In a mathematical exploration, Alex wants to find the greatest prime divisor of a given positive integer. Create a program that accomplishes this task where Alex has a positive integer, represented as a long long int. The program should find and display the greatest prime divisor of this integer.Write a program that displays the greatest prime divisor for various positive integers.Input format :The input consists of a long long integer n, representing the encryption number.Output format :The output prints the greatest prime divisor as a long long integer.Refer to the sample output for formatting specifications.Code constraints :In this scenario, the given test cases fall under the following constraints:1 ≤ n ≤ 100
Question
In a mathematical exploration, Alex wants to find the greatest prime divisor of a given positive integer. Create a program that accomplishes this task where Alex has a positive integer, represented as a long long int. The program should find and display the greatest prime divisor of this integer.Write a program that displays the greatest prime divisor for various positive integers.Input format :The input consists of a long long integer n, representing the encryption number.Output format :The output prints the greatest prime divisor as a long long integer.Refer to the sample output for formatting specifications.Code constraints :In this scenario, the given test cases fall under the following constraints:1 ≤ n ≤ 100
Solution
Sure, here is a simple C++ program that accomplishes this task:
#include<bits/stdc++.h>
using namespace std;
long long greatestPrimeDivisor(long long n) {
long long maxPrime = -1;
while (n % 2 == 0) {
maxPrime = 2;
n >>= 1;
}
for (int i = 3; i <= sqrt(n); i += 2) {
while (n % i == 0) {
maxPrime = i;
n = n / i;
}
}
if (n > 2)
maxPrime = n;
return maxPrime;
}
int main() {
long long n;
cin >> n;
cout << greatestPrimeDivisor(n) << endl;
return 0;
}
This program works by repeatedly dividing the input number by 2 until it is no longer divisible, then it does the same for all odd numbers up to the square root of the number. If the number is still greater than 2 after this process, it means the number itself is a prime number and thus the greatest prime divisor. The greatest prime divisor found during this process is stored in the maxPrime variable and returned by the function.
Similar Questions
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Single File Programming QuestionProblem StatementSam is on a quest to decipher enchanted scrolls in an ancient temple that requires unveiling the HCF (Greatest Common Divisor) and LCM (Least Common Multiple) of two numbers. Write a program where Sam enters two positive integers, calculates the HCF and LCM using a for loop, and displays them.ExampleInput:6 3Output:HCF = 3LCM = 6Explanation:HCF (Greatest Common Divisor) of two numbers is the largest number that divides both of them without leaving a remainder.For 6 and 3, the common divisors are 1 and 3. The largest of these is 3, so the HCF is 3.LCM (Least Common Multiple) of two numbers is the smallest number that is a multiple of both.To find the LCM, we typically use the formula: LCM(a, b) = (a * b) / HCF(a, b).For 6 and 3, (6 * 3) / 3 = 18 / 3 = 6. So, the LCM is 6.Note: This question helps in clearing AMCAT exam.Input format :The input consists of two space-separated integers: n1 and n2.Output format :The first line prints "HCF = " followed by an integer which is the HCF of n1 and n2.The second line prints "LCM = " followed by an integer which is the LCM of n1 and n2.Refer to the sample outputs for the formatting specifications.Code constraints :In this scenario, the test cases fall under the following constraints:1 ≤ n1, n2 ≤ 1000Sample test cases :Input 1 :6 3Output 1 :HCF = 3LCM = 6Input 2 :1000 1Output 2 :HCF = 1LCM = 1000
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