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Last digit of sum of numbers in the given range in the Fibonacci series

  • Difficulty Level : Medium
  • Last Updated : 26 Mar, 2021

Given two non-negative integers M, N which signifies the range [M, N] where M ≤ N, the task is to find the last digit of the sum of FM + FM+1… + FN where FK is the Kth Fibonacci number in the Fibonacci series
 

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, … 
 

Examples: 
 

Input: M = 3, N = 9 
Output:
Explanation: 
We need to find F3 + F4 + F5 + F6 + F7 + F8 + F9 
=> 2 + 3 + 5 + 8 + 13 + 21 + 34 = 86. 
Clearly, the last digit of the sum is 6.
Input: M = 3, N = 7 
Output:
Explanation: 
We need to find F3 + F4 + F5 + F6 + F7 
=> 2 + 3 + 5 + 8 + 13 = 31. 
Clearly, the last digit of the sum is 1. 
 

 



Naive Approach: The naive approach for this problem is to one by one find the sum of all Kth Fibonacci Numbers where K lies in the range [M, N] and return the last digit of the sum in the end. The time complexity for this approach is O(N) and this method fails for higher-ordered values of N. 
Efficient Approach: An efficient approach for this problem is to use the concept of Pisano Period
 

  • The idea is to calculate the sum of (M – 1) and N Fibonacci numbers respectively, and subtracting the last digit of the computed values.
  • This is because the last digit of the sum of all the Kth Fibonacci numbers such that K lies in the range [M, N] is equal to the difference of the last digits of the sum of all the Kth Fibonacci numbers in the range [0, N] and the sum of all the Kth Fibonacci numbers in the range [0, M – 1].
  • These values can respectively be calculated by the concept of the Pisano period in a very short time.
  • Let’s understand how the Pisano period works. The following table illustrates the first 10 Fibonacci numbers along with its values obtained when modulo 2 is performed on the numbers.
     
i012345678910
Fi011235813213455
Fi mod 2011011011010
  •  
  • Clearly, the Pisano period for (Fi mod 2) is 3 since 011 repeat itself and length(011) = 3.
  • Now, lets observe the following identity: 
     

7 = 2 * 3 + 1 
Dividend = (Quotient × Divisor) + Remainder 
=> F7 mod 2 = F1 mod 2 = 1. 
 

  • Therefore, instead of calculating the last digit of the sum of all numbers in the range [0, N], we simply calculate the sum until the remainder given that the Pisano period for Fi mod 10 is 60.

Below is the implementation of the above approach:
 

C++




// C++ program to calculate
// last digit of the sum of the
// fibonacci numbers from M to N
#include<bits/stdc++.h>
using namespace std;
 
// Calculate the sum of the first
// N Fibonacci numbers using Pisano
// period
long long fib(long long n)
{
     
    // The first two Fibonacci numbers
    long long f0 = 0;
    long long f1 = 1;
 
    // Base case
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    else
    {
        // Pisano period for % 10 is 60
        long long rem = n % 60;
 
        // Checking the remainder
        if(rem == 0)
           return 0;
 
        // The loop will range from 2 to
        // two terms after the remainder
        for(long long i = 2; i < rem + 3; i++)
        {
           long long f = (f0 + f1) % 60;
           f0 = f1;
           f1 = f;
        }
         
        long long s = f1 - 1;
        return s;
    }
}
 
// Driver Code
int main()
{
    long long m = 10087887;
    long long n = 2983097899;
 
    long long final = abs(fib(n) - fib(m - 1));
    cout << final % 10 << endl;
}
 
// This code is contributed by Bhupendra_Singh

Java




// Java program to calculate
// last digit of the sum of the
// fibonacci numbers from M to N
import java.util.*;
 
class GFG{
 
// Calculate the sum of the first
// N Fibonacci numbers using Pisano
// period
static int fib(long n)
{
     
    // The first two Fibonacci numbers
    int f0 = 0;
    int f1 = 1;
 
