# Uses python3
# Given two integers n and m, output Fn mod m (that is, the remainder of Fn when divided by m
def Huge_Fib(n,m):
if n == 0 : return 0
elif n == 1: return 1
else:
a,b = 0,1
for i in range(1,n):
a, b = b, (a+b) % m
print(b);
n,m = map(int, input().split());
Huge_Fib(n,m);
The code works very well. However, when I run a case as n = 99999999999999999, m = 2, it takes me much time. Do you have any better solutions?
Here is my solution, you don't have to go through 99999999999999999 iterations if you find the pisano period.
I also recommend that you watch this video: https://www.youtube.com/watch?v=Nu-lW-Ifyec
# Uses python3
import sys
def get_fibonacci_huge(n, m):
if n <= 1:
return n
arr = [0, 1]
previousMod = 0
currentMod = 1
for i in range(n - 1):
tempMod = previousMod
previousMod = currentMod % m
currentMod = (tempMod + currentMod) % m
arr.append(currentMod)
if currentMod == 1 and previousMod == 0:
index = (n % (i + 1))
return arr[index]
return currentMod
if __name__ == '__main__':
input = sys.stdin.read();
n, m = map(int, input.split())
print(get_fibonacci_huge(n,m))
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