Count ways to express even number ‘n’ as sum of even integers

• Difficulty Level : Easy
• Last Updated : 27 Apr, 2021

Given an positive even integer ‘n’. Count total number of ways to express ‘n’ as sum of even positive integers. Output the answer in modulo 109 + 7
Examples:

Input: 6
Output: 4

Explanation
There are only four ways to write 6
as sum of even integers:
1. 2 + 2 + 2
2. 2 + 4
3. 4 + 2
4. 6
Input: 8
Output: 8

Approach is to find pattern or recursive function whichever is possible. The approach would be the same as already discussed in “Count ways to express ‘n’ as sum of odd integers“. Here the given number is even that means even sum can only be achieved by adding the (n-2)th number as two times. We can notice that (by taking some examples) adding a 2 to a number doubles the count. Let the total number of ways to write ‘n’ be ways(n). The value of ‘ways(n)’ can be written by formula as follows:

ways(n) = ways(n-2) + ways(n-2)
ways(n) = 2 * ways(n-2)

ways(2) = 1 = 20
ways(4) = 2 = 21
ways(6) = 4 = 22
ways(8) = 8 = 23
''
''
''
ways(2 * n) = 2n-1

Replace n by (m / 2)
=> ways(m) = 2m/2 - 1

C++

 // C++ program to count ways to write// number as sum of even integers#includeusing namespace std; // Initialize mod variable as constantconst int MOD = 1e9 + 7; /* Iterative Function to calculate (x^y)%p in O(log y) */int power(int x, unsigned int y, int p){    int res = 1;      // Initialize 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)            res = (1LL * res * x) % p;         // y must be even now        y = y>>1; // y = y/2        x = (1LL * x * x) % p;    }    return res;} // Return number of ways to write 'n'// as sum of even integersint countEvenWays(int n){  return power(2, n/2 - 1, MOD);} // Driver codeint main(){    int n = 6;    cout << countEvenWays(n) << "\n";     n = 8;    cout << countEvenWays(n);   return 0;}

Java

 // JAVA program to count ways to write// number as sum of even integers class GFG {         // Initialize mod variable as constant    static int MOD = 1000000007;          /* Iterative Function to calculate    (x^y)%p in O(log y) */    static int power(int x, int y, int p)    {          // Initialize result        int res = 1;                      // Update x if it is more        // than or equal to p        x = x % p;               while (y > 0)        {            // If y is odd, multiply x            // with result            if (y % 2 == 1)                res = (1 * res * x) % p;                  // y must be even now            y = y >> 1; // y = y/2            x = (1 * x * x) % p;        }        return res;    }          // Return number of ways to write    // 'n' as sum of even integers    static int countEvenWays(int n)    {      return power(2, n/2 - 1, MOD);    }          // Driver code    public static void main(String args[])    {        int n = 6;        System.out.println(countEvenWays(n));        n = 8;        System.out.println(countEvenWays(n));    }} /* This code is contributed by Nikita Tiwari. */

Python

 # PYTHON program to count ways to write# number as sum of even integers # Initialize mod variable as constantMOD = 1e9 + 7 # Iterative Function to calculate# (x^y)%p in O(log y)def power(x, y, p) :    res = 1      # Initialize 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) :            res = (1 * res * x) % p                 # y must be even now        y = y >> 1  # y = y/2        x = (1 * x * x) % p                      return res   # Return number of ways to write 'n'# as sum of even integersdef countEvenWays(n) :    return power(2, n/2 - 1, MOD) # Driver coden = 6print (int(countEvenWays(n)))n = 8print (int(countEvenWays(n))) # This code is contributed by Nikita Tiwari.

C#

 // C# program to count ways to write// number as sum of even integersusing System; class GFG {         // Initialize mod variable as constant    static int MOD = 1000000007;         /* Iterative Function to calculate    (x^y)%p in O(log y) */    static int power(int x, int y, int p)    {                 // Initialize result        int res = 1;                     // Update x if it is more        // than or equal to p        x = x % p;             while (y > 0)        {                         // If y is odd, multiply x            // with result            if (y % 2 == 1)                res = (1 * res * x) % p;                 // y must be even now            y = y >> 1; // y = y/2            x = (1 * x * x) % p;        }                 return res;    }         // Return number of ways to write    // 'n' as sum of even integers    static int countEvenWays(int n)    {        return power(2, n/2 - 1, MOD);    }         // Driver code    public static void Main()    {        int n = 6;        Console.WriteLine(countEvenWays(n));                 n = 8;        Console.WriteLine(countEvenWays(n));    }} // This code is contributed by vt_m.

PHP

 0)    {        // If y is odd, multiply        // x with result        if (\$y & 1)            \$res = (1 * \$res *                        \$x) % \$p;         // y must be even now        \$y = \$y >> 1; // y = y/2        \$x = (1 * \$x *                  \$x) % \$p;    }    return \$res;} // Return number of ways// to write 'n' as sum of// even integersfunction countEvenWays(\$n){    global \$MOD;    return power(2, \$n /                 2 - 1, \$MOD);} // Driver code\$n = 6;echo countEvenWays(\$n), "\n"; \$n = 8;echo countEvenWays(\$n); // This code is contributed// by ajit?>

Javascript



Output:

4
8

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