Large Fibonacci Number (N can be upto 10^15000000) -SPOJ

I’m trying to solve FIBHARD from spoj. Here N goes very high. My first thought was that Euler totient function will work but doesn’t seems to be working. Any suggestion how to deal with such a large N. @vijju123 @galencolin @everule1


Will matrix exponentation work?

check N, how it will work? I have no idea, If you had any please share.

Matrix exponentation is of O(lgn). Btw fibonacci series has formula Fn = {[(√5 + 1)/2] ^ n} / √5

Please check N. Maybe we need to apply some maths on N. After that we can use some other technique. I think provided N neither suitable for Binet’s formula, nor for matrix expo.

Though i tried with this formula in python, It doen’t work.

On codeforces, there is a blog for calculating fib(n) for 10^18, but that’s not enough for this problem.

I think there is something to do with modulo 998244353 and FFT for this problem.

I hope, it will be useful to you.

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Maybe this is correct solution. I had no idea what FFT is.
Have a look This might help.
This is still not the complete solution(we have to still find the period of the function)

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Maybe you are right, but at present i’m trying to figure out what the heck is FFT.

This code is giving the desired output having input as large as 10^18.

You still need to read the large value of n and convert that into
long req = (n-1)%(mod*mod-1);
Here, req is < 10^18, so it wont be a problem, once req is obtained correctly from n.

#include <map>
#include <bits/stdc++.h>
#include <iostream>
using namespace std;

#define long long long
const long M = 998244353; // modulo
map<long, long> F;

long f(long n) {
	if (F.count(n)) return F[n];
	long k=n/2;
	if (n%2==0) { // n=2*k
		return F[n] = (f(k)*f(k) + f(k-1)*f(k-1)) % M;
	} else { // n=2*k+1
		return F[n] = (f(k)*f(k+1) + f(k-1)*f(k)) % M;

	long n;
	while (cin >> n) {
	    long req = (n-1)%(M*M-1);   // get req correct for large values of n
	    cout << (n==0 ? 0 : f(req)) << endl;

I used this article from codeforces and the comment of this question to arrive to this solution.

We can use fast doubling method using the formulas:

For nth Fibonacci, if n is :-

Even -> F(2k) = F(k)[2F(k+1)-F(k)]
Odd -> F(2k+1) = F(k+1)*F(k+1) + F(k)*F(k)

Also using BigInteger and modular arithmetic as well.

@rryadav please explain it?

Binet’s formula? Not sure how to apply it here though.

What about using the fact that the Fibonacci sequence modulo p is periodic? If you find the period, let’s call it P, then you just need to calculate the (N%P)-th term. P is always smaller than p^2 + 1.

[EDIT] I just submited this approach and worked. I couldn’t figure out a fast way to find P so I did an offline brute force search.


Yeah buddy i know last digit is repeating after 60 'th term similarly last two digits repeat after 300 terms , last three digit is repeating after 1500 terms and so on but how i will find period for mod m ?

As I said, I did it locally (offline) and then used it in my submition.

Brute Force
const unsigned int MOD=998244353;
unsigned int a=1,b=1;
long long P=1;
while (a!=0 || b!=1){
    ui tmp=a;
    b += tmp;
    if (b>=MOD) b-=MOD;

The value of P I found was 1996488708


Yeah same :slight_smile: Thanks (actually i forgot this is called Pisano Period )

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Good to know, I hadn’t heard of that before. Anyway, I will almost certainly forget this by tomorrow :confused:

Lmao … There is no need to learn the name , u know algorithm and that’s enough :slight_smile:

BTW here is my code It will give TLE …how can i optimize ?

the N%P in your code has many % operations, you can reduce that. There’s many resources of bigint clases implementations that you can check. Most of them have a definition of modulus operation.

Once you have found N%P you calculate the fibo of that number using a log(N%P) approach: matrix exponentiation or fast doubling.

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