Find the last digit of given series

Given an integer n, find the last digit of this sequence,\displaystyle S =\sum_{i=0,\ 2^i\le n}^\infty \sum_{j=0}^n 2^{2^i + 2j}
i.e. Summation of F(n) from i = 0 to 2i ≤ n, where F(n) is the summation of 22i+2j Where j varies from 0 to n. n can varies from 0 to 1017
Examples:

Input: 2
Output: 6
Explanation:
After computing the above expression, the value
obtained is 216. Hence the last digit is 6.

Input: 3
Output: 0

A Naive approach is to run two loops, one for ‘i’ and other for ‘j’ and compute the value after taking (modulo 10) with every calculated value. But this approach will definitely time out for large value of n.

An efficient approach is to expand the above expression in a general form that can easily be calculated.



\displaystyle S =\sum_{i=0,\ 2^i\le n}^\infty \sum_{j=0}^n 2^{2^i + 2j}

\implies \displaystyle \sum_{i=0,\ 2^i\le n}^\infty \sum_{j=0}^n 2^{2^i}\cdot 2^{2j}

\implies\displaystyle\sum_{i=0,\ 2^i\le n}^\infty 2^{2^i} \cdot \sum_{j=0}^n 2^{2j}

\implies\displaystyle\sum_{i=0,\ 2^i\le n}^\infty 2^{2^i} \cdot \sum_{j=0}^n 4^j

  1. First expression \sum_{i=0,\ 2^i\le n}^\infty 2^{2^i} can be calculated directly by iterating a loop for all values of ‘i’ till 2i ≤ n.
  2. Second expression \sum_{j=0}^n 4^j can be calculated easily by using Geometric series formula i.e.,
     \sum_{j=0}^n r^k = \dfrac{r^{n+1} - 1}{r-1}
  3. Final answer will be the product of both these calculated result in both the steps. But after performing any calculation part of these expression, we have to take modulo with 10 to avoid overflow.
  4. See below programs to understand more.

    C++

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // C++ program to calculate to find last
    // digit of above expression
    #include <bits/stdc++.h>
    using namespace std;
      
    /* Iterative Function to calculate (x^y)%p in O(log y) */
    long long powermod(long long x, long long y, long long p)
    {
        long long res = 1; // Initialise result
      
        x = x % p; // Update x if it is more than or
                  // equal to p
      
        while (y > 0) {
      
            // If y is odd, multiply x with result
            if (y & 1LL)
                res = (res * x) % p;
      
            // y must be even now
            y = y >> 1LL; // y = y/2
            x = (x * x) % p;
        }
        return res;
    }
      
    // Returns modulo inverse of a with respect to m 
    // using extended Euclid Algorithm
    long long modInverse(long long a, long long m)
    {
        long long m0 = m, t, q;
        long long x0 = 0, x1 = 1;
      
        if (m == 1)
            return 0;
      
        while (a > 1) {
      
            // q is quotient
            q = a / m;
      
            t = m;
      
            // m is remainder now, process same as
            // Euclid's algo
            m = a % m, a = t;
      
            t = x0;
      
            x0 = x1 - q * x0;
      
            x1 = t;
        }
      
        // Make x1 positive
        if (x1 < 0)
            x1 += m0;
      
        return x1;
    }
      
    // Function to calculate the above expression
    long long evaluteExpression(long long& n)
    {
        // Initialize the result
        long long firstsum = 0, mod = 10;
      
        // Compute first part of expression
        for (long long i = 2, j = 0; (1LL << j) <= n; i *= i, ++j)
            firstsum = (firstsum + i) % mod;
      
        // Compute second part of expression
        // i.e., ((4^(n+1) - 1) / 3) mod 10
        // Since division of 3 in modulo can't
        // be performed directly therefore we
        // need to find it's modulo Inverse
        long long secondsum = (powermod(4LL, n + 1, mod) - 1) * 
                               modInverse(3LL, mod);
      
        return (firstsum * secondsum) % mod;
    }
      
    // Driver code
    int main()
    {
        long long n = 3;
        cout << evaluteExpression(n) << endl;
      
        n = 10;
        cout << evaluteExpression(n);
      
        return 0;
    }

    chevron_right

    
    

    Java

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // Java program to calculate to find last 
    // digit of above expression 
      
    class GFG{
    /* Iterative Function to calculate (x^y)%p in O(log y) */
    static long powermod(long x, long y, long p) 
        long res = 1; // Initialise result 
      
        x = x % p; // Update x if it is more than or 
                // equal to p 
      
        while (y > 0) { 
      
            // If y is odd, multiply x with result 
            if ((y & 1L)>0
                res = (res * x) % p; 
      
