Is there any way to optimize the following code without changing the logic and language ? Reason : Time limit exceeded in UVa

Question

10110 - Light, more light

There is man named ”mabu” for switching on-off light in our University. He switches on-off the lights in a corridor. Every bulb has its own toggle switch. That is, if it is pressed then the bulb turns on. Another press will turn it off. To save power consumption (or may be he is mad or something else) he does a peculiar thing. If in a corridor there is n bulbs, he walks along the corridor back and forth n times and in i-th walk he toggles only the switches whose serial is divisable by i. He does not press any switch when coming back to his initial position. A i-th walk is defined as going down the corridor (while doing the peculiar thing) and coming back again. Now you have to determine what is the final condition of the last bulb. Is it on or off?

Input

The input will be an integer indicating the n’th bulb in a corridor. Which is less then or equals 2³² −1. A zero indicates the end of input. You should not process this input.

Output

Output ‘yes’ if the light is on otherwise ‘no’, in a single line.

Sample Input

3
6241
8191
0

Sample Output

no
yes
no

Here is my code:

#include <stdio.h>

int main()
{
    unsigned int n, a, d, i, j;
    while (1)
    {
        scanf("%u", &n);
        if (n == 0)
        {
            break;
        }

        a = n / 2;

        d = 0;

        for (i = 1; i <= a; i++)
        {
            if (n % i == 0)
            {
                d++;
            }
        }
        if (d % 2 == 0)
        {
            printf("yes\n");
        }
        else
        {
            printf("no\n");
        }
    }
    return 0;
}

Is that even possible in C ? Not allowed to use any other language :( Language they used : ANSI C 5.3.0 - GNU C Compiler with options: -lm -lcrypt -O2 -pipe -ansi -DONLINE_JUDGE

The run time should not exceed 3 seconds.

Answer 1

Looping to n/2 is very slow as code only needs to loop to sqrt(n).

    a=n/2;
    d=0;
    for (i=1; i<=a; i++) {
        if (n%i==0) {
          d++;
        }
    }

is replaceable with below which loops up to sqrt(n) times.

    d=0;
    if (n > 1) { 
      d += 1;
    }
    for (i=2; i < n/i; i++) {
        if (n%i==0) {
          d += 2;
        }
    }
    if (i == n/i) {
        if (n%i==0) {
          d += 1;
        }
    }

And since later code only needs the even-ness of d, the for (i=1; i < a/i; i++) { loop serves little other than to get i up to near the square root of n. Consider testing if n is a perfect square instead and watch out for edge cases when n <= 1. (Careful with double sqrt(double) as that may incur unexpected rounding.)

Note: do not use ...; i*i < n; i++ as i*i may overflow when n is a large value near UINT_MAX.

Pedantically, use a wider type that unsigned, which may have less range than needed. Suggest unsigned long.