## Minimum count of numbers in the Fibonacci series

### Question:

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

## Logic:

• Generate the Fibonacci series to find the count.
• Pass the series and n to the function.
• Find the nearest smaller Fibonacci number 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 become 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,i,count=0,n;
printf("Enter the number:");
scanf("%d",&n);
a=0,a=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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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.