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Count total number of even sum sequences
  • Difficulty Level : Hard
  • Last Updated : 15 Oct, 2019

Given an integer N, the task is to count all possible sequences of length N such that all the elements of the sequence are from the range [1, N] and the sum of the elements of the sequence is even. As the answer coulb be very large so print the answer modulo 109 + 7.

Examples:

Input: N = 3
Output: 13
All possible sequences of length 3 will be (1, 1, 2), (1, 3, 2),
(3, 1, 2), (3, 3, 2), (1, 2, 1), (1, 2, 3), (3, 2, 1), (3, 2, 3),
(2, 1, 1), (2, 1, 3), (2, 3, 1), (2, 3, 3) and (2, 2, 2).

Input: N = 5
Output: 1562

Approach: To get even sum for any sequence, the number of odd elements must be even. Let’s choose to put x number of odd elements in the sequence where x is even. The total number of ways to put these odd numbers will be C(N, x) and in each position, y number of elements can be put where y is the count of odd numbers from 1 to N and the remaining positions can be filled with even numbers in the same way. So if x odd numbers are to be taken then their contribution will be C(N, x) * yx * (N – y)(N – x).



Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
#define M 1000000007
#define ll long long int
  
// Iterative function to calculate
// (x^y)%p in O(log y)
ll power(ll x, ll y, ll p)
{
  
    // Initialize result
    ll res = 1;
  
    // Update x if it is greater
    // than or equal to p
    x = x % p;
  
    while (y > 0) {
  
        // If y is odd then multiply
        // x with the result
        if (y & 1)
            res = (res * x) % p;
  
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
  
// Function to return n^(-1) mod p
ll modInverse(ll n, ll p)
{
    return power(n, p - 2, p);
}
  
// Function to return (nCr % p) using
// Fermat's little theorem
ll nCrModPFermat(ll n, ll r, ll p)
{
    // Base case
    if (r == 0)
        return 1;
  
    // Fill factorial array so that we
    // can find all factorial of r, n
    // and n-r
    ll fac[n + 1];
    fac[0] = 1;
    for (ll i = 1; i <= n; i++)
        fac[i] = fac[i - 1] * i % p;
  
    return (fac[n] * modInverse(fac[r], p) % p
            * modInverse(fac[n - r], p) % p)
           % p;
}
  
// Function to return the count of
// odd numbers from 1 to n
ll countOdd(ll n)
{
    ll x = n / 2;
    if (n % 2 == 1)
        x++;
    return x;
}
  
// Function to return the count of
// even numbers from 1 to n
ll counteEven(ll n)
{
    ll x = n / 2;
    return x;
}
  
// Function to return the count
// of the required sequences
ll CountEvenSumSequences(ll n)
{
  
    ll count = 0;
  
    for (ll i = 0; i <= n; i++) {
  
        // Take i even and n - i odd numbers
        ll even = i, odd = n - i;
  
        // Number of odd numbers must be even
        if (odd % 2 == 1)
            continue;
  
        // Total ways of placing n - i odd
        // numbers in the sequence of n numbers
        ll tot = (power(countOdd(n), odd, M)
                  * nCrModPFermat(n, odd, M))
                 % M;
        tot = (tot * power(counteEven(n), i, M)) % M;
  
        // Add this number to the final answer
        count += tot;
        count %= M;
    }
    return count;
}
  
// Driver code
int main()
{
  
    ll n = 5;
  
    cout << CountEvenSumSequences(n);
  
    return 0;
}

Java




// Java implementation of the above approach 
import java.util.*;
  
class GFG 
{
    static final int M = 1000000007;
  
    // Iterative function to calculate 
    // (x^y)%p in O(log y) 
    static long power(long x, int y, int p) 
    
      
        // Initialize result 
        long res = 1
      
        // Update x if it is greater 
        // than or equal to p 
        x = x % p; 
      
        while (y > 0)
        
      
            // If y is odd then multiply 
            // x with the result 
            if ((y & 1) == 1
                res = (res * x) % p; 
      
            // y must be even now 
            y = y >> 1; // y = y/2 
            x = (x * x) % p; 
        
        return res; 
    
      
    // Function to return n^(-1) mod p 
    static long modInverse(long n, int p) 
    
        return power(n, p - 2, p); 
    
      
    // Function to return (nCr % p) using 
    // Fermat's little theorem 
    static long nCrModPFermat(long n, int r, int p) 
    
        // Base case 
        if (r == 0
            return 1
      
        // Fill factorial array so that we 
        // can find all factorial of r, n 
        // and n-r 
        long fac[] = new long[(int)n + 1]; 
        fac[0] = 1
        int i ;
        for ( i = 1; i <= n; i++) 
            fac[i] = fac[i - 1] * i % p; 
      
        return (fac[(int)n] * modInverse(fac[r], p) % p * 
                              modInverse(fac[(int)n - r], p) % p) % p; 
    
      
    // Function to return the count of 
    // odd numbers from 1 to n 
    static long countOdd(long n) 
    
        long x = n / 2
        if (n % 2 == 1
            x++; 
        return x; 
    
      
    // Function to return the count of 
    // even numbers from 1 to n 
    static long counteEven(long n) 
    
        long x = n / 2
        return x; 
    
      
    // Function to return the count 
    // of the required sequences 
    static long CountEvenSumSequences(long n) 
    
        long count = 0
      
        for (int i = 0; i <= n; i++)
        
      
            // Take i even and n - i odd numbers 
            int even = i, odd = (int)n - i; 
      
            // Number of odd numbers must be even 
            if (odd % 2 == 1
                continue
      
