#include<iostream>
#include<cmath>
#include<utility>
bool is_prime (const unsigned n) {
if (n<2) return false;
if (!(n%2)) return n==2;
unsigned max_factor=(unsigned)sqrt((double)n)+1;
for (unsigned factor=3; factor<max_factor; ++factor) {
if (!(n%factor)) return false;
}
return true;
}
void print_primes(const unsigned low, const unsigned high) {
for (int i=low; i<=high; ++i) {
if (is_prime (i)) std::cout<<i<<std::endl;
}
}
int main () {
unsigned n, x;
std::cout<<"Print prime numbers\nFrom: ";
std::cin>>n;
std::cout<<"To: ";
std::cin>>x;
if (x<n) std::swap(x,n);
print_primes (n, x);
}
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Loop through some numbers - for example, 2 through 100 - and check each one whether it is a prime number (write a second loop to test whether it is divisible by any number between 2 and the number minus 1). If, in this second loop, you find a factor that is greater than 1 and less than the number, it is not a prime, and you can print it out.
PRINT 2,3,5,7,11,13,17,19,23,29,31,37
If you just want a hint: One way to check whether a number is prime is by dividing it by any number between 2 and the square root of your number. If the number divides by any of these, it is not prime. If you want the code: import math for num in range(1,101): if all(num%i!=0 for i in range(2,int(math.sqrt(num))+1)): print num
To write a C program to find prime numbers between 1 to 500, you can use a nested loop structure. In the outer loop, iterate from 2 to 500, and in the inner loop, check if the number is divisible by any number from 2 to the square root of the number. If it is not divisible by any number other than 1 and itself, then it is a prime number. Print out all prime numbers found within the specified range. Remember to include necessary header files, such as <stdio.h>, and use appropriate logic to implement the program efficiently.
Yes. All non-prime numbers are composite numbers, which simply means a number has one or more prime factors other than 1 and the number itself.