仅使用 1 个循环语句查找最多 n 位的素数 [关闭]

Question

我是编程新手并试图解决一些问题，在这个问题中，我试图仅使用 1 个循环语句来查找最多 n 位的素数。

Answer 1

一个循环和递归：

#include <iostream>
#include <cmath>

#define NUDIGS 3 //number of digits

bool TestPrime(int n, int ndiv)
{
    if ( n == ndiv )
        return true;
    if ( (n % ndiv) == 0 )
        return false;
    return TestPrime (n, ++ndiv) ;
}

int main()
{
    //calculate the max number with BUDIGITS
    int max = pow(10, NUDIGS) - 1;
    std::cout << "testing numbers from 2 to " << max << std::endl;
    int ntot = 0;
    
    //Loop testing every number. '1' is not considered as a prime number.
    for (int i=2; i <= max; i++)
    {
         //call a recursive test
         if ( TestPrime(i, 2) )
         {
             std::cout << i << " is prime " << std::endl;
             ntot++;
         }
    }
    std::cout << ntot << " prime numbers found" << std::endl;
}

Answer 2

//this program will print prime number from 2 to N . and break loop 
without using break statment

#include <iostream>

using namespace std;
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int range;
    cin>>range;
    int num=2;
    for(int i=2;i<=num;i++){
        if(num>range){
            i=num+1;
        }
        else{
            if(i==num){
                cout<<num<<" ";
                i=1;
                num=num+1;
            }
            else if(num%i==0){
                i=1;
                num=num+1;
            }
        }
    }
    cout<<endl;
    return 0;
}