Program to find last two digits of Nth Fibonacci number (original) (raw)
Last Updated : 24 Mar, 2023
Given a number ‘n’, write a function that prints the last two digits of n-th (‘n’ can also be a large number) Fibonacci number.
Examples:
Input : n = 65 Output : 65
Input : n = 365 Output : 65
Recommended: Please solve it on “_PRACTICE_” first, before moving on to the solution.
A simple solution is to find n-th Fibonacci number and print its last two digit. But N can be very large, so it wouldn't work.
A better solution is to use the fact that after 300-th Fibonacci number last two digits starts repeating.
- Find m = n % 300.
- Return m-th Fibonacci number.
C++ `
// Program to find last two digits of n-th // Fibonacci number #include<bits/stdc++.h> using namespace std; typedef long long int ll;
// Fills f[] with first 300 fibonacci numbers void precomput(ll f[]) { /* 0th and 1st number of the series are 0 and 1*/ f[0] = 0; f[1] = 1;
/* Add the previous 2 numbers in the series
and store last two digits of result */
for (ll i = 2; i < 300; i++)
f[i] = (f[i-1] + f[i-2])%100;}
// Returns last two digits of n'th Fibonacci // Number int findLastDigit(ll f[], int n) { return f[n%300]; }
/* Driver program to test above function */ int main () { // Precomputing units digit of first 300 // Fibonacci numbers ll f[300] = {0}; precomput(f);
ll n = 1;
cout << findLastDigit(f, n) << endl;
n = 61;
cout << findLastDigit(f, n) << endl;
n = 7;
cout << findLastDigit(f, n) << endl;
n = 67;
cout << findLastDigit(f, n) << endl;
return 0;}
Java
// Program to find last two digits of // n-th Fibonacci number import java.util.Arrays; class GFG {
// Fills f[] with first 300
// fibonacci numbers
static void precomput(long f[])
{
/* 0th and 1st number of the
series are 0 and 1*/
f[0] = 0;
f[1] = 1;
/* Add the previous 2 numbers in
the series and store last two
digits of result */
for (int i = 2; i < 300; i++)
f[i] = (f[i-1] + f[i-2]) % 100;
}
// Returns last two digits of n'th
// Fibonacci Number
static long findLastDigit(long f[], int n)
{
return (f[(n%300)]);
}
/* Driver program to test above function */
public static void main (String args[])
{
// Precomputing units digit of
// first 300 Fibonacci numbers
long f[] = new long[300];
Arrays.fill(f,0);
precomput(f);
int n = 1;
System.out.println(findLastDigit(f, n));
n = 61;
System.out.println(findLastDigit(f, n));
n = 7;
System.out.println(findLastDigit(f, n));
n = 67;
System.out.println(findLastDigit(f, n));
}}
/This code is contributed by Nikita Tiwari./
Python3
Python code to find last two
digits of n-th Fibonacci number
def precomput(f):
# 0th and 1st number of the series
# are 0 and 1
f.append(0)
f.append(1)
# Add the previous 2 numbers in the series
# and store last two digits of result
for i in range(2,300):
f.append((f[i-1] + f[i-2]) % 100)Returns last two digits of
n'th Fibonacci Number
def findLastDigit(f,n): return f[n%300]
driver code
f = list() precomput(f) n = 1 print(findLastDigit(f, n)) n = 61 print(findLastDigit(f, n)) n = 7 print(findLastDigit(f, n)) n = 67 print(findLastDigit(f, n))
This code is contributed by "Abhishek Sharma 44"
C#
// Program to find last two digits of // n-th Fibonacci number using System;
class GFG {
// Fills f[] with first 300
// fibonacci numbers
static void precomput(long []f)
{
/* 0th and 1st number of the
series are 0 and 1*/
f[0] = 0;
f[1] = 1;
/* Add the previous 2 numbers in
the series and store last two
digits of result */
for (int i = 2; i < 300; i++)
f[i] = (f[i-1] + f[i-2]) % 100;
}
// Returns last two digits of n'th
// Fibonacci Number
static long findLastDigit(long []f, int n)
{
return (f[(n % 300)]);
}
/* Driver program to test above function */
public static void Main ()
{
// Precomputing units digit of
// first 300 Fibonacci numbers
long []f = new long[300];
precomput(f);
int n = 1;
Console.WriteLine(findLastDigit(f, n));
n = 61;
Console.WriteLine(findLastDigit(f, n));
n = 7;
Console.WriteLine(findLastDigit(f, n));
n = 67;
Console.WriteLine(findLastDigit(f, n));
}}
// This code is contributed by anuj_67.