            // Total ways of placing n - i odd 
            // numbers in the sequence of n numbers 
            long tot = (power(countOdd(n), odd, M) * 
                          nCrModPFermat(n, odd, M)) % M; 
            tot = (tot * power(counteEven(n), i, M)) % M; 
      
            // Add this number to the final answer 
            count += tot; 
            count %= M; 
        
        return count; 
    
      
    // Driver code 
    public static void main (String[] args) 
    
      
        long n = 5
      
        System.out.println(CountEvenSumSequences(n)); 
    
}
  
// This code is contributed by AnkitRai01

Python3




# Python3 implementation of the approach 
M = 1000000007
  
# Iterative function to calculate 
# (x^y)%p in O(log y) 
def power(x, y, p): 
  
    # Initialize result 
    res = 1
  
    # Update x if it is greater 
    # than or equal to p 
    x = x %
  
    while (y > 0) : 
  
        # If y is odd then multiply 
        # x with the result 
        if (y & 1) : 
            res = (res * x) %
  
        # y must be even now 
        y = y >> 1 # y = y/2 
        x = (x * x) %
          
    return res 
  
# Function to return n^(-1) mod p 
def modInverse(n, p) : 
  
    return power(n, p - 2, p) 
  
# Function to return (nCr % p) using 
# Fermat's little theorem 
def nCrModPFermat(n, r, p) : 
      
    # Base case 
    if (r == 0) : 
        return 1
  
    # Fill factorial array so that we 
    # can find all factorial of r, n 
    # and n-r 
    fac = [0] * (n+1
    fac[0] = 1
    for i in range(1, n+1) : 
        fac[i] = fac[i - 1] * i %
  
    return (fac[n] * modInverse(fac[r], p) % p * 
                     modInverse(fac[n - r], p) % p) %
  
# Function to return the count of 
# odd numbers from 1 to n 
def countOdd(n) : 
  
    x = n // 2
    if (n % 2 == 1) : 
        x += 1
    return
  
# Function to return the count of 
# even numbers from 1 to n 
def counteEven(n) : 
  
    x = n // 2
    return x
  
# Function to return the count 
# of the required sequences 
def CountEvenSumSequences(n) : 
  
    count = 0
  
    for i in range(n + 1) : 
  
        # Take i even and n - i odd numbers 
        even = i
        odd = n -
  
        # Number of odd numbers must be even 
        if (odd % 2 == 1) : 
            continue
  
        # Total ways of placing n - i odd 
        # numbers in the sequence of n numbers 
        tot = (power(countOdd(n), odd, M) * 
                 nCrModPFermat(n, odd, M)) %
        tot = (tot * power(counteEven(n), i, M)) %
  
        # Add this number to the final answer 
        count += tot 
        count %=
      
    return count 
  
# Driver code 
n = 5
print(CountEvenSumSequences(n)) 
  
# This code is contributed by
# divyamohan123

C#




// C# implementation of the above approach 
using System;         
  
class GFG 
{
    static readonly int M = 1000000007;
  
    // Iterative function to calculate 
    // (x^y)%p in O(log y) 
    static long power(long x, int y, int p) 
    
      
        // Initialize result 
        long res = 1; 
      
        // Update x if it is greater 
        // than or equal to p 
        x = x % p; 
      
        while (y > 0)
        
      
            // If y is odd then multiply 
            // x with the result 
            if ((y & 1) == 1) 
                res = (res * x) % p; 
      
            // y must be even now 
            y = y >> 1; // y = y/2 
            x = (x * x) % p; 
        
        return res; 
    
      
    // Function to return n^(-1) mod p 
    static long modInverse(long n, int p) 
    
        return power(n, p - 2, p); 
    
      
    // Function to return (nCr % p) using 
    // Fermat's little theorem 
    static long nCrModPFermat(long n, int r, int p) 
    
        // Base case 
        if (r == 0) 
            return 1; 
      
        // Fill factorial array so that we 
        // can find all factorial of r, n 
        // and n-r 
        long []fac = new long[(int)n + 1]; 
        fac[0] = 1; 
        int i;
        for (i = 1; i <= n; i++) 
            fac[i] = fac[i - 1] * i % p; 
      
        return (fac[(int)n] * modInverse(fac[r], p) % p * 
                              modInverse(fac[(int)n - r], p) % p) % p; 
    
      
    // Function to return the count of 
    // odd numbers from 1 to n 
    static long countOdd(long n) 
    
        long x = n / 2; 
        if (n % 2 == 1) 
            x++; 
        return x; 
    
      
    // Function to return the count of 
    // even numbers from 1 to n 
    static long counteEven(long n) 
    
        long x = n / 2; 
        return x; 
    
      
    // Function to return the count 
    // of the required sequences 
    static long CountEvenSumSequences(long n) 
    
        long count = 0; 
      
        for (int i = 0; i <= n; i++)
        
      
            // Take i even and n - i odd numbers 
            int even = i, odd = (int)n - i; 
      
            // Number of odd numbers must be even 
            if (odd % 2 == 1) 
                continue
      
            // Total ways of placing n - i odd 
            // numbers in the sequence of n numbers 
            long tot = (power(countOdd(n), odd, M) * 
                        nCrModPFermat(n, odd, M)) % M; 
            tot = (tot * power(counteEven(n), i, M)) % M; 
      
            // Add this number to the final answer 
            count += tot; 
            count %= M; 
        
        return count; 
    
      
    // Driver code 
    public static void Main (String[] args) 
    
        long n = 5; 
      
        Console.WriteLine(CountEvenSumSequences(n)); 
    
}
  
// This code is contributed by Rajput-Ji
Output:
1562

Time Complexity: O(N*logN)

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