            // y must be even now 
            y = y >> 1L; // y = y/2 
            x = (x * x) % p; 
        
        return res; 
      
    // Returns modulo inverse of a with respect to m 
    // using extended Euclid Algorithm 
    static long modInverse(long a, long m) 
        long  m0 = m, t, q; 
        long  x0 = 0, x1 = 1
      
        if (m == 1
            return 0
      
        while (a > 1) { 
      
            // q is quotient 
            q = a / m; 
      
            t = m; 
      
            // m is remainder now, process same as 
            // Euclid's algo 
            m = a % m;
            a = t; 
      
            t = x0; 
      
            x0 = x1 - q * x0; 
      
            x1 = t; 
        
      
        // Make x1 positive 
        if (x1 < 0
            x1 += m0; 
      
        return x1; 
      
    // Function to calculate the above expression 
    static long evaluteExpression(long n) 
        // Initialize the result 
        long firstsum = 0, mod = 10
      
        // Compute first part of expression 
        for (long i = 2, j = 0; (1L << j) <= n; i *= i, ++j) 
            firstsum = (firstsum + i) % mod; 
      
        // Compute second part of expression 
        // i.e., ((4^(n+1) - 1) / 3) mod 10 
        // Since division of 3 in modulo can't 
        // be performed directly therefore we 
        // need to find it's modulo Inverse 
        long secondsum = (powermod(4L, n + 1, mod) - 1) * 
                            modInverse(3L, mod); 
      
        return (firstsum * secondsum) % mod; 
      
    // Driver code 
    public static void main(String[] args) 
        long n = 3
        System.out.println(evaluteExpression(n)); 
      
        n = 10
        System.out.println(evaluteExpression(n)); 
      
    }
    // This code is contributed by mits

    chevron_right

    
    

    Python3

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    # Python3 program to calculate to find last 
    # digit of above expression 
      
    # Iterative Function to calculate (x^y)%p in O(log y) 
    def powermod(x, y, p): 
       
        res = 1; # Initialise result 
      
        x = x % p; # Update x if it is more than or 
                # equal to p 
      
        while (y > 0):
      
            # If y is odd, multiply x with result 
            if ((y & 1)>0): 
                res = (res * x) % p; 
      
            # y must be even now 
            y = y >> 1; # y = y/2 
            x = (x * x) % p;
              
        return res; 
      
    # Returns modulo inverse of a with respect to m 
    # using extended Euclid Algorithm 
    def modInverse(a, m):
       
        m0 = m; 
        x0 = 0;
        x1 = 1
      
        if (m == 1): 
            return 0
      
        while (a > 1): 
      
            # q is quotient 
            q = int(a / m); 
      
            t = m; 
      
            # m is remainder now, process same as 
            # Euclid's algo 
            m = a % m;
            a = t; 
      
            t = x0; 
      
            x0 = x1 - q * x0; 
      
            x1 = t; 
      
        # Make x1 positive 
        if (x1 < 0): 
            x1 += m0; 
      
        return x1; 
      
    # Function to calculate the above expression 
    def evaluteExpression(n): 
       
        # Initialize the result 
        firstsum = 0;
        mod = 10
      
        # Compute first part of expression
        i=2;
        j=0;
        while ((1 << j) <= n): 
            firstsum = (firstsum + i) % mod;
            i *= i;
            j+=1;
      
        # Compute second part of expression 
        # i.e., ((4^(n+1) - 1) / 3) mod 10 
        # Since division of 3 in modulo can't 
        # be performed directly therefore we 
        # need to find it's modulo Inverse 
        secondsum = (powermod(4, n + 1, mod) - 1) * modInverse(3, mod); 
      
        return (firstsum * secondsum) % mod; 
      
    # Driver code 
      
    n = 3
    print(evaluteExpression(n)); 
      
    n = 10
    print(evaluteExpression(n)); 
      
    # This code is contributed by mits

    chevron_right

    
    

    C#

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    // C# program to calculate to find last 
    // digit of above expression 
      
    class GFG{
    /* Iterative Function to calculate (x^y)%p in O(log y) */
    static long powermod(long x, long y, long p) 
        long res = 1; // Initialise result 
      
        x = x % p; // Update x if it is more than or 
                // equal to p 
      
        while (y > 0) { 
      
            // If y is odd, multiply x with result 
            if ((y & 1)>0) 
                res = (res * x) % p; 
      
