先决条件：打印所有质因数及其幂
给定自然数N和P ，任务是在N的因式分解中找到P的幂！ .
例子
Input: N = 4, P = 2
Output: 3
Explanation:
Power of 2 in the prime factorization of 4! = 24 is 3
Input: N = 24, P = 4
Output: 11
朴素的方法：这个想法是找到从1 到 N 的每个数字的P的幂，然后将它们相加，正如我们在乘法幂相加过程中所知道的那样。
时间复杂度： O(N*P)
有效的方法：
求N中数字P的幂！请执行下列操作：
- 使用本文中讨论的方法找出数字P 的所有质因数及其频率。将主要因子及其频率存储在地图中。
- 在N的因式分解中找出P的每个素因数的幂！通过使用本文中讨论的方法。
- 将上述步骤中获得的每个功率除以它们在图中对应的频率。
- 将上述步骤的结果存储在一个数组中，这些元素中的最小值将在 N! 的因式分解中给出P的幂。
下面是上述方法的实现：
C++
// C++ program to find the power
// of P in N!
#include
using namespace std;
// Map to store all the prime
// factors of P
unordered_map Map;
// Function to find the prime
// factors of N im Map
void findPrimeFactors(int N)
{
int i;
// Clear map
Map.clear();
// Check for factors of 2
while (N % 2 == 0) {
Map[2] += 1;
N /= 2;
}
// Find all the prime factors
for (i = 3; i <= sqrt(N); i += 2) {
// If i is a factors
// then increase the
// frequency of i
while (N % i == 0) {
Map[i] += 1;
N /= i;
}
}
if (N > 2) {
Map[N] += 1;
}
}
// Function to find the power
// of prime number P in N!
int PowInFactN(int N, int P)
{
int ans = 0;
int temp = P;
// Loop until temp <= N
while (temp <= N) {
// Add the number of
// numbers divisible
// by N
ans += N / temp;
// Each time multiply
// temp by P
temp = temp * P;
}
// Returns ans
return ans;
}
// Function that find the
// powers of any P in N!
int findPowers(int N, int P)
{
// Find all prime factors
// of number P
findPrimeFactors(P);
// To store the powers of
// all prime factors
vector Powers;
// Traverse the map
for (auto& it : Map) {
// Prime factor and
// corres. powers
int primeFac = it.first;
int facPow = it.second;
// Find power of prime
// factor primeFac
int p = PowInFactN(N,
primeFac);
// Divide frequency by
// facPow
p /= facPow;
// Store the power of
// primeFac^facPow
Powers.push_back(p);
}
// Return the minimum
// element in Power array
return *min_element(Powers.begin(),
Powers.end());
}
// Driver's Code
int main()
{
int N = 24, P = 4;
// Function to find power of
// P in N!
cout << findPowers(N, P);
return 0;
} Java
// Java program to find the power
// of P in N!
import java.util.*;
class GFG{
// Map to store all the prime
// factors of P
static HashMap Map = new HashMap();
// Function to find the prime
// factors of N im Map
static void findPrimeFactors(int N)
{
int i;
// Clear map
Map.clear();
// Check for factors of 2
while (N % 2 == 0) {
if(Map.containsKey(2))
Map.put(2, Map.get(2) + 1);
else
Map.put(2, 1);
N /= 2;
}
// Find all the prime factors
for (i = 3; i <= Math.sqrt(N); i += 2) {
// If i is a factors
// then increase the
// frequency of i
while (N % i == 0) {
if(Map.containsKey(i))
Map.put(i, Map.get(i) + 1);
else
Map.put(i, 1);
N /= i;
}
}
if (N > 2) {
if(Map.containsKey(N))
Map.put(N, Map.get(N) + 1);
else
Map.put(N, 1);
}
}
// Function to find the power
// of prime number P in N!
static int PowInFactN(int N, int P)
{
int ans = 0;
int temp = P;
// Loop until temp <= N
while (temp <= N) {
// Add the number of
// numbers divisible
// by N
ans += N / temp;
// Each time multiply
// temp by P
temp = temp * P;
}
// Returns ans
return ans;
}
// Function that find the
// powers of any P in N!
static int findPowers(int N, int P)
{
// Find all prime factors
// of number P
findPrimeFactors(P);
// To store the powers of
// all prime factors
Vector Powers = new Vector();
// Traverse the map
for (Map.Entry it : Map.entrySet()) {
// Prime factor and
// corres. powers
int primeFac = it.getKey();
int facPow = it.getValue();
// Find power of prime
// factor primeFac
int p = PowInFactN(N,
primeFac);
// Divide frequency by
// facPow
p /= facPow;
// Store the power of
// primeFac^facPow
Powers.add(p);
}
// Return the minimum
// element in Power array
return Collections.min(Powers);
}
// Driver's Code
public static void main(String[] args)
{
int N = 24, P = 4;
// Function to find power of
// P in N!
