Calculate Huge Fibonacci number modulo M in C++ - c++

Question

Problem statement : Given two integers n and m, output Fn mod m (that is, the remainder of Fn when divided by m).
Input Format. The input consists of two integers n and m given on the same line (separated by a space).
Constraints. 1 ≤ n ≤ 10^18, 2 ≤ m ≤ 10^5
Output Format. Output Fn mod m.
I tried the following program and it didn't work. The method pi is returning the right Pisano period though for any number as per http://webspace.ship.edu/msrenault/fibonacci/fiblist.htm
#include <iostream>
long long pi(long long m) {
long long result = 2;
for (long long fn2 = 1, fn1 = 2 % m, fn = 3 % m;
fn1 != 1 || fn != 1;
fn2 = fn1, fn1 = fn, fn = (fn1 + fn2) % m
) {
result++;
}
return result;
}
long long get_fibonaccihuge(long long n, long long m) {
long long periodlength = pi(m);
int patternRemainder = n % periodlength;
long long *sum = new long long[patternRemainder];
sum[0] = 0;
sum[1] = 1;
for (int i = 2; i <= patternRemainder; ++i)
{
sum[i] = sum[i - 1] + sum[i - 2];
}
return sum[patternRemainder] % m;
}
int main() {
long long n, m;
std::cin >> n >> m;
std::cout << get_fibonaccihuge(n, m) << '\n';
}
The exact program/logic is working well in python as expected. What's wrong withthis cpp program ? Is it the data types ?

Answer 1

Performing 10^18 additions isn't going to be very practical. Even on a teraflop computer, 10^6 seconds is still 277 hours.
But 10^18 ~= 2^59.8 so there'll be up to 60 halving steps.
Calculate (a,b) --> (a^2 + b^2, 2ab + b^2) to go from (n-1,n)th to (2n-1,2n)th consecutive Fibonacci number pairs in one step.
At each step perform the modulus calculation for each operation. You'll need to accommodate integers up to 3*1010 ≤ 235 in magnitude (i.e. up to 35 bits).
(cf. a related older answer of mine).

Answer 2

This was my solution for this problem, it works well and succeeded in the submission test ...
i used a simpler way to get the pisoano period ( pisano period is the main tricky part in this problem ) ... i wish to be helpful
#include <iostream>
using namespace std;
unsigned long long get_fibonacci_huge_naive(unsigned long long n, unsigned long long m)
{
if (n <= 1)
return n;
unsigned long long previous = 0;
unsigned long long current = 1;
for (unsigned long long i = 0; i < n - 1; ++i)
{
unsigned long long tmp_previous = previous;
previous = current;
current = tmp_previous + current;
}
return current % m;
}
long long get_pisano_period(long long m)
{
long long a = 0, b = 1, c = a + b;
for (int i = 0; i < m * m; i++)
{
c = (a + b) % m;
a = b;
b = c;
if (a == 0 && b == 1)
{
return i + 1;
}
}
}
unsigned long long get_fibonacci_huge_faster(unsigned long long n, unsigned long long m)
{
n = n % get_pisano_period(m);
unsigned long long F[n + 1] = {};
F[0] = 0;
F[-1] = 1;
for (int i = 1; i <= n; i++)
{
F[i] = F[i - 1] + F[i - 2];
F[i] = F[i] % m;
}
return F[n];
}
int main()
{
unsigned long long n, m;
std::cin >> n >> m;
std::cout << get_fibonacci_huge_faster(n, m) << '\n';
}