C How can I insert another loop or method to make 100 equivalent to 541?-CodePudding

My goal is to display the first list of prime numbers depending on user's input, when the user input 10, the program should display first 10 prime numbers which are 2 3 5 7 11 13 17 19 23 29. I'm thinking of having generated list from 2 which is the first prime number to 541 which is the 100th prime number then if the user input for example 100 then it will be equivalent to 2 upto 541. Will my algorithm works? currently I'm able to display prime numbers from 2 - 541 and stucked on implementing another loop for 1-100 input.

#include <iostream>

using namespace std;
   
int main() {
    int N, counter, i, j, isPrime, n;
    cout << "List of prime numbers from 1-100: ";
    cin >> N; // if the user input 10 then the first 10 prime numbers are "2 3 5 7 11 13 17 19 23 29"
 
    // list of prime from 2-541 (or first list from 1-100 prime numbers)
    for (i = 2; i <= 541; i  ) {
        isPrime = 0;
        
        for (j = 2; j <= i/2; j  ) {
            if (i % j == 0) {
                isPrime = 1;
                break;
            }
        }
        
        if (isPrime == 0 && N != 101)
            cout << i << " ";
       
    }
}

CodePudding user response：

Your current algorithm doesn't work, because N is never modified. If the user inputs 101 then nothing will be printed, because if (isPrime == 0 && N != 101) will always be false. If the user inputs anything else then it will always print the first 100 primes.

The key idea will be to count how many primes were found and break after the Nth prime. Or count down, how many primes we still need t ofind. The rough outline will be as follows:

for (i = 2; N > 0; i  ) {  // Loop as long as we need to find more primes
    ...

    if(isPrime == 0) {
        std::cout << ...
        N--;               // <- Found one prime, N-1 to go
    }
}