给定一个整数N ，任务是对长度为N的所有可能的序列进行计数，以使该序列的所有元素都在[1，N]范围内，并且该序列的元素之和为偶数。由于答案库很大，因此请以10 9 + 7为模输出答案。
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的奇数计数，其余位置可以用偶数以相同的方式。因此，如果要使用x个奇数，则它们的贡献将为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 AnkitRai01Python3
# 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
# divyamohan123C#
// 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输出：
1562
时间复杂度： O(N * logN)