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Home/ Questions/Q 6878137
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T04:41:11+00:00 2026-05-27T04:41:11+00:00

I wrote this code to show the primes between 1 and 100. The only

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I wrote this code to show the primes between 1 and 100. The only condition is to do not use functions, whole code should be inline. I would ask if I can improve (optimize) it much more?

#include<iostream>

using namespace std;

int main() {

    int i=2,j=2;

    cout<<"Prime numbers between 1 and 100 are:"<<endl;
    cout<<"2"<<"\t";
    while(i!=100) {
        for(int j=2;j<i;j++) {
            if(i%j==0)
            break;

            if(j==i-1)
            cout<<i<<"\t";
        }

        i++;
    }

    cout<<endl;
    system("pause");
    return 0;
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T04:41:12+00:00Added an answer on May 27, 2026 at 4:41 am

    You are checking every number from 2 to 100. But since 2 is the only even prime number, you can skip every even number after 2. This goes for both i and j. So start i and j at 3, and increment them by 2.

    #include<iostream>
    
    using namespace std;
    
    int main() {
        cout<<"Prime numbers between 1 and 100 are:"<<endl;
        cout<<"2"<<"\t";
        for (int i=3; i<100;i+=2) {
            // This loop stops either when j*j>i or when i is divisible by j.
            // The first condition means prime, the second, not prime.
            int j=3;
            for(;j*j<=i && i%j!=0; j+=2); // No loop body
    
            if (j*j>i) cout << i << "\t";
        }
        cout<<endl;
        return 0;
    }
    

    In addition to the trick mentioned above, I’ve added the condition j*j<=i which logically is the exact same as j<=sqrt(i). There’s no need to compute the square root when you can do a simple multiplication.

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