Given an integer n, find the last digit of this sequence,
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. [Tex]\implies \displaystyle \sum_{i=0,\ 2^i\le n}^\infty \sum_{j=0}^n 2^{2^i}\cdot 2^{2j} [/Tex]
[Tex]\implies\displaystyle\sum_{i=0,\ 2^i\le n}^\infty 2^{2^i} \cdot \sum_{j=0}^n 4^j [/Tex]
- First expression
can be calculated directly by iterating a loop for all values of ‘i’ till 2i ? n. - Second expression
can be calculated easily by using Geometric series formula i.e., 
- 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.
can be calculated directly by iterating a loop for all values of ‘i’ till 2i ? n.
can be calculated easily by using Geometric series formula i.e., 
C++
// 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 Algorithmlong 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 expressionlong 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 codeint main(){ long long n = 3; cout << evaluteExpression(n) << endl; n = 10; cout << evaluteExpression(n); return 0;} |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
PHP
<?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?> |
Javascript
// JavaScript 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) { let 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) { let m0 = m, t, q; let x0 = 0, x1 = 1; if (m == 1) return 0; while (a > 1) { // q is quotient q = Math.floor(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 let firstsum = 0, mod = 10; // Compute first part of expression for (var 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 let secondsum = (powermod(4, n + 1, mod) - 1) * modInverse(3, mod); return (firstsum * secondsum) % mod; } // Driver code let n = 3; console.log(evaluteExpression(n)); n = 10; console.log(evaluteExpression(n)); // This code is contributed by phasing17 |
Output:
0
8
Time complexity: O(log(n))
Auxiliary space: O(1)
Note: Asked in TCS Code vita contest.
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