给定一个数字
。任务是检查N是否为特洛伊木马编号。
Trojan Number是一个数字，它是一个强数，但不是一个完美的幂。如果对于每个主除数或N的因子p，p2也是一个除数，则数字N被称为强数。换句话说，每个素因数至少出现两次。
所有特洛伊木马数字都很强。但是，并非所有强数都是Trojan数：只有那些不能表示为m k的数字，其中m和k是大于1的正整数。
例子：
Input : N = 108
Output : YES
Input : N = 8
Output : NO
这个想法是存储每个素数的计数，并检查该计数是否大于2，那么它将是一个强数。
该部分可以通过筛的质因数分解很容易地计算出来。
下一步是检查给定的数字是否不能表示为x y 。要检查数字是否是完美的幂，请参阅本文。
下面是上述问题的实现：
C++
// CPP program to check if a number is
// Trojan Number or not
#include
using namespace std;
// Function to check if a number
// can be expressed as x^y
bool isPerfectPower(int n)
{
if (n == 1)
return true;
// Try all numbers from 2 to sqrt(n) as base
for (int x = 2; x <= sqrt(n); x++) {
int y = 2;
int p = pow(x, y);
// Keep increasing y while power 'p'
// is smaller than n.
while (p <= n && p > 0) {
if (p == n)
return true;
y++;
p = pow(x, y);
}
}
return false;
}
// Function to check if a number is Strong
bool isStrongNumber(int n)
{
unordered_map count;
while (n % 2 == 0) {
n = n / 2;
count[2]++;
}
// count the number for each prime factor
for (int i = 3; i <= sqrt(n); i += 2) {
while (n % i == 0) {
n = n / i;
count[i]++;
}
}
if (n > 2)
count[n]++;
int flag = 0;
for (auto b : count) {
// minimum number of prime divisors
// should be 2
if (b.second == 1) {
flag = 1;
break;
}
}
if (flag == 1)
return false;
else
return true;
}
// Function to check if a number
// is Trojan Number
bool isTrojan(int n)
{
if (!isPerfectPower(n) && isStrongNumber(n))
return true;
else
return false;
}
// Driver Code
int main()
{
int n = 108;
if (isTrojan(n))
cout << "YES";
else
cout << "NO";
return 0;
} Java
// Java program to check if a number is
// Trojan Number or not
import java.util.*;
class GFG
{
// Function to check if a number
// can be expressed as x^y
static boolean isPerfectPower(int n)
{
if (n == 1)
{
return true;
}
// Try all numbers from 2 to sqrt(n) as base
for (int x = 2; x <= Math.sqrt(n); x++)
{
int y = 2;
int p = (int) Math.pow(x, y);
// Keep increasing y while power 'p'
// is smaller than n.
while (p <= n && p > 0)
{
if (p == n)
{
return true;
}
y++;
p = (int) Math.pow(x, y);
}
}
return false;
}
// Function to check if a number is Strong
static boolean isStrongNumber(int n)
{
HashMap count = new HashMap();
while (n % 2 == 0)
{
n = n / 2;
if (count.containsKey(2))
{
count.put(2, count.get(2) + 1);
}
else
{
count.put(2, 1);
}
}
// count the number for each prime factor
for (int i = 3; i <= Math.sqrt(n); i += 2)
{
while (n % i == 0)
{
n = n / i;
if (count.containsKey(i))
{
count.put(i, count.get(i) + 1);
}
else
{
count.put(i, 1);
}
}
}
if (n > 2)
{
if (count.containsKey(n))
{
count.put(n, count.get(n) + 1);
}
else
{
count.put(n, 1);
}
}
int flag = 0;
for (Map.Entry b : count.entrySet())
{
// minimum number of prime divisors
// should be 2
if (b.getValue() == 1)
{
flag = 1;
break;
}
}
if (flag == 1)
{
return false;
}
else
{
return true;
}
}
// Function to check if a number
// is Trojan Number
static boolean isTrojan(int n)
{
if (!isPerfectPower(n) && isStrongNumber(n))
{
return true;
}
else
{
return false;
}
}
// Driver Code
public static void main(String[] args)
{
int n = 108;
if (isTrojan(n))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python 3 program to check if a number
# is Trojan Number or not
from math import sqrt, pow
# Function to check if a number
# can be expressed as x^y
def isPerfectPower(n):
if n == 1:
return True
# Try all numbers from 2 to
# sqrt(n) as base
for x in range(2, int(sqrt(n)) + 1):
y = 2
p = pow(x, y)
# Keep increasing y while power
# 'p' is smaller than n.
while p <= n and p > 0:
if p == n:
return True
y += 1
p = pow(x, y)
return False
# Function to check if a number
# is Strong
def isStrongNumber(n):
count = {i:0 for i in range(n)}
while n % 2 == 0:
n = n // 2
count[2] += 1
# count the number for each
# prime factor
for i in range(3,int(sqrt(n)) + 1, 2):
while n % i == 0:
n = n // i
count[i] += 1
if n > 2:
count[n] += 1
flag = 0
for key,value in count.items():
# minimum number of prime
# divisors should be 2
if value == 1:
flag = 1
break
if flag == 1:
return False
return True
# Function to check if a number
# is Trojan Number
def isTrojan(n):
return isPerfectPower(n) == False and isStrongNumber(n)
# Driver Code
if __name__ == '__main__':
n = 108
if (isTrojan(n)):
print("YES")
else:
print("NO")
# This code is contributed by
# Surendra_GangwarC#
// C# program to check if a number is
// Trojan Number or not
using System;
using System.Collections.Generic;
class GFG
{
// Function to check if a number
// can be expressed as x^y
static bool isPerfectPower(int n)
{
if (n == 1)
{
return true;
}
// Try all numbers from 2 to sqrt(n) as base
for (int x = 2; x <= Math.Sqrt(n); x++)
{
int y = 2;
int p = (int) Math.Pow(x, y);
// Keep increasing y while power 'p'
// is smaller than n.
while (p <= n && p > 0)
{
if (p == n)
{
return true;
}
y++;
p = (int) Math.Pow(x, y);
}
}
return false;
}
// Function to check if a number is Strong
static bool isStrongNumber(int n)
{
Dictionary count = new Dictionary();
while (n % 2 == 0)
{
n = n / 2;
if (count.ContainsKey(2))
{
count[2] = count[2] + 1;
}
else
{
count.Add(2, 1);
}
}
// count the number for each prime factor
for (int i = 3; i <= Math.Sqrt(n); i += 2)
{
while (n % i == 0)
{
n = n / i;
if (count.ContainsKey(i))
{
count[i] = count[i] + 1;
}
else
{
count.Add(i, 1);
}
}
}
if (n > 2)
{
if (count.ContainsKey(n))
{
count[n] = count[n] + 1;
}
else
{
count.Add(n, 1);
}
}
int flag = 0;
foreach(KeyValuePair b in count)
{
// minimum number of prime divisors
// should be 2
if (b.Value == 1)
{
flag = 1;
break;
}
}
if (flag == 1)
{
return false;
}
else
{
return true;
}
}
// Function to check if a number
// is Trojan Number
static bool isTrojan(int n)
{
if (!isPerfectPower(n) &&
isStrongNumber(n))
{
return true;
}
else
{
return false;
}
}
// Driver Code
public static void Main(String[] args)
{
int n = 108;
if (isTrojan(n))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by Princi Singh 输出：
YES