可以做一個vector直接排除已確認是質數者的倍數

然後偶數除了2可以不用看

邊界特殊情況要處理好

要加速的話可以不要像我一樣用sqrt()

#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
using std::vector;
int check_prime(int num);

int main(){
	
	unsigned int start, end;
	
	while(cin >> start >> end){
		vector<unsigned int> prime_list;
		
		if (start==1){
			if (end == 1){
				cout << 0 << endl;
				continue;
			}
			else{
				start++;
			}
			
		}
		
		if ((start % 2) == 0){
			if (start == 2){
				prime_list.push_back(2);
				start++;
			}
			else{
				start++;
			}
		}
		

		for(int count = start; count <= end; count = count + 2){
			
			if (prime_list.empty()){
				
				if (check_prime(count)){
					prime_list.push_back(count);
				}
				else{
					continue;
				}
			}
			else{
				
				int flag1 = 1;
				for(int j = 0; j < prime_list.size(); j++){
					if ((count % prime_list[j]) == 0){
						flag1 = 0;
						break; 
					}
				
				}
				
				if (flag1 == 0){
					continue;
				}
				else{
					if (check_prime(count)){
						prime_list.push_back(count);
					}
					else{
						continue;
					}
				}
				
			}
		}
		/*for(int k = 0; k < prime_list.size(); k++){
			cout << prime_list[k] << endl;
		}*/
		cout << prime_list.size() << endl; 
		
	}
}

int check_prime(int num){
	int flag = 1;
	
	for(int i = 3; i <= sqrt(num); i=i+2){
		
		if ((num%i) == 0){
			flag = 0;
			break;
		}
	
	}
	
	return flag;
}