System.out.print(findPowers(N, P));
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 program to find the power
# of P in N!
import math
# Map to store all
# the prime factors of P
Map = {}
# Function to find the prime
# factors of N im Map
def findPrimeFactors(N):
# Clear map
Map.clear()
# Check for factors of 2
while (N % 2 == 0):
if 2 in Map:
Map[2] += 1
else:
Map[2] = 1
N = N // 2
# Find all the prime factors
for i in range(3, int(math.sqrt(N)) + 1, 2):
# If i is a factors
# then increase the
# frequency of i
while (N % i == 0):
if i in Map:
Map[i] += 1
else:
Map[i] = 1
N = N // i
if (N > 2):
if N in Map:
Map[N] += 1
else:
Map[N] = 1
# Function to find the power
# of prime number P in N!
def PowInFactN(N, P):
ans = 0
temp = P
# Loop until temp <= N
while (temp <= N):
# Add the number of
# numbers divisible
# by N
ans = ans + (N // temp)
# Each time multiply
# temp by P
temp = temp * P
# Returns ans
return ans
# Function that find the
# powers of any P in N!
def findPowers(N, P):
# Find all prime factors
# of number P
findPrimeFactors(P)
# To store the powers of
# all prime factors
Powers = []
# Traverse the map
for it1, it2 in Map.items():
# Prime factor and
# corres. powers
primeFac = it1
facPow = it2
# Find power of prime
# factor primeFac
p = PowInFactN(N, primeFac)
# Divide frequency by
# facPow
p = p // facPow
# Store the power of
# primeFac^facPow
Powers.append(p)
# Return the minimum
# element in Power array
return min(Powers)
N, P = 24, 4
# Driver code
if __name__ == "__main__":
# Function to find
# power of P in N!
print(findPowers(N, P))
# This code is contriibuted by divyeshrabadiya07C#
// C# program to find the power
// of P in N!
using System;
using System.Linq;
using System.Collections.Generic;
class GFG{
// Map to store all the prime
// factors of P
static Dictionary Map = new Dictionary();
// Function to find the prime
// factors of N im Map
static void findPrimeFactors(int N)
{
int i;
// Clear map
Map.Clear();
// Check for factors of 2
while (N % 2 == 0) {
if(Map.ContainsKey(2))
Map[2] = Map[2] + 1;
else
Map.Add(2, 1);
N /= 2;
}
// Find all the prime factors
for (i = 3; i <= Math.Sqrt(N); i += 2) {
// If i is a factors
// then increase the
// frequency of i
while (N % i == 0) {
if(Map.ContainsKey(i))
Map[i] = Map[i] + 1;
else
Map.Add(i, 1);
N /= i;
}
}
if (N > 2) {
if(Map.ContainsKey(N))
Map[N] =Map[N] + 1;
else
Map.Add(N, 1);
}
}
// Function to find the power
// of prime number P in N!
static int PowInFactN(int N, int P)
{
int ans = 0;
int temp = P;
// Loop until temp <= N
while (temp <= N) {
// Add the number of
// numbers divisible
// by N
ans += N / temp;
// Each time multiply
// temp by P
temp = temp * P;
}
// Returns ans
return ans;
}
// Function that find the
// powers of any P in N!
static int findPowers(int N, int P)
{
// Find all prime factors
// of number P
findPrimeFactors(P);
// To store the powers of
// all prime factors
List Powers = new List();
// Traverse the map
foreach (KeyValuePair it in Map) {
// Prime factor and
// corres. powers
int primeFac = it.Key;
int facPow = it.Value;
// Find power of prime
// factor primeFac
int p = PowInFactN(N,
primeFac);
// Divide frequency by
// facPow
p /= facPow;
// Store the power of
// primeFac^facPow
Powers.Add(p);
}
// Return the minimum
// element in Power array
return Powers.Min();
}
// Driver's Code
public static void Main(String[] args)
{
int N = 24, P = 4;
// Function to find power of
// P in N!
Console.Write(findPowers(N, P));
}
}
// This code is contributed by sapnasingh4991 输出：
11
时间复杂度： O(sqrt(P)*(log P N))
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