What's the matter? Afraid I might roll into your interview

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*fizzbuzzes*

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>>59740723

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#include <stdio.h>
#include <math.h>

#define BELOW 2000000

int isaprime (int num);

int main (void) {

int i;
float sum = 0;

for (i = 2; i < BELOW; i++) {

if (isaprime(i) == 1) {
sum = sum + i;
printf ("\n%d\t%.1f", i, sum);
}
}

getch();
return 0;
}

int isaprime (int num) {

int i;

for (i = 2; i <= sqrt(num); i++) {
if (num % i == 0) {
return 0;
}
else {
;
}
}

return 1;
}

>>59741064
how do you explain the sqrt(num)

>>59741107
if the root is smaller than i it isn't a prime. this is due to the definition of primes.

>>59741064
>for (i = 2; i < BELOW; i++) {
Wrong. Start from 3, increment in 2. Start sum from 2. Which will also speedup

(i = 2; i <= sqrt(num); i++) {

, as you can also start from 3, and increment in 2 to sqrt(n) because you're only testing odds and even numbers won't divide odd.
>>59741126
>if the root is smaller than i it isn't a prime. this is due to the definition of primes.
No, it's because you'd be repeating already found pair factors, e.g. 2*4 = 8, 4*2 = 8, so you should have limited divisors to sqrt(8), as 2, 4 has already been found.

#include <iostream>

using namespace std;

int main()
{
    int j , i;
    int flag = 0;
    int sum = 3;
    for(i = 2; i <= 2000000;i++)
    {
        cout << i << endl; 
        for(j = 2; j < i; j++)
        {
            cout << i <<" % " << j << " = " << i % j << endl; 
            if((i % j) == 0)
            {
                flag = 0;
                break;  
            }else
            {
                flag = 1;
                continue;        
            }
        
        }
        cout << flag << endl;
        if(flag == 1)
        {
            cout << i << " Is prime\n";
            sum = sum + i;
        }else if(flag == 0)
        {
            cout << i << " Is not prime\n";
        }else
        {
            cout << "IDK...\n";
        }
    }
    cout << sum << endl;
}