这段代码的复杂程度（Big O）是多少

Question

I initially thought the complexity is O(n^3) because the inner loop goes to i*i, but now I think the complexity is O(n^2) because of the "break" statement.

What's your thought?

#include <stdio.h>

int main()
{
    int i,k,l,m;
    unsigned int j;
    l=0;
    for (i=3; i<65535;i+= 2) {
        k=1;
        for (j=3; j <= i*i; j += 2) {
            if (j==i) continue;
            if(i%j ==0) {
                k=0;
                break;
            }
         }
         if (k) { l++; }
     }
     printf("%i\n", l);
     return 0;
}

Answer 1

The inner loop is O(N^2) for prime numbers but fast for non-primes (worst case O(N^1/2) because you only have to search up to sqrt(N)).

The number of prime numbers, however, is small compared to the number of non-primes. An approximation of the number of primes to up X is: X / log(X), as found in this reference link.

So throwing out the non-primes as inconsequential, there are N / log(N) primes and each inner loop takes O(N^2) time, so the total time is O(N^3 / log(N)).

Answer 2

for (j=3; j <= i*i; j += 2) { if (j==i) continue; if(i%j ==0) { k=0; break; } }

Without j == i

here j=3,5,7,9,11,13,15,17....65535. That means j will contain all the prime numbers upto 65535 other than 2. Lets have a Math here. j< i*i . if i is a power of 2 or a prime number, then j loop will get executed completely, but we can neglect power of 2 as i is always odd.so for any i , at some point i=j [ O(n) complexity ]

If i is a consonant then i % j == 0 will happen quickly . As pointed out by JS1 , the j loop will take n/logn by neglecting j == i .

So total time complexity = O(N*2 / logn )