判断质数的方法

穷举法

#include <iostream>
using namespace std;

int main()
{
	int m, i;
	cin >> m;
	for (i = 2; i <= m; i++){
		if (m % i == 0)
			break;
	}
	if (i > m){
		cout << m << " is a prime number." << endl;
	}
	else{
		cout << m << " isn't a prime number." << endl;
	}
	return 0;
}

对于上述方法可以优化 对于一个数m的两个不同因子一个必定小于$\sqrt{m}$m​另一个必定大于$\sqrt{m}$m​

#include <iostream>
#include<cmath>
using namespace std;

int main()
{
	int m, i;
	cin >> m;
	for (i = 2; i <= sqrt(m); i++){
		if (m % i == 0)
			break;
	}
	if (i > sqrt(m)){
		cout << m << " is a prime number." << endl;
	}
	else{
		cout << m << " isn't a prime number." << endl;
	}
	return 0;
}