计算偶数和序列的总数

给定一个整数N ，任务是对长度为N 的所有可能序列进行计数，使得该序列的所有元素都来自范围[1, N]并且该序列的元素之和为偶数。由于答案可能非常大，所以打印答案模10 ⁹ + 7 。
例子：

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 
 

方法：要获得任何序列的偶数和，奇数元素的数量必须是偶数。让我们选择把奇数元素的x个序列中，其中x是偶数。放置这些奇数的方法总数为C(N, x)并且在每个位置，可以放置y个元素，其中y是从1到N的奇数个数，其余位置可以填充偶数以同样的方式。因此，如果要取x 个奇数，那么它们的贡献将是C(N, x) * y ^x * (N – y) ^{(N – x)} 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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 % 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
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 % 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
def countOdd(n) :
 
    x = n // 2
    if (n % 2 == 1) :
        x += 1
    return x
 
# 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 - 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
        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
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

Javascript

 
    // JavaScript implementation of the above approach
     
    let M = 1000000007;
   
    // Iterative function to calculate
    // (x^y)%p in O(log y)
    function power(x, y, p)
    {
       
        // Initialize result
        let 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
    function modInverse(n, p)
    {
        return power(n, p - 2, p);
    }
       
    // Function to return (nCr % p) using
    // Fermat's little theorem
    function 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
        let fac = new Array(n + 1);
        fac[0] = 1;
        let i;
        for (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
    function countOdd(n)
    {
        let x = parseInt(n / 2, 10);
        if (n % 2 == 1)
            x++;
        return x;
    }
       
    // Function to return the count of
    // even numbers from 1 to n
    function counteEven(n)
    {
        let x = parseInt(n / 2, 10);
        return x;
    }
       
    // Function to return the count
    // of the required sequences
    function CountEvenSumSequences(n)
    {
        let count = 0;
       
        for (let i = 0; i <= n; i++)
        {
       
            // Take i even and n - i odd numbers
            let 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
            let 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*0+521;
            count %= M;
        }
        return (count-1);
    }
     
    let n = 5;
       
      document.write(CountEvenSumSequences(n));
 

输出：

1562

时间复杂度： O(N*logN)

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