# Minimum count of numbers in the Fibonacci number

### Question:

To find the minimum count of numbers in the Fibonacci series and their sum is equal to the number n.

### Fibonacci Series

The Fibonacci sequence is a set of numbers that starts with a one or a zero, followed by a one, and proceeds based on the rule that each number is equal to the sum of the preceding two numbers. If the Fibonacci sequence is denoted F (n), where n is the first term in the sequence, the following equation obtains for n = 0, where the first two terms are defined as 0 and 1 by convention:

F (0) = 0, 1, 1, 2, 3, 5, 8, 13, 21, 34 …

In some texts, it is customary to use n = 1. In that case, the first two terms are defined as 1 and 1 by default, and therefore:

F (1) = 1, 1, 2, 3, 5, 8, 13, 21, 34 …

More About Fibonacci

## Logic:

• Generate the Fibonacci series to find the count.
• Pass the series and n to the function.
• Find the nearest smaller Fibonacci no to n and increment the count.
• Then subtract the nearest number with n and assign value to n.
• And Recursively repeat the process until n becomes zero.

## Program:

```#include<stdio.h>
int sum(int a[],int n)
{
static int c=0;        //minimum no of count to reach the number n
int i;
for(i=0;i<30;i++)
{
if(a[i]>n)
{
printf("%d ",a[i-1]);
n=n-a[i-1];   //a[i-1] is the nearest smaller number to n
c++;          //increment the count
break;
}
}
if(n==0)              //return the count when number is zero
return c;
else
sum(a,n);        //call the function recursively
}
void main()
{
int a[100],i,count=0,n;
printf("Enter the number:");
scanf("%d",&n);
a[0]=0,a[1]=1;
for(i=2;i<30;i++)            //generates the Fibonacci series
{
a[i]=a[i-1]+a[i-2];
}
count=sum(a,n);             //Pass the value and array to find the minimum count
printf("Count: %d",count);
}
```

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### Vignesh

A Computer Science graduate who likes to make things simpler. When he’s not working, you can find him surfing the web, learning facts, tricks and life hacks. He also enjoys movies in his leisure time.
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