HackerRank Fundamentals – Is Fibo

Challenge: Is Fibo

Subdomeniu: Fundamentals (fundamentals)

Scor cont: 20.0 / 20

Submission status: Accepted

Submission score: 1.0

Submission ID: 464721252

Limbaj: python3

Link challenge: https://www.hackerrank.com/challenges/is-fibo/problem

Cerinta

You are given an integer, $N$. Write a program to determine if $N$ is an element of the _Fibonacci sequence_.  

The first few elements of the Fibonacci sequence are $0,1,1,2,3,5,8,13, \cdots$. A Fibonacci sequence is one where every element is a sum of the previous two elements in the sequence. The first two elements are $0$ and $1$. 

Formally:   
$$\begin{align*}
fib_{0} &= 0 \\\  
fib_{1} &= 1 \\\  
\vdots \\\
fib_{n} &= fib_{n-1} + fib_{n-2} \forall n > 1
\end{align*}$$

**Function Description**   

Complete the *isFibo* function in the editor below.   

*isFibo* has the following parameters:   
- *int n:* the number to check   

**Returns**   
- *string:* either `IsFibo` or `IsNotFibo`

**Input Format**  
The first line contains $t$, number of test cases.   
$t$ lines follow. Each line contains an integer $n$.

**Constraints**  
$1 \le t \le 10^5$  
$1 \le n \le 10^{10}$

**Sample Input**  
<pre>
STDIN	Function
-----	--------
3		t = 3
5		n = 5
7		n = 7
8		n = 8
</pre>

**Sample Output**  

	IsFibo
    IsNotFibo
    IsFibo

**Explanation**  
$5$ is a Fibonacci number given by $\text{fib}_5  = 3 + 2$  
$7$ is not a Fibonacci number  
$8$ is a Fibonacci number given by $\text{fib}_6  = 5 + 3$  

**Time Limit**  
The time limit for this challenge is given [here](https://www.hackerrank.com/environment).

Cod sursa

#!/bin/python3

import os
import sys

# Complete the solve function below.
def solve(n):
    if n in q:
        return "IsFibo"
    else:
        return "IsNotFibo"
q=[0]
a=0
b=1
for  i in range(0,10000):
    c=a+b
    a=b
    b=c
    q.append(c)
if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    t = int(input())

    for t_itr in range(t):
        n = int(input())

        result = solve(n)

        fptr.write(result + '\n')

    fptr.close()