给定一个整数N ，任务是计算将N表示为指数的方式的数量，即x y ，其中x和y是正整数。
例子：
Input: N = 64
Output: 4
Explanation: 64 can be expressed as 26, 43, 82 and 641
Input: N = 27
Output: 2
方法：解决给定问题的想法是找到数N的素数分解，然后找到给定数N的素数的指数的GCD的素数。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to calculate GCD of a
// and b using Euclidean Algorithm
long long int gcd(long long int a,
long long int b)
{
// Iterate until b is non-zero
while (b > 0) {
long long int rem = a % b;
a = b;
b = rem;
}
// Return the GCD
return a;
}
// Function to count the number of
// ways N can be expressed as x^y
int countNumberOfWays(long long int n)
{
// Base Case
if (n == 1)
return -1;
// Stores the gcd of powers
long long int g = 0;
int power = 0;
// Calculate the degree of 2 in N
while (n % 2 == 0) {
power++;
n /= 2;
}
g = gcd(g, power);
// Calculate the degree of prime numbers in N
for (int i = 3; i <= sqrt(n); i += 2) {
power = 0;
// Calculate the degree of
// prime 'i' in N
while (n % i == 0) {
power++;
n /= i;
}
g = gcd(g, power);
}
// If N is a prime, g becomes 1.
if (n > 2)
g = gcd(g, 1);
// Stores the number of ways
// to represent N as x^y
int ways = 1;
// Find the number of Factors of g
power = 0;
while (g % 2 == 0) {
g /= 2;
power++;
}
// Update the count of ways
ways *= (power + 1);
// Iterate to find rest of the prime numbers
for (int i = 3; i <= sqrt(g); i += 2) {
power = 0;
// Find the power of i
while (g % i == 0) {
power++;
g /= i;
}
// Update the count of ways
ways *= (power + 1);
}
// If g is prime
if (g > 2)
ways *= 2;
// Return the total number of ways
return ways;
}
// Driver Code
int main()
{
int N = 64;
cout << countNumberOfWays(N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to calculate GCD of a
// and b using Euclidean Algorithm
static int gcd(int a,
int b)
{
// Iterate until b is non-zero
while (b > 0) {
int rem = a % b;
a = b;
b = rem;
}
// Return the GCD
return a;
}
// Function to count the number of
// ways N can be expressed as x^y
static int countNumberOfWays(int n)
{
// Base Case
if (n == 1)
return -1;
// Stores the gcd of powers
int g = 0;
int power = 0;
// Calculate the degree of 2 in N
while (n % 2 == 0) {
power++;
n /= 2;
}
g = gcd(g, power);
// Calculate the degree of prime numbers in N
for (int i = 3; i <= (int)Math.sqrt(n); i += 2) {
power = 0;
// Calculate the degree of
// prime 'i' in N
while (n % i == 0) {
power++;
n /= i;
}
g = gcd(g, power);
}
// If N is a prime, g becomes 1.
if (n > 2)
g = gcd(g, 1);
// Stores the number of ways
// to represent N as x^y
int ways = 1;
// Find the number of Factors of g
power = 0;
while (g % 2 == 0) {
g /= 2;
power++;
}
// Update the count of ways
ways *= (power + 1);
// Iterate to find rest of the prime numbers
for (int i = 3; i <= (int)Math.sqrt(g); i += 2) {
power = 0;
// Find the power of i
while (g % i == 0) {
power++;
g /= i;
}
// Update the count of ways
ways *= (power + 1);
}
// If g is prime
if (g > 2)
ways *= 2;
// Return the total number of ways
return ways;
}
// Driver Code
public static void main(String[] args)
{
int N = 64;
System.out.print(countNumberOfWays(N));
}
}
// This code is contributed by code_hunt.Python3
# Python3 program for the above approach
import math
# Function to calculate GCD of a
# and b using Euclidean Algorithm
def gcd(a, b) :
# Iterate until b is non-zero
while (b > 0) :
rem = a % b
a = b
b = rem
# Return the GCD
return a
# Function to count the number of
# ways N can be expressed as x^y
def countNumberOfWays(n) :
# Base Case
if (n == 1) :
return -1
# Stores the gcd of powers
g = 0
power = 0
# Calculate the degree of 2 in N
while (n % 2 == 0) :
power += 1
n //= 2
g = gcd(g, power)
# Calculate the degree of prime numbers in N
for i in range(3, int(math. sqrt(g)) + 1, 2):
power = 0
# Calculate the degree of
# prime 'i' in N
while (n % i == 0) :
power += 1
n //= i
g = gcd(g, power)
# If N is a prime, g becomes 1.
if (n > 2) :
g = gcd(g, 1)
# Stores the number of ways
# to represent N as x^y
ways = 1
# Find the number of Factors of g
power = 0
while (g % 2 == 0) :
g //= 2
power += 1
# Update the count of ways
ways *= (power + 1)
# Iterate to find rest of the prime numbers
for i in range(3, int(math. sqrt(g)) + 1, 2):
power = 0
# Find the power of i
while (g % i == 0) :
power += 1
g /= i
# Update the count of ways
ways *= (power + 1)
# If g is prime
if (g > 2) :
ways *= 2
# Return the total number of ways
return ways
# Driver Code
N = 64
print(countNumberOfWays(N))
# This code is contributed by sanjoy_62.C#
// C# program for the above approach
using System;
public class GFG
{
// Function to calculate GCD of a
// and b using Euclidean Algorithm
static int gcd(int a,
int b)
{
// Iterate until b is non-zero
while (b > 0)
{
int rem = a % b;
a = b;
b = rem;
}
// Return the GCD
return a;
}
// Function to count the number of
// ways N can be expressed as x^y
static int countNumberOfWays(int n)
{
// Base Case
if (n == 1)
return -1;
// Stores the gcd of powers
int g = 0;
int power = 0;
// Calculate the degree of 2 in N
while (n % 2 == 0)
{
power++;
n /= 2;
}
g = gcd(g, power);
// Calculate the degree of prime numbers in N
for (int i = 3; i <= (int)Math.Sqrt(n); i += 2)
{
power = 0;
// Calculate the degree of
// prime 'i' in N
while (n % i == 0)
{
power++;
n /= i;
}
g = gcd(g, power);
}
// If N is a prime, g becomes 1.
if (n > 2)
g = gcd(g, 1);
// Stores the number of ways
// to represent N as x^y
int ways = 1;
// Find the number of Factors of g
power = 0;
while (g % 2 == 0)
{
g /= 2;
power++;
}
// Update the count of ways
ways *= (power + 1);
// Iterate to find rest of the prime numbers
for (int i = 3; i <= (int)Math.Sqrt(g); i += 2)
{
power = 0;
// Find the power of i
while (g % i == 0)
{
power++;
g /= i;
}
// Update the count of ways
ways *= (power + 1);
}
// If g is prime
if (g > 2)
ways *= 2;
// Return the total number of ways
return ways;
}
// Driver Code
public static void Main(String[] args)
{
int N = 64;
Console.Write(countNumberOfWays(N));
}
}
// This code is contributed by shikhasingrajput输出：
4时间复杂度： O(√N)
辅助空间： O(1)