 
           
        // Calculate dp1[] 
        dp1[i] = dp1[i - 1] + dp2[i - 1]; 
   
        // Calculate dp2[] 
        dp2[i] = ((v[i] - v[i - 1]) / 2) * 
                 dp2[i - 1] +
                 ((v[i] - v[i - 1] - 1) / 2) * 
                 dp1[i - 1]; 
    } 
   
    // Return final ans 
    return (dp1[cnt - 1] + dp2[cnt - 1]); 
} 
 
// Driver code
static void Main()
{
     
    // Function call to generate the 
    // fibonacci numbers 
    fibonacci(); 
       
    // Given number 
    int num = 13; 
       
    // Function call 
    Console.Write(find(num));
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript

 
// Javascript program for the above approach
var fib = Array(101).fill(0);
var dp1 = Array(101).fill(0);
var dp2 = Array(101).fill(0);
var v = Array(101).fill(0);
 
// Function to generate the
// fibonacci number
function fibonacci()
{
     
    // First two number of
    // fibonacci sqequence
    fib[1] = 1;
    fib[2] = 2;
 
    for(i = 3; i <= 87; i++)
    {
        fib[i] = fib[i - 1] + fib[i - 2];
    }
}
 
// Function to find maximum ways to
// represent num as the sum of
// fibonacci number
function find(num)
{
    var cnt = 0;
     
    // Generate the Canonical form
    // of given number
    for(i = 87; i > 0; i--)
    {
        if (num >= fib[i])
        {
            v[cnt++] = i;
            num -= fib[i];
        }
    }
 
    // Reverse the number
    for(i = 0; i < cnt / 2; i++)
    {
        var t = v[i];
        v[i] = v[cnt - i - 1];
        v[cnt - i - 1] = t;
    }
 
    // Base condition of dp1 and dp2
    dp1[0] = 1;
    dp2[0] = parseInt((v[0] - 1) / 2);
 
    // Iterate from 1 to cnt
    for(i = 1; i < cnt; i++)
    {
         
        // Calculate dp1
        dp1[i] = dp1[i - 1] + dp2[i - 1];
 
        // Calculate dp2
        dp2[i] = parseInt((v[i] - v[i - 1]) /
                           2) * dp2[i - 1] +
                 parseInt((v[i] - v[i - 1] - 1) /
                           2) * dp1[i - 1];
    }
 
    // Return final ans
    return (dp1[cnt - 1] + dp2[cnt - 1]);
}
 
// Driver code
 
// Function call to generate the
// fibonacci numbers
fibonacci();
 
// Given number
var num = 13;
 
// Function call
document.write(find(num));
 
// This code is contributed by todaysgaurav
 

输出：

3

时间复杂度： O(log N)
辅助空间： O(log N)