给定数字N。任务是检查给定数N是否具有唯一素数。如果是，则打印“是”，否则打印“否” 。
例子：
Input: N = 30
Output: YES
Explanation:
N = 30 = 2*3*5
As all the prime factors of 30 are unique.
Input: N = 100
Output: NO
Explanation:
N = 100 = 2*2*5*5
As all the prime factors of 100 are not unique because 2 and 5 are repeated twice.
方法：
- 使用筛网筛查找到给定数N的所有素数。
- 如果获得的所有素数的乘积等于N，则所有素数都是唯一的，因此请打印YES。
- 其他打印编号。
下面是上述方法的实现：
CPP
// C++ program for the above approach
#include
using namespace std;
// Function that returns the all the
// distinct prime factors in a vector
vector primeFactors(int n)
{
int i, j;
vector Prime;
// If n is divisible by 2
if (n % 2 == 0) {
Prime.push_back(2);
}
// Divide n till all factors of 2
while (n % 2 == 0) {
n = n / 2;
}
// Check for the prime numbers other
// than 2
for (i = 3; i <= sqrt(n); i = i + 2) {
// Store i in Prime[] i is a
// factor of n
if (n % i == 0) {
Prime.push_back(i);
}
// Divide n till all factors of i
while (n % i == 0) {
n = n / i;
}
}
// If n is greter than 2, then n is
// prime number after n divided by
// all factors
if (n > 2) {
Prime.push_back(n);
}
// Returns the vector Prime
return Prime;
}
// Function that check whether N is the
// product of distinct prime factors
// or not
void checkDistinctPrime(int n)
{
// Returns the vector to store
// all the distinct prime factors
vector Prime = primeFactors(n);
// To find the product of all
// distinct prime factors
int product = 1;
// Find the product
for (auto i : Prime) {
product *= i;
}
// If product is equals to N,
// print YES, else print NO
if (product == n)
cout << "YES";
else
cout << "NO";
}
// Driver Code
int main()
{
int N = 30;
checkDistinctPrime(N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function that returns the all the
// distinct prime factors in a vector
static Vector primeFactors(int n)
{
int i, j;
Vector Prime = new Vector();
// If n is divisible by 2
if (n % 2 == 0) {
Prime.add(2);
}
// Divide n till all factors of 2
while (n % 2 == 0) {
n = n / 2;
}
// Check for the prime numbers other
// than 2
for (i = 3; i <= Math.sqrt(n); i = i + 2) {
// Store i in Prime[] i is a
// factor of n
if (n % i == 0) {
Prime.add(i);
}
// Divide n till all factors of i
while (n % i == 0) {
n = n / i;
}
}
// If n is greter than 2, then n is
// prime number after n divided by
// all factors
if (n > 2) {
Prime.add(n);
}
// Returns the vector Prime
return Prime;
}
// Function that check whether N is the
// product of distinct prime factors
// or not
static void checkDistinctPrime(int n)
{
// Returns the vector to store
// all the distinct prime factors
Vector Prime = primeFactors(n);
// To find the product of all
// distinct prime factors
int product = 1;
// Find the product
for (int i : Prime) {
product *= i;
}
// If product is equals to N,
// print YES, else print NO
if (product == n)
System.out.print("YES");
else
System.out.print("NO");
}
// Driver Code
public static void main(String[] args)
{
int N = 30;
checkDistinctPrime(N);
}
}
// This code is contributed by sapnasingh4991 Python3
# Python3 program for the above approach
# Function that returns the all the
# distinct prime factors in a vector
def primeFactors(n) :
Prime = [];
# If n is divisible by 2
if (n % 2 == 0) :
Prime.append(2);
# Divide n till all factors of 2
while (n % 2 == 0) :
n = n // 2;
# Check for the prime numbers other
# than 2
for i in range(3, int(n ** (1/2)),2) :
# Store i in Prime[] i is a
# factor of n
if (n % i == 0) :
Prime.append(i);
# Divide n till all factors of i
while (n % i == 0) :
n = n // i;
# If n is greter than 2, then n is
# prime number after n divided by
# all factors
if (n > 2) :
Prime.append(n);
# Returns the vector Prime
return Prime;
# Function that check whether N is the
# product of distinct prime factors
# or not
def checkDistinctPrime(n) :
# Returns the vector to store
# all the distinct prime factors
Prime = primeFactors(n);
# To find the product of all
# distinct prime factors
product = 1;
# Find the product
for i in Prime :
product *= i;
# If product is equals to N,
# print YES, else print NO
if (product == n) :
print("YES");
else :
print("NO");
# Driver Code
if __name__ == "__main__" :
N = 30;
checkDistinctPrime(N);
# This code is contributed by Yash_RC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function that returns the all the
// distinct prime factors in a vector
static List primeFactors(int n)
{
int i;
List Prime = new List();
// If n is divisible by 2
if (n % 2 == 0) {
Prime.Add(2);
}
// Divide n till all factors of 2
while (n % 2 == 0) {
n = n / 2;
}
// Check for the prime numbers other
// than 2
for (i = 3; i <= Math.Sqrt(n); i = i + 2) {
// Store i in Prime[] i is a
// factor of n
if (n % i == 0) {
Prime.Add(i);
}
// Divide n till all factors of i
while (n % i == 0) {
n = n / i;
}
}
// If n is greter than 2, then n is
// prime number after n divided by
// all factors
if (n > 2) {
Prime.Add(n);
}
// Returns the vector Prime
return Prime;
}
// Function that check whether N is the
// product of distinct prime factors
// or not
static void checkDistinctPrime(int n)
{
// Returns the vector to store
// all the distinct prime factors
List Prime = primeFactors(n);
// To find the product of all
// distinct prime factors
int product = 1;
// Find the product
foreach (int i in Prime) {
product *= i;
}
// If product is equals to N,
// print YES, else print NO
if (product == n)
Console.Write("YES");
else
Console.Write("NO");
}
// Driver Code
public static void Main(String[] args)
{
int N = 30;
checkDistinctPrime(N);
}
}
// This code is contributed by sapnasingh4991 输出：
YES
时间复杂度： O(N * log(log N))，其中N是给定的数字。