Prime Number C++ using "While, For, Do...While" Loop?

The Sieve of Eratosthenes is the classic way to solve this problem. And with the way I've done it below, you get all the kinds of loops you need.

#include <iostream>

#include <sstream>

#include <deque>

using namespace std;

void drainSieve(deque<bool> &);

int main(int argc, char *argv[]) {

    size_t n,x;

    string in;

    stringstream ss;

    do {

        cout << "Enter max value for prime search : ";

        getline(cin,in);

        ss.clear(); ss.str(in);

    } while ((!(ss >> n)) || (ss.good()));

    // Find primes

    deque<bool> sieve(n+1,true);

    drainSieve(sieve);

    // Prompt user to query for primes

    cout << "Enter numbers [0 .. " << n << "] to check." << endl;

    while (true) {

        cout << "> ";

        getline(cin,in);

        ss.clear(); ss.str(in);

        if ((ss >> x) && (x <= n) && !ss.good()) {

            cout << x << " is ";

            if (sieve[x] == false) cout << "not ";

            cout << "prime." << endl;

        }

    }

    return 0;

}

void drainSieve(deque<bool> &ps) {

    size_t N = ps.size();

    ps[0] = ps[1] = false;

    for (size_t i = 2; i*i < N; i++) {

        if (ps[i] == true) {

            for (size_t j = i+i; j < N; j += i) {

                ps[j] = false;

            }

        }

    }

}

#if 0

Sample run:

Enter max value for prime search : 12000

Enter numbers [0 .. 12000] to check.

> 71

71 is prime.

> 10711

10711 is prime.

> 8

8 is not prime.

>

#endif

oftentimes conversing, a for loop of here kind: for(statement1 ; statement2 ; statement3) { statement4; } might properly be rewritten as a on an identical time as loop of here kind: statement1; on an identical time as(statement2) { statement4; statement3; } in case you exchange your loop in this way, you will in all probability discover it works.

#include <iostream>

using namespace std;

int main()

{

long long int number=1,b;

cout<<"please insert two numbers to find out wheather this number is prime or not:";

cin >> number >>b;

while (number <b)

{

for (long long int i=2;i<number;i++)

{

if (number%i==0)

break;

if (number==(i+1))

cout << "number" << number << " is odd \n";

}

number++;

}

return 0;

}

using do while?