    // Base case
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    else
    {
         
        // Pisano period for % 10 is 60
        int rem = (int) (n % 60);
 
        // Checking the remainder
        if(rem == 0)
        return 0;
 
        // The loop will range from 2 to
        // two terms after the remainder
        for(int i = 2; i < rem + 3; i++)
        {
           int f = (f0 + f1) % 60;
           f0 = f1;
           f1 = f;
        }
         
        int s = f1 - 1;
        return s;
    }
}
 
// Driver Code
public static void main(String args[])
{
    int m = 10087887;
    long n = 2983097899L;
    int Final = (int)Math.abs(fib(n) -
                              fib(m - 1));
     
    System.out.println(Final % 10);
}
}
 
// This code is contributed by AbhiThakur

Python3




# Python3 program to calculate
# Last Digit of the sum of the
# Fibonacci numbers from M to N
 
# Calculate the sum of the first
# N Fibonacci numbers using Pisano
# period
def fib(n):
 
    # The first two Fibonacci numbers
    f0 = 0
    f1 = 1
 
    # Base case
    if (n == 0):
        return 0
    if (n == 1):
        return 1
    else:
 
        # Pisano Period for % 10 is 60
        rem = n % 60
 
        # Checking the remainder
        if(rem == 0):
            return 0
 
        # The loop will range from 2 to
        # two terms after the remainder
        for i in range(2, rem + 3):
            f =(f0 + f1)% 60
            f0 = f1
            f1 = f
 
        s = f1-1
        return(s)
 
# Driver code
if __name__ == '__main__':
     
    m = 10087887
    n = 2983097899
 
    final = fib(n)-fib(m-1)
 
    print(final % 10)

C#




// C# program to calculate
// last digit of the sum of the
// fibonacci numbers from M to N
using System;
 
class GFG{
 
// Calculate the sum of the first
// N fibonacci numbers using Pisano
// period
static int fib(long n)
{
     
    // The first two fibonacci numbers
    int f0 = 0;
    int f1 = 1;
 
    // Base case
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    else
    {
         
        // Pisano period for % 10 is 60
        int rem = (int)(n % 60);
 
        // Checking the remainder
        if(rem == 0)
           return 0;
 
        // The loop will range from 2 to
        // two terms after the remainder
        for(int i = 2; i < rem + 3; i++)
        {
           int f = (f0 + f1) % 60;
           f0 = f1;
           f1 = f;
        }
         
        int s = f1 - 1;
        return s;
    }
}
 
// Driver Code
public static void Main()
{
    int m = 10087887;
    long n = 2983097899L;
    int Final = (int)Math.Abs(fib(n) -
                              fib(m - 1));
     
    Console.WriteLine(Final % 10);
}
}
 
// This code is contributed by Code_Mech

Javascript




<script>
// javascript program to calculate
// last digit of the sum of the
// fibonacci numbers from M to N
 
    // Calculate the sum of the first
    // N Fibonacci numbers using Pisano
    // period
    function fib(n) {
 
        // The first two Fibonacci numbers
        var f0 = 0;
        var f1 = 1;
 
        // Base case
        if (n == 0)
            return 0;
        if (n == 1)
            return 1;
        else {
 
            // Pisano period for % 10 is 60
            var rem = parseInt( (n % 60));
 
            // Checking the remainder
            if (rem == 0)
                return 0;
 
            // The loop will range from 2 to
            // two terms after the remainder
            for (i = 2; i < rem + 3; i++) {
                var f = (f0 + f1) % 60;
                f0 = f1;
                f1 = f;
            }
 
            var s = f1 - 1;
            return s;
        }
    }
 
    // Driver Code
    var m = 10087887;
    var n = 2983097899;
    var Final = parseInt( Math.abs(fib(n) - fib(m - 1)));
 
    document.write(Final % 10);
 
// This code is contributed by Rajput-Ji
</script>
Output: 
5

 

Time Complexity: O(1), because this code runs almost 60 times for any input number. 
 

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