给定两个整数N和M ，任务是找到长度为N的可能序列a 1 ，a 2 …的计数，以使该序列所有元素的乘积为M。
例子：
Input: N = 2, M = 6
Output: 4
Possible sequences are {1, 6}, {2, 3}, {3, 2} and {6, 1}
Input: N = 3, M = 24
Output: 30
方法：
- 考虑M的素因式分解为p 1 k 1 p 2 k 2 …p z k z 。
- 对于每个素数因子，其指数必须分布在N个不同的组中，因为在乘以这N个项时，素数因子的指数将相加。可以使用此处说明的公式完成此操作。
- 对于每个素数因子，将其指数分布在N个序列中的方式数量等于
每1≤i≤z N + K i -1 C N-1 。 - 使用乘法的基本原理，将所有主要因子的结果相乘以获得答案。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to calculate the
// value of ncr effectively
int ncr(int n, int r)
{
// Initializing the result
int res = 1;
for (int i = 1; i <= r; i += 1) {
// Multiply and divide simultaneously
// to avoid overflow
res *= (n - r + i);
res /= i;
}
// Return the answer
return res;
}
// Function to return the number of sequences
// of length N such that their product is M
int NoofSequences(int N, int M)
{
// Hashmap to store the prime factors of M
unordered_map prime;
// Calculate the prime factors of M
for (int i = 2; i <= sqrt(M); i += 1) {
// If i divides M it means it is a factor
// Divide M by i till it could be
// divided to store the exponent
while (M % i == 0) {
// Increase the exponent count
prime[i] += 1;
M /= i;
}
}
// If the number is a prime number
// greater than sqrt(M)
if (M > 1) {
prime[M] += 1;
}
// Initializing the ans
int ans = 1;
// Multiply the answer for every prime factor
for (auto it : prime) {
// it.second represents the
// exponent of every prime factor
ans *= (ncr(N + it.second - 1, N - 1));
}
// Return the result
return ans;
}
// Driver code
int main()
{
int N = 2, M = 6;
cout << NoofSequences(N, M);
return 0;
} Java
// Java implementation of the above approach
import java.util.HashMap;
class geeks
{
// Function to calculate the
// value of ncr effectively
public static int nCr(int n, int r)
{
// Initializing the result
int res = 1;
for (int i = 1; i <= r; i++)
{
// Multiply and divide simultaneously
// to avoid overflow
res *= (n - r + i);
res /= i;
}
// Return the answer
return res;
}
// Function to return the number of sequences
// of length N such that their product is M
public static int NoofSequences(int N, int M)
{
// Hashmap to store the prime factors of M
HashMap prime = new HashMap<>();
// Calculate the prime factors of M
for (int i = 2; i <= Math.sqrt(M); i++)
{
// If i divides M it means it is a factor
// Divide M by i till it could be
// divided to store the exponent
while (M % i == 0)
{
// Increase the exponent count
if (prime.get(i) == null)
prime.put(i, 1);
else
{
int x = prime.get(i);
prime.put(i, ++x);
}
M /= i;
}
}
// If the number is a prime number
// greater than sqrt(M)
if (M > 1)
{
if (prime.get(M) != null)
{
int x = prime.get(M);
prime.put(M, ++x);
}
else
prime.put(M, 1);
}
// Initializing the ans
int ans = 1;
// Multiply the answer for every prime factor
for (HashMap.Entry entry : prime.entrySet())
{
// entry.getValue() represents the
// exponent of every prime factor
ans *= (nCr(N + entry.getValue() - 1, N - 1));
}
// Return the result
return ans;
}
// Driver code
public static void main(String[] args)
{
int N = 2, M = 6;
System.out.println(NoofSequences(N, M));
}
}
// This code is contributed by
// sanjeev2552 Python3
# Python3 implementation of the above approach
# Function to calculate the
# value of ncr effectively
def ncr(n, r):
# Initializing the result
res = 1
for i in range(1,r+1):
# Multiply and divide simultaneously
# to avoid overflow
res *= (n - r + i)
res //= i
# Return the answer
return res
# Function to return the number of sequences
# of length N such that their product is M
def NoofSequences(N, M):
# Hashmap to store the prime factors of M
prime={}
# Calculate the prime factors of M
for i in range(2,int(M**(.5))+1):
# If i divides M it means it is a factor
# Divide M by i till it could be
# divided to store the exponent
while (M % i == 0):
# Increase the exponent count
prime[i]= prime.get(i,0)+1
M //= i
# If the number is a prime number
# greater than sqrt(M)
if (M > 1):
prime[M] =prime.get(M,0) + 1
# Initializing the ans
ans = 1
# Multiply the answer for every prime factor
for it in prime:
# it.second represents the
# exponent of every prime factor
ans *= (ncr(N + prime[it] - 1, N - 1))
# Return the result
return ans
# Driver code
N = 2
M = 6
print(NoofSequences(N, M))
# This code is contributed by mohit kumar 29C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
public class geeks
{
// Function to calculate the
// value of ncr effectively
public static int nCr(int n, int r)
{
// Initializing the result
int res = 1;
for (int i = 1; i <= r; i++)
{
// Multiply and divide simultaneously
// to avoid overflow
res *= (n - r + i);
res /= i;
}
// Return the answer
return res;
}
// Function to return the number of sequences
// of length N such that their product is M
public static int NoofSequences(int N, int M)
{
// Hashmap to store the prime factors of M
Dictionaryprime = new Dictionary();
// Calculate the prime factors of M
for (int i = 2; i <= Math.Sqrt(M); i++)
{
// If i divides M it means it is a factor
// Divide M by i till it could be
// divided to store the exponent
while (M % i == 0)
{
// Increase the exponent count
if (!prime.ContainsKey(i))
prime.Add(i, 1);
else
{
int x = prime[i];
prime.Remove(i);
prime.Add(i, ++x);
}
M /= i;
}
}
// If the number is a prime number
// greater than sqrt(M)
if (M > 1)
{
if (prime.ContainsKey(M))
{
int x = prime[M];
prime.Remove(M);
prime.Add(M, ++x);
}
else
prime.Add(M, 1);
}
// Initializing the ans
int ans = 1;
// Multiply the answer for every prime factor
foreach(KeyValuePair entry in prime)
{
// entry.getValue() represents the
// exponent of every prime factor
ans *= (nCr(N + entry.Value - 1, N - 1));
}
// Return the result
return ans;
}
// Driver code
public static void Main(String[] args)
{
int N = 2, M = 6;
Console.WriteLine(NoofSequences(N, M));
}
}
// This code is contributed by Princi Singh 输出：
4