Write a ‘C’ program to check if a number is Keith Number.

- A Keith number is an n-digit integer (NUM>9) such that if a Fibonacci-like sequence (in which each term in the sequence is the sum of the n previous terms) is formed with the first n terms taken as the decimal digits of the number NUM, then NUM itself occurs as a term in the sequence.

**Example:**

**75 – Keith number**7,5

7+5=12

5+12=17

12+17=29

17+29=46

29+46=75

75 Occurs in the generated sequence and so it is a Keith number.

**341 – Not a Keith number**3, 4, 1

3+4+1=8,

4+1+8=13,

1+8+13=22,

8+13+22=43,

13+22+43=78,

22+43+78=143

78+43+143=264

43+143+264=450

341 is not a Keith number since it will not occur in the generated sequence.

**Input** : ‘n’**Output** : Yes / No**Constraints**:

1 <= ‘n’ <= 10000000

**Code:**

```
# include <stdio.h>
# include <stdlib.h>
int lenCount(int nm)
{ int ctr=0;
while(nm>0)
{
nm=nm/10;
ctr++;
}
return ctr;
}
int main()
{
int num1=0,arr1[10],num2=0,flg=0,i=0,sum=0;
scanf("%d",&num1);
num2=num1;
for(i=lenCount(num2)-1;i>=0;i--)
{
arr1[i]=num1 % 10;
num1/=10;
}
while(flg==0)
{
for(i=0;i<lenCount(num2);i++)
sum+=arr1[i];
if(sum==num2)
{
printf("Yes");
flg=1;
}
if(sum>num2)
{
printf("No");
flg=1;
}
for(i=0;i<lenCount(num2);i++)
{
if(i!=lenCount(num2)-1)
arr1[i]=arr1[i+1];
else
arr1[i]=sum;
}
sum=0;
}
}
```

Code language: C/AL (cal)