PHP
JavaScript
`
Output:
1 61 13 53
Time Complexity: O(300), it will run in constant time.
Auxiliary Space: O(300), no extra space is required, so it is a constant.
Approach 2: Iterative Approach:
In this implementation, we simply iterate through the Fibonacci sequence up to the n-th number, keeping track of the last two digits of each number using the modulo operator. This allows us to compute the last two digits of even large Fibonacci numbers without needing to store all of the previous numbers in the sequence.
Here Are steps of this approach:
- We start by initializing the first two terms of the sequence to 0 and 1 respectively. Then, for each i from 2 to n, we calculate the ith Fibonacci number by adding the (i-1)th and (i-2)th Fibonacci numbers.
- In each iteration, we keep track of the previous two Fibonacci numbers using two variables, prev and curr. Initially, prev is set to 0 and curr is set to 1. In each iteration, we update prev to curr and curr to the sum of prev and curr. After n iterations, the value of curr is the nth Fibonacci number.
- This approach has a time complexity of O(n) and a space complexity of O(1) since we only need to store two variables at any given time. This makes it more efficient than the recursive approach, which has a time complexity of O(2^n) and a space complexity of O(n). However, the iterative approach can be harder to understand than the recursive approach for some people, as it involves more explicit manipulation of variables and requires a good understanding of loops and basic arithmetic operations. C++ `
#include <bits/stdc++.h> using namespace std;
// Function to find last two digits of the n-th Fibonacci number int findLastTwoDigits(int n) { int a = 0, b = 1; if (n == 0) return a; if (n == 1) return b;
for (int i = 2; i <= n; i++) {
int c = (a + b) % 100;
a = b;
b = c;
}
return b;}
int main() { int n = 1; cout << findLastTwoDigits(n) << endl; n = 61; cout << findLastTwoDigits(n) << endl; n = 7; cout << findLastTwoDigits(n) << endl; n = 67; cout << findLastTwoDigits(n) << endl; return 0; }
Java
import java.util.*;
public class LastTwoDigitsFibonacci {
// Function to find last two digits of n-th Fibonacci number
public static int findLastDigit(int n) {
int f1 = 0, f2 = 1, f3;
// Iterate till n-1
for (int i = 1; i < n; i++) {
f3 = (f1 + f2) % 100;
f1 = f2;
f2 = f3;
}
// Return last two digits of n-th Fibonacci number
return f2;
}
// Driver code
public static void main(String[] args) {
int n = 1;
System.out.println(findLastDigit(n));
n = 61;
System.out.println(findLastDigit(n));
n = 7;
System.out.println(findLastDigit(n));
n = 67;
System.out.println(findLastDigit(n));
}}
Python3
Program to find last two digits of n-th
Fibonacci number
import math
Returns last two digits of n'th Fibonacci
Number
def findLastDigit(n): a = 0 b = 1 if (n <= 1): return n else: for i in range(2, n + 1): c = (a + b) % 100 a = b b = c return b
Driver program to test above function
n = 1 print(findLastDigit(n)) n = 61 print(findLastDigit(n)) n = 7 print(findLastDigit(n)) n = 67 print(findLastDigit(n))
C#
using System;
class Program { static int FindLastDigit(int n) { int a = 0, b = 1, c; for (int i = 2; i <= n; i++) { c = (a + b) % 100; a = b; b = c; } return b; }
static void Main(string[] args)
{
int n = 1;
Console.WriteLine(FindLastDigit(n));
n = 61;
Console.WriteLine(FindLastDigit(n));
n = 7;
Console.WriteLine(FindLastDigit(n));
n = 67;
Console.WriteLine(FindLastDigit(n));
}}
JavaScript
function findLastTwoDigits(n) { let a = 0, b = 1; if (n === 0) return a; if (n === 1) return b;
for (let i = 2; i <= n; i++) {
let c = (a + b) % 100;
a = b;
b = c;
}
return b;}
console.log(findLastTwoDigits(1)); console.log(findLastTwoDigits(61)); console.log(findLastTwoDigits(7)); console.log(findLastTwoDigits(67));
`
Time Complexity: O(N).
Auxiliary Space: O(1), no extra space is required, so it is a constant.