C++ find prime numbers

C++ find all prime numbers between 2-200

Prime number: refers to natural numbers greater than 1, except for 1 and itself (2 factors), which have no other factors

Problem solving ideas

Let $m$ is 2~$\sqrt{m}$Divide if $m$ cannot be 2~$\sqrt{m}$Divide any integer between, you can determine $m$ is a prime number.

Some friends may ask why it is to $\sqrt{m}$Not to mmWhat about$m$ ?

Then I will explain to my friends, the
formula ① $n$=$\sqrt{m}$ ×$\sqrt{m}$
Formula ② $n$=$n$ ×1
herennThe factor of$n$ is$\sqrt{m}$And 1 and itself, a total of 3 factors, so we only need to calculate to $\sqrt{m}$That's it!

Problem solving ideas

To record mmWhether$m$ is a prime number can be represented by a boolean variable (bool)pr. Set pr to true (true) at the beginning of the loop, if$m$ is divisible by an integer, it means$m$ is not a prime number, and the value of pr becomes false (false) at this time. Finally, according to whether the value of pr is true, decide whether to output$m$ . Where$n\; is$ used to represent the number of prime numbers.

#include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
int main()
{
    
    
	int m, k, i, n = 0;
	bool pr;
	for (m = 2; m <= 200; m = m + 1)
	{
    
    
		pr = true;
		k = int(sqrt(m));
		for(i=2;i<=k;i++)
			if (m % i == 0)
			{
    
    
				pr = false;
				break;
			}
		if (pr)
		{
    
    
			cout << setw(5) << m;
			n = n + 1;
			if (n % 5 == 0)cout << endl;
		}
	}
	cout << "\n1-200的素数有：" << n << "个！" << endl;
	cout << endl;
	system("pause");
	return 0;
}

Program analysis

When found $After\; m$ is divisible by an integer,mmcan be judged$m$ is not a prime number, it is not necessary to continue to checkmmWhether$m$ will be divisible by other integers, so use break to end the loop early. The if statement on line 19 checks whether pr is true, ifmm inthis loop$m$ is never divisible by any integer, pr keeps its true at the beginning of the loop, so the prime numbermmshould be output$m$ . Then we know that even numbers are not prime numbers (including at least 1 and itself, but also 2 and another factor), so at the end of this loop letmmThe value of$m$ +2, then use the same method to detect the newmmWhether$m$ is a prime number.

operation result