# Check if a number is a Trojan Numbers in C++

C++Server Side ProgrammingProgramming

## Concept

With respect of given number n, the task is to verify whether n is a Trojan Number or not. Trojan Number is defined as a number that is a strong number without a perfect power. We can say that a number n is treated as a strong number if, for every prime divisor or factor p of n, p^2 is also a divisor. We can say in another way, every prime factor appears at least twice. We should remember that all Trojan numbers are strong. But it is not true for vice-versa that means, not all strong numbers are Trojan numbers: only those that cannot be represented as a^b, where a and b are positive integers greater than 1.

Input

n = 72
72 is expressed as 6×6×2 i.e. (6^2)×2 i.e. Strong Number but without perfect power.

Output

YES

Input

n = 16
16 is expressed as 2×2×2×2 i.e. 2^4 i.e. Strong number with perfect power.

Output

NO

## Approach

At first, we have to store the count of each prime factor and verify if the count is greater than 2 then it will be a Strong Number.

In case of next step, we have to verify whether the given number is expressed as a^b or not. If it is not expressed as a^b, we can say that it is perfect power; otherwise it is perfect power. Finally, at the last step, we can conclude that if this given number is strong without perfect power, the number is treated as Trojan Number.

## Example

Live Demo

// CPP program to check if a number is
// Trojan Number or not
#include <bits/stdc++.h>
using namespace std;
bool isPerfectPower1(int n1){
if (n1 == 1)
return true;
for (int x1 = 2; x1 <= sqrt(n1); x1++) {
int y1 = 2;
int p1 = pow(x1, y1);
while (p1 <= n1 && p1 > 0) {
if (p1 == n1)
return true;
y1++;
p1 = pow(x1, y1);
}
}
return false;
}
bool isStrongNumber1(int n1){
unordered_map<int, int> count1;
while (n1 % 2 == 0) {
n1 = n1 / 2;
count1[2]++;
}
for (int i1 = 3; i1 <= sqrt(n1); i1 += 2) {
while (n1 % i1 == 0) {
n1 = n1 / i1;
count1[i1]++;
}
}
if (n1 > 2)
count1[n1]++;
int flag1 = 0;
for (auto b : count1) {
if (b.second == 1) {
flag1 = 1;
break;
}
}
if (flag1 == 1)
return false;
else
return true;
}
bool isTrojan1(int n1){
if (!isPerfectPower1(n1) && isStrongNumber1(n1))
return true;
else
return false;
}
// Driver Code
int main(){
int n1 = 72;
if (isTrojan1(n1))
cout << "YES";
else
cout << "NO";
return 0;
}

## Output

Yes
Published on 23-Jul-2020 07:59:17