            // y must be even now 
            y = y >> 1; // y = y/2 
            x = (x * x) % p; 
        
        return res; 
      
    // Returns modulo inverse of a with respect to m 
    // using extended Euclid Algorithm 
    static long modInverse(long a, long m) 
        long  m0 = m, t, q; 
        long  x0 = 0, x1 = 1; 
      
        if (m == 1) 
            return 0; 
      
        while (a > 1) { 
      
            // q is quotient 
            q = a / m; 
      
            t = m; 
      
            // m is remainder now, process same as 
            // Euclid's algo 
            m = a % m;
            a = t; 
      
            t = x0; 
      
            x0 = x1 - q * x0; 
      
            x1 = t; 
        
      
        // Make x1 positive 
        if (x1 < 0) 
            x1 += m0; 
      
        return x1; 
      
    // Function to calculate the above expression 
    static long evaluteExpression(long n) 
        // Initialize the result 
        long firstsum = 0, mod = 10; 
      
        // Compute first part of expression 
        for (int i = 2, j = 0; (1 << j) <= n; i *= i, ++j) 
            firstsum = (firstsum + i) % mod; 
      
        // Compute second part of expression 
        // i.e., ((4^(n+1) - 1) / 3) mod 10 
        // Since division of 3 in modulo can't 
        // be performed directly therefore we 
        // need to find it's modulo Inverse 
        long secondsum = (powermod(4L, n + 1, mod) - 1) * 
                            modInverse(3L, mod); 
      
        return (firstsum * secondsum) % mod; 
      
    // Driver code 
    public static void Main() 
        long n = 3; 
        System.Console.WriteLine(evaluteExpression(n)); 
      
        n = 10; 
        System.Console.WriteLine(evaluteExpression(n)); 
      
    }
    // This code is contributed by mits

    chevron_right

    
    

    PHP

    filter_none

    edit
    close

    play_arrow

    link
    brightness_4
    code

    <?php
    // PHP program to calculate to find 
    // last digit of above expression 
      
    /* Iterative Function to calculate
       (x^y)%p in O(log y) */
    function powermod($x, $y, $p
        $res = 1; // Initialise result 
      
        $x = $x % $p; // Update x if it is more 
                      // than or equal to p 
      
        while ($y > 0)
        
      
            // If y is odd, multiply 
            // x with result 
            if (($y & 1) > 0) 
                $res = ($res * $x) % $p
      
            // y must be even now 
            $y = $y >> 1; // y = y/2 
            $x = ($x * $x) % $p
        
        return $res
      
    // Returns modulo inverse of a 
    // with respect to m using 
    // extended Euclid Algorithm 
    function modInverse($a, $m
        $m0 = $m
        $x0 = 0;
        $x1 = 1; 
      
        if ($m == 1) 
            return 0; 
      
        while ($a > 1) 
        
      
            // q is quotient 
            $q = (int)($a / $m); 
      
            $t = $m
      
            // m is remainder now, process 
            // same as Euclid's algo 
            $m = $a % $m;
            $a = $t
      
            $t = $x0
      
            $x0 = $x1 - $q * $x0
      
            $x1 = $t
        
      
        // Make x1 positive 
        if ($x1 < 0) 
            $x1 += $m0
      
        return $x1
      
    // Function to calculate the
    // above expression 
    function evaluteExpression($n
        // Initialize the result 
        $firstsum = 0;
        $mod = 10; 
      
        // Compute first part of expression 
        for ($i = 2, $j = 0; (1 << $j) <= $n;
                              $i *= $i, ++$j
            $firstsum = ($firstsum + $i) % $mod
      
        // Compute second part of expression 
        // i.e., ((4^(n+1) - 1) / 3) mod 10 
        // Since division of 3 in modulo can't 
        // be performed directly therefore we 
        // need to find it's modulo Inverse 
        $secondsum = (powermod(4, $n + 1, $mod) - 1) * 
                      modInverse(3, $mod); 
      
        return ($firstsum * $secondsum) % $mod
      
    // Driver code 
    $n = 3; 
    echo evaluteExpression($n) . "\n"
      
    $n = 10; 
    echo evaluteExpression($n); 
      
    // This code is contributed by mits
    ?>

    chevron_right

    
    

    Output:

    0
    8
    

    Time complexity: O(log(n))
    Auxiliary space: O(1)

    Note: Asked in TCS Code vita contest.

    This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.



    My Personal Notes arrow_drop_up

    Improved By : Mithun Kumar



    Article Tags :
    Practice Tags :


    Be the First to upvote.